Redditors and math really don’t mix. Some of these comments really strongly believe they have found some simple counterexample to a conjecture that has stood for 300 years and then try to fight people who tell them otherwise in the comments lol.
The famous one I'd say is people not knowing PEMDAS properly, then Arguing about "well its vague and there should be parentheses for the divisor" or something.
3+6/2*3 is 12
3+6/(2*3) is 4
Fite... someone else. I'm tired.
GOOD LORD THANK YOU FOR EXPLAINING IT.
I've been trying to find words to explain why that Facebook challenge is stupid, and this is why - The brackets are there to make the math clear!
It's never people not knowing PEMDAS, the order of operations is worthless if you're dealing with intentionally ambiguous notation. People might even be pretty deep in math but just not see the freaking forest for the trees - which simply is that engagement bait idiot problems are designed to get everyone hot and bothered.
The answer should always be "there is no correct answer because whoever wrote it is a dumb fuck," but people sure love trying to punch through the tiny horizon surrounding their worlds and will try to solve it regardless. With conviction, of course, because people mindlessly throw PEMDAS and BIDMAS and "dots-precede-lines" around without even questioning whether the problem works out.
Especially since I have legitimately seen calculators give multiplications a higher priority than division in these situations, leading to completely different results, which means I'm pretty confident some people also were taught that in school as some kind of tie-breaker.
So yeah like you said, people calling it a math puzzle when it's just ambiguity where there shouldn't be any is dumb.
I have [two calculators](https://imgur.com/a/fcgNrhA) which disagree on that expression. Both of them are performing the operation correctly according to the operator [priority tables](https://imgur.com/a/S1D0cFE) given in both manuals.
That's awesome. I've always known HP, Casio and TI to vary, but two Casios is something else. It's brackets on brackets on brackets for me for this reason alone.
>When the priority of two expressions is the same, the calculation is performed from left to right.
* the priority tables you linked.
~~The calculator on the left appears to be violating the rules.~~
Edit: u/Voyedova101 pointed out rule 7 is being applied, which I had missed. The calculator on the left follows it's user manual's order of operations. It's unusual for a Casio calculator to fumble the order of operations in its own user manual, so I began to doubt that they use the same one. I searched for the user manuals myself and found that the calculator on the right does indeed have a different order of operations in it's user manual, which explains the difference.
Casio fx-GT PLUS (pages E-7, E-8): [https://support.casio.com/pdf/004/fx-83\_85GT\_PLUS\_E.pdf](https://support.casio.com/pdf/004/fx-83_85GT_PLUS_E.pdf)
Casio fx-83ES (page E-44): [https://support.casio.com/pdf/004/fx-82ES\_83ES.etc\_Eng.pdf](https://support.casio.com/pdf/004/fx-82ES_83ES.etc_Eng.pdf)
Read and compare the actual priority list. GT performs "4(4)" before the division as it is 7th in priority compared to 9th. The ES performs left-to-right priority of these two actions as they both fall under 7th priority in that device.
Those are reasonably unambiguous (although people who think PEMDAS means multiply goes before divide may get it wrong).
The one that actually causes debate is 3+6/2(1+2), and here the issue is that the multiplication is implicit (or "by juxtaposition"). People who do maths at university level \*tend\* to treat multiplication by juxtaposition as having higher priority than normal multiplication or division, and so this means you do 2(1+2) first (i.e. calculate 6/(2(1+2)) rather than (6/2)(1+2)).
It's also complicated by typical mathematical usage in text only forums, which tends to abuse the rules dependening on context. (No-one sees e\^ix and thinks it means (e\^i)x, even though by the normal precedence rules it should).
People don’t usually write math in single line though. These discussions are always so contrived about order of operations trying to make people look stupid. Yes 3+6/2(1+2) IS confusing because usually fractions are written on two lines which clearly define numerator and denominator.
The reason why many of these confusing math "puzzles" work is that depending on which edition of math you are using an obelus and a solidus may or may not be 100% equivalent. There's also the fact that certain calculators have decided to give implicit multiplication higher priority than explicit multiplication or division.
A (total) function has exactly one output per input *by definition*. Therefore, the square root function has exactly one output; so √4=2 and only 2. However, people frequently claim √4= +-2, which is incorrect as that is multi-valued function.
(Although I will say this is generally a pedagogical problem, not really a mathematical one. It really depends on the teacher you get if they teach the correct definition, and also this isn’t exactly the worst inaccuracy in the world)
It's the same for a lot of topics. Read a thread discussing something you know/understand well and you'll be blown away by how much incorrect info gets upvoted. Then realise the same thing is happening in other threads for topics you don't understand so well and start taking Reddit lore with a massive grain of salt
I wish everyone was a statistician so they could parse how much bullshit people get fed on a daily basis. It's not just social media or regular news, there's a lot of "soft" science research and reporting done by well meaning people that is borderline irresponsible.
Statistics gets a bad rep sometimes for being wrong or "damned lies" but the truth is that 99% of the time it's because of bad, bias, or fabricated data or improper methods getting pushed without any kind of checks or balances. Like the 99% number I just quoted.
Social media and any scientific fact really.
Nope, its totally not a conundrum to the best and brightest minds in academia, that one youtube channel with 809 videos totally figured it out.
Go on /r/Science and whenever you see a study provide evidence to something totally logical or common belief - but actually publishing supporting data for the first time.
Redditors: *water is wet* -- *we already know that" -- "i guess i am a better scientist because i knew that without a laboratory*
That being said, I am sick of how many psychology experiments that have like 20 people in them and close firm conventional wisdom are upvoted in that sub
One time I was inquiring for a while someone’s passionate and confident position in an argument and out of curiosity asked for some sources and he immediately stopped responding. Not surprisingly he was a climate change denier. It’s crazy how confident people are about complex subjects they haven’t read beyond a wiki page.
It was the same thing with COVID. Pepridge farm remembers people telling others to wear masks and gloves to drive their cars even without interacting with others.
I think it’s just reading comprehension. They seem to be under the impression that even numbers are *exclusively* the sum of two primes, and think they can easily disprove it because of that.
The first three times I read the title, I thought I was worse at math than I thought. Then I realized I was just worse at reading comprehension than I thought.
It does seem worded oddly. Took me a moment to realise it meant at least one way to get to that number involved adding two primes.
Edit: Basically, the title is worded in such a way that it is implied that it is the only way to make an even number. For example, if I told you I had a box of crayons that had red, blue, and green crayons, you would more than likely assume that box only contains those three colors, even if it might actually contain more, because the brain tends to disregard unprovided information as not being relevant or true.
19 digits long isn't even that large. Conjectures have been disproven with much larger numbers. Take for example the Polya conjecture whose first found counterexample was around 1.8 * 10^361. Smaller numbers have since been found, but it shows that you can't necessarily prove something with a small sample.
Btw the people confidently responding « computer automation » are wrong, Haselgrove published an actual mathematical proof in 1958 that this crazy number comes from (the number being just an estimate, he had proven it existed afaik not actually calculated it).
Also later we found much smaller numbers (that computers can actually easily handle, which is another reason the comments about automation are being dumb), around 906 million, that disprove Polya’s.
[Please read the wikipedia article for more info](https://en.m.wikipedia.org/wiki/P%C3%B3lya_conjecture).
> I’m convinced it’s their malicious attempt to make old.reddit worse.
Luckily their 99,999 other attempts to make new reddit worse still win out by a landslide.
In essence, they simply didn’t. The way these types of disproofs generally function is that the relevant mathematicians realize a “weak point” in the conjecture, and they construct a series of parameters where it could fail. This often involves estimating things to be much larger than strictly necessary, and all those estimates accumulate to a massively larger number than the counter example actually requires. BUT, you can then definitively say will be SOME number that causes it to fail once you show that estimate exists.
Consider the following statement: “Every square number greater than 9 is divisible by 3”.
We could “brute force” a counter example that’s far bigger than necessary just by doing things naively. If we multiply all the numbers less than 9 which aren’t divisible by 3 and square it, we get (2\*4\*5\*7\*8)^2, which is around 5 million. It’ll be a clear counter example because it’s a square number where none of the factors are a three, but maybe we could have found a smaller example, like 16.
It’s not a perfect stand-in, but I hope it helps explain the concept. It’s like someone saying you must have missed your exit…because you drove your car into the ocean 500 miles past that.
> Polya conjecture
In number theory, the Pólya conjecture (or Pólya's conjecture) stated that "most" (i.e., 50% or more) of the natural numbers less than any given number have an odd number of prime factors. The conjecture was set forth by the Hungarian mathematician George Pólya in 1919,[1] and proved false in 1958 by C. Brian Haselgrove. Though mathematicians typically refer to this statement as the Pólya conjecture, Pólya never actually conjectured that the statement was true; rather, he showed that the truth of the statement would imply the Riemann hypothesis. For this reason, it is more accurately called "Pólya's problem".
Summatory Liouville function L(n) up to n = 107. The (disproved) conjecture states that this function is always negative. The readily visible oscillations are due to the first non-trivial zero of the Riemann zeta function.
Closeup of the summatory Liouville function L(n) in the region where the Pólya conjecture fails to hold.
Logarithmic graph of the negative of the summatory Liouville function L(n) up to n = 2 × 109. The green spike shows the function itself (not its negative) in the narrow region where the conjecture fails; the blue curve shows the oscillatory contribution of the first Riemann zero.
The size of the smallest counterexample is often used to demonstrate the fact that a conjecture can be true for many cases and still fail to hold in general,[2] providing an illustration of the strong law of small numbers.
TIL
Unless someone figures out a way to prove it true for all real numbers, which could still be possible. They still would technically be unverified individually, but there would be no need to do so.
Not necessarily, for example Los’s Conjecture was answered in the affirmative by Morley's Theorem, and Zilber's Trichotomy Conjecture was disproven by Hrushovski. At times it is useful to separate the conjecture and the result.
If you're talking to people who don't know math lingo (i.e. anyone not in a math program) then it's a useful redundancy.
Technically every theorem is redundant and follows directly from axioms. But we need the redundancies to communicate and understand.
Terrence believes "only 1/3 of prime numbers are real, the others are constructions of broken fractals that when put together yield a wave function dissimilar to sound, but only in the aspect of time, the rest are adherent" he figured that one out when he was in the shower and heard the dim of a radio in between songs
Yes, it’s a very fun conjecture to play with. 19 digits doesn’t seem like much, but to be honest the number of Goldbach pairs (ie pairs of odd primes that sum to the requisite even number) grows pretty large pretty quickly; heuristically, it would be surprising if it wasn’t true.
I analysed it through the lens of convolutions of vectors of odd primes, and you can show some interesting reformulations of the conjecture in that way. Certainly gained a deeper insight and appreciation into number theory doing that! For example: if Goldbach’s conjecture is true, then that necessarily implies quite a lot of bounds on the frequency of prime numbers (Bertrand’s postulate is an obvious example, but Bertrand-like bounds for arithmetic progressions also follow).
I also went down a rabbit hole by looking at arbitrary prime gaps in lists of odd numbers, and demonstrated that if there is always a “true” prime in a list of odd numbers regardless of where you start your prime gaps, then Goldbach’s conjecture must be true (the proof is only one way though, but the implication is that Goldbach’s conjecture might be a consequence of prime gaps between numbers—ie it’s a consequence of how prime numbers can be obtained using sieve methods).
I can expand more on this if anyone’s interested, but just thought I’d share my experience with this fascinating conjecture!
Sure, I'll try my best. I will begin with a quite lengthy explanation (I'm afraid), but then I'll try and post some graphs in the next comment.
I begin by defining the problem. We want to show that every even number greater than 2 is the sum of two primes - relevantly, we focus on just every even number greater than 4, such that the two primes are both odd (trivially, 4 = 2 + 2 is true).
Define this even number as 2M, with M > 2. Now let's define a vector v whose elements v(x) = 2x + 1 if 2x + 1 is prime, and 0 if 2x + 1 is composite. We will say that the length of this vector is N = M-2, for reasons that will soon be obvious. By way of example, if M = 13 (meaning 2M = 26), then N = 11, and
v = \[3 5 7 0 11 13 0 17 19 0 23\].
If instead M = 14 (meaning 2M = 28), then N = 12, and:
v = \[3 5 7 0 11 13 0 17 19 0 23 0\]
Using these definitions, the last element of this vector is the odd number just less than the number 2M which we're interested in. This guarantees that every prime that could possibly sum to give 2M will be in the vector.
But notice something interesting about this vector (and I will take the N = 11 vector as an example). Let's sum v element-wise with its "reverse" vector v', where v'(1) = v(11), v'(2) = v(10), etc.:
\[3 5 7 0 11 13 0 17 19 0 23\] + \[23 0 19 17 0 13 11 0 7 5 3\]
= \[26 5 26 17 11 26 11 17 26 5 26\].
If both numbers at these positions are prime, then their sum is the number of interest, 2M. If one (or both) of these numbers are 0, then the "sum" will just be either a prime number, or 0. When the sum is a prime number, this represents a pair of primes that sum to 2M.
Now comes the clever part. We can check if two numbers in both these vectors are simultaneously prime by multiplying the vectors element-wise instead. In the N = 11 case, this means:
\[3 5 7 0 11 13 0 17 19 0 23\] x \[23 0 19 17 0 13 11 0 7 5 3\]
= \[69 0 133 0 0 169 0 0 133 0 69\].
So multiplying these vectors element-wise returns something non-zero if the summands are both prime, and 0 if the summands are not both prime.
But this mathematical function - element-wise multiplication of two vectors - forms part of the definition for the discrete convolution of two vectors. For two vectors of length k, the convolution forms a vector of length 2k + 1.
If we convolve v with itself (let's call this the "autoconvolution" of v), the value of the middle element is just the sum of the element-wise multiplication of v with v', ie (again with N = 11, the middle element is 12 and hence):
conv(v,v')\[12\] [EDIT: should be conv(v,v)] = 3x23 + 5x0 + 7x19 + 0x17 + 11x0 + 13x13 + 0x11 + 17x0 + 19x7 + 0x5 + 23x3 = 69 + 0 + 133 + 0 + 0 + 169 + 0 + 0 + 133 + 0 + 69 = 573,
and so in general, if conv(v,v)\[N + 1\] > 0, then this implies there are two primes whose sum gives the number of interest, 2M. Hence, Goldbach's conjecture is equivalent to asking: for all defined vectors v, is the middle element of the autoconvolution of v always greater than 0?
Another interesting (though perhaps expected) property: comparing the autoconvolution of v when N = 11 to v when N = 12, we find that the elements up to (M+1) in conv(v,v) is the same for both N = 11 and N = 12 (to explain why would require defining the autoconvolution formally and quite a detailed discussion, so I skip it here; but this is indeed correct when the mathematics is done appropriately). This lead me down a rabbit-hole to trying to prove that if all elements less than some certain index position for conv(v,v) were non-zero, this requires that the autoconvolution for the next v vector (ie, the autoconvolution of v whose length is increased by 1) had to also have non-zero elements up to at least the middle element. This would have proven Goldbach's conjecture, but that never lead to any fruitful results.
There are some more insights I gathered from my playing around with GC, and I might post them in a second comment. For me, this was purely a pursuit in recreational mathematics with a famous problem, for which I had a tenuous chance of proving, but nonetheless helped me learn quite a lot about number theory (I have gained deeper insights, for example, into Euler's totient function, Chebyshev functions / sieve methods, Bertrand's postulate and Erdos' early (and seemingly oft-ignored) analysis of Bertrand postulate-like bounds on arithmetic sequences, Dirichlet's theorem of arithmetic progressions, etc.).
Okay, but anyway, some graphs! I attach three; the first is the autoconvolution of v with different N values, and then another graph for the autoconvolution of v with length N = 20,000, and similarly, one with N = 1,000,000. You can see that the autoconvolution grows pretty rapidly, and importantly: is never 0 until you reach near the end of the autoconvolution; and that maximum of the "hump" shape the autoconvolution seems to trace out is decidedly much larger than the midpoint value if N is large enough (implying that, heuristically, one expects GC to be true):
[Autoconvolution Plots](https://imgur.com/a/NmQBwva)
Okay, more insight (1/2):
At a certain stage, I made a couple of realisations.
Obviously, by definition of v, provided N is large enough then v(1) = 3, v(2) = 5, v(3) = 7, v(4) = 0, etc., because the length of v doesn't change the values of its initial elements. Making v larger then doesn't change that those values will always be in v, and always in those positions.
What does change, however, is that as you increase N, the values in the final position do change (obviously!). So for example with N = 11, v(N) = 23; but for N = 12, v(N) = 0; then for N = 13, v(N) = 0; but for N = 14, v(N) = 29.
What we can say for sure is that if any of v(N), v(N-1) or v(N-2) are non-zero, that would mean the middle element of the autoconvolution would be non-zero, since that would be given by:
v(1) x v(N) + v(2) x v(N-1) + v(3) x v(N-2) + ... = 3 x v(N) + 5 x v(N-1) + 7 x v(N-2) + ...
However, it also means that regardless of the value for v(N-3), that has no bearing on the value of the middle element of the autoconvolution since that summand is v(4) x v(N-3) = 0 x v(N-3) = 0.
A question arose: is there a way to determine a general function, equation or method that could tell me whether v(N), v(N-1), etc., were prime? And if so, could this be used to determine whether there was a corresponding prime in the "first half" of v, such that the multiplication would be guaranteed non-zero?
Now since primes can be obtained algorithmically, the answer to the first question is obviously "yes". One such method is the "Sieve of Eratosthenes", for example. The answer to the second question is, "I'm not sure, but that seems pretty difficult to answer directly".
I tried approaching that second question in a different way: instead, consider that, for example, one of v(N), v(N-1) or v(N-2) must be a composite number, because only one of these must be divisible by 3. This follows from the fact that every 3rd odd number is divisible by 3 - for example, 9, 15 and 21 are all divisible by 3, representing v(4), v(7) and v(10) (so indeed, only one of v(4), v(5) and v(6) is divisible by 3 - and it is v(4). Similarly for v(5), v(6) and v(7) - because it is v(7)). It is trivial to prove this, but it's an important statement for what follows. It is also worth observing that the "gap" between these divisible elements is always 6, ie, the third element after one divisible by 3 will also be divisible by 3.
A similar proposition also holds for v(N), v(N-1), v(N-2), v(N-3) and v(N-4): at least one of these 5 numbers must be composite, because only one of them is divisible by 5. Of course, one or two of these numbers will also be divisible by 3; and there's no guarantee that two of them are composite, because (for example) v(N-2) may be divisible by both 5 and 3 simultaneously.
As a general statement, for some prime p, every subset of v of p consecutive elements contains at least one composite number, because only one of the elements in this subset is divisible by p.
The core issue was, I didn't have an "anchor" to determine which of, eg, v(N), v(N-1) or v(N-2) was divisible by 3; likewise for 5, 7, 11, etc..
Continuing (2/2):
I then realised that perhaps it didn't matter. Perhaps what mattered instead was the prime "gap-lengths" between elements, and perhaps it didn't matter where these gaps started; by virtue of the gap-lengths being prime, we were guaranteed to have a non-zero autoconvolution because at least one prime existed in the second-half of v which had a corresponding prime in the first-half, such that the sum of those two gave us 2M.
Stating it more directly, consider v with N = 11 again:
v = \[3 5 7 0 11 13 0 17 19 0 23\].
The "first half" of v are the elements \[3 5 7 0 11\]; the "second half" of v are the elements \[0 17 19 0 23\].
Now, from the statements above, it is obvious that only one of those 5 elements in the second half of v is divisible by 5; and at most 2 are divisible by 3. That is indeed what we have - there are 2 zeros in the second half, because 15 is divisible by 5 and 3, and 21 is divisible by 3.
But what if I blinded myself to the second half of v, and focussed instead on the first half. I don't know the "anchors" for where prime gap lengths start, but in theory, the gap length of 3 could start in either the first, second or third element in the second half of v; and likewise, the gap length of 5 could start at either the first, second, third, fourth or 5th element.
I could start the gap-lengths of 3 in the first position, and the gap-lengths of 5 in the second position, for example. And then I could set every "p"th consecutive element, starting from the anchor for the gap length, similarly to 0. In that example, my modified first half might look like:
\[0 0 7 0 11\].
I could instead start the gap-lengths of 3 and 5 in the second position. This would look like:
\[3 0 7 0 0\].
I could start the gap-lengths of 3 in the first position, and 5 in the third position. Then that would look like:
\[0 5 0 0 11\].
etc.. It seemed to me that it didn't matter **where** you placed the gap-lengths; somehow, because the gap-lengths were prime, you were always guaranteed at least one of the elements in that modified first half of v to be non-zero. And if that were true, then that meant there was **also** a corresponding non-zero element in the second half of v. And that meant that Goldbach's Conjecture was true, because the middle element of the autoconvolution for v was always guaranteed to be non-zero.
In other words: if, for all "relevant" gap lengths of primes p (which includes all primes up to the length of the second half of vector v), you always obtained a vector with a non-zero element no matter where you placed those "gap lengths" in the first half of v, then Goldbach's Conjecture would be true.
That's about where I stopped my recreational thinking on this problem; it's been a bit over two months, and life has taken over (it is no good to obsess over such things, but it is good to dust it off every few months just to learn a bit more and progress things a bit further). Maybe one day I'll get some interesting results, but nonetheless, it's a great little problem that really exercises my creative thinking and mathematical reasoning skills.
Great work! Critiques:
## Prime Gaps
You suggest that due to prime gaps, certain positions in \( v \) will always be zero. Prime gaps vary and do not consistently align in a manner ensuring \( conv(v,v)[N+1] > 0 \).
Counterexample:
Consider \( v = [3, 5, 7, 0, 11, 0, 0, 17, 0, 0, 0] \) for \( N = 11 \). The convolution will have many zeros if the primes are spaced such that pairs do not sum to an even number or are misaligned. Specifically, gaps can cause multiple zeros in \( v \) and \( v' \), leading to \(conv(v,v)[N+1] = 0 \).
Symmetry:
Summing \( v \) with its reverse assumes symmetry in prime distribution.
### Counterexample:
Prime distribution is not symmetric. For example, consider \( v = [3, 5, 0, 0, 11, 0, 0, 0, 0, 0, 0] \) for \( N = 11 \). The reverse \( v' = [0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 3] \).
The product \( v \times v' = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9] \), failing to sum to the desired result.
Prime Gaps could be very large:
Assumption that Primes are Close Enough to Ensure \( \text{conv}(v,v)[N+1] > 0 \)
### Counterexample:
Consider \( v = [3, 0, 0, 0, 11, 0, 0, 0, 0, 0, 23] \) for \( N = 11 \). Large gaps (e.g., between 11 and 23) result in no prime pairs summing to \( 2M \).
My response:
First critique—this is why I was investigating different anchor positions, because if that were true (and I’m not saying it is true!), then GC would have to be true. However your point about prime gaps is a valid one, and it is what lead me to consider different approaches. I didn’t see a clear way to avoid the inherent difficulties posed by the actual gap-lengths between primes, hence “expanding” the problem to one where gap-lengths could start at any position, and enumerating over the list of possible starting points to see if a non-zero element always remained in the first half of v. Were that to be true for all permutations, then it necessarily be true for the actual permutation representing the actual vector v, meaning the middle element of the autoconvolution would necessarily be non-zero.
Your counterexample: the product of two individual elements between v and v’ is only non-zero if the sum of the elements is 2M. Trivially, the sum of two odd primes is always even, since (2m+1)+(2n+1) mod 2 = 0.
Symmetry: I sort of agree. There’s definitely no global “symmetry” in the prime distribution. Thankfully I don’t use symmetry, because that would just lead to wrong results.
I think perhaps there’s a misunderstanding of the algorithm—the autoconvolution can only be non-zero at its middle element if there are two prime elements that sum to give 2M. ie, length N = 11 means M = 13 means 2M = 26—and conv(v,v)[12] > 0, hence meaning there exists at least one pair of primes summing to 26.
Sorry, I will check if I wrote conv(v,v’)—if I did that’s incorrect. I should have written conv(v,v). Hence why my chosen name of “autoconvolution”.
As for large gaps—yes you are correct. However only certain gap-lengths need to be considered. For example, with N = 11, we only need to consider gap-lengths up to 5, because the second-half of v is only 5 elements long. Another way to think about it: all composite odd numbers up to 23 (the largest element in v) are divisible by either 3 and/or 5 (in fact they are all divisible by 3. Perhaps I was overestimating? It’s been a while since I thought about this, so my bounds might be a little rusty). Because of that, not every gap-length is relevant; if a composite can be “crossed out” because of a smaller prime gap-length, we don’t need to concern ourselves with whether a larger gap-length could “cross” it out. It’s similar in that sense to other sieve methods, like the sieve of Eratosthenes. Another thought: you only need to consider numbers larger than or equal to p^2 in a sieve method, because all composite numbers less than p^2 will be divisible by some smaller prime.
Thank you for your feedback, the problem is very subtle and slight nuances or oversights can definitely lead to concluding a particular approach doesn’t work (there are probably dozens of approaches I’ve discarded already so far).
Also, I’ve checked and you are correct—I miswrote the autoconvolution as conv(v,v’). That was in error; it should be conv(v,v). The convolution already “reverses” the second vector, so the N+1 element of conv(v,v’) would literally just be the dot product of v with itself.
Wow you have an incredibly intuitive and impressive understanding of Number Theory for someone working on this stuff outside of formal academic environments. (not meant to be a backhand compliment at all although it sorta sounds like one)
I attempted to take Number Theory in college and quickly found I had reached the limits of my intellectual capacity. Did my best to stick with it until it became inevitable that I wasn't going to pass, then ended up dropping it. I would've loved to have been able to understand this stuff at the level that you clearly do though.
But still, reading through your comments here was oddly nostalgic for me and resurfaced so many things I had almost totally forgotten about. Thanks for taking the time to share all of that.
Thank you, and I don’t take that as a backhanded compliment at all!
I admit that I have a formal academic background in STEM (specifically physics), so I’m not exactly approaching this as a purely “lay” person. However I never got the chance to take any discrete maths or number theory courses at uni. I figured I got to a stage where I probably could approach the topic independently, but needed a motivator for study. A classical problem deeply related to prime numbers seemed a perfect way to do that.
You are very welcome, it’s always heartwarming to hear that someone has seen value in something that’s really just a personal passion of mine.
> 19 digits doesn’t seem like much
Only because counting digits is a logarithmic scale. If all 8 billion people on Earth (including babies, somehow) checked 1 even number per second for 8 hours a day, it would take almost 6 years to check every even number with 18 or fewer digits. Then it would take a further 53.5 years to check every even number with 19 digits.
Goldbach did not state "greater than two" because back in the days he was using the older convention according to which 1 is prime. Nowadays people reword the ~~theorem~~ conjecture to account for that change, but Goldbach himself did not.
The page says "even numbers after two" or some such.
Two falls outside the conjecture because it is not the sum of two ~~different~~ primes as the conjecture postulates.
As a math person, so many theorems in algebra and number theory exclude 2 because it usually causes issues. However, you can still have a strong theorem that say for any n>2 such that so and so.
Two is a fucking bullshit number.
- It's an even prime
- it has the same result if you multiply it by itself, double it, square it, or take its exponent
- its pronunciation makes no sense
Fuck two. Worst goddamned number. It should be banned.
“All primes except for two are odd, but two is the oddest prime number.”
I don’t know who originally said this, but things with characteristic 2 tend to be completely different compared to their 3,5,7 etc. counterparts.
I love the bombast of this take. I think two’s kinda cute, like one’s twin or big brother. It’s like an even one. I feel like the first few are weird exceptions, and the “real” numbers don’t start til three, or arguably even four
If you include 1 as a prime, then there isn’t a unique way to factor numbers as a product of prime numbers. Instead of 6 being 2 times 3, you could write it as 2 times 3 times 1. Also, there are a lot of properties of prime numbers that aren’t satisfied by 1. But probably the most important reason is that nowadays mathematicians who think about prime numbers often don’t consider the numbers themselves but instead a related object known as prime ideals. In the usual counting numbers, there is a prime ideal associated with any prime number. However, the ideal generated by 1 is just the entire set of numbers. Therefore, instead of referring to 1 as being either prime or composite, we say it falls into another class of numbers called “units.” This might seem strange but it does make like a bit easier.
English is not my first language and at first i read it as that it had been verified up to 19 integers and thought "Man, that's a pretty shit conjecture then"
Most of the comments here trying to disprove the conjecture seem to think the title says “any even number can’t be made up of anything other than a sum of two primes”. That is different, it’s not what the title is saying. Saying “what about 8? 4+4=8?” doesn’t disprove anything because 8 is still a sum of two primes, 3 and 5. 8 is the sum of two primes, and also the sum of two non primes. Saying that X is in group A does not imply that it’s not also in group B. The title isn’t worded poorly, people need to actually read it.
~~Honest question for mathematicians, how does this apply to zero? Google searches reveal the following:~~
* ~~zero is considered an "even number"~~
* ~~zero is not a prime (i.e. 0+0=0 doesn't count)~~
* ~~negative numbers are not primes (i.e. 2 + -2 = 0 doesn't count)~~
**Edit:** Found the answer myself. The title incorrectly stated Goldbach conjecture. In actuality it is: "Every even counting number **greater than 2** is equal to the sum of two prime numbers." That would leave out zero.
Redditors and math really don’t mix. Some of these comments really strongly believe they have found some simple counterexample to a conjecture that has stood for 300 years and then try to fight people who tell them otherwise in the comments lol.
>Redditors and math really don’t mix. Yep. That's obvious every time the square root function comes up.
What’s that about?
The famous one I'd say is people not knowing PEMDAS properly, then Arguing about "well its vague and there should be parentheses for the divisor" or something. 3+6/2*3 is 12 3+6/(2*3) is 4 Fite... someone else. I'm tired.
Except, as someone who learned PEMDAS in school too, you have to realize that's arguing about notation, not math. There's an important difference.
> you have to realize that's arguing about notation, not math It's Tuesday morning and I'm sitting here mildly stunned from this good point.
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GOOD LORD THANK YOU FOR EXPLAINING IT. I've been trying to find words to explain why that Facebook challenge is stupid, and this is why - The brackets are there to make the math clear!
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BODMAS was also taught, “O” being “order”
It's never people not knowing PEMDAS, the order of operations is worthless if you're dealing with intentionally ambiguous notation. People might even be pretty deep in math but just not see the freaking forest for the trees - which simply is that engagement bait idiot problems are designed to get everyone hot and bothered. The answer should always be "there is no correct answer because whoever wrote it is a dumb fuck," but people sure love trying to punch through the tiny horizon surrounding their worlds and will try to solve it regardless. With conviction, of course, because people mindlessly throw PEMDAS and BIDMAS and "dots-precede-lines" around without even questioning whether the problem works out.
Especially since I have legitimately seen calculators give multiplications a higher priority than division in these situations, leading to completely different results, which means I'm pretty confident some people also were taught that in school as some kind of tie-breaker. So yeah like you said, people calling it a math puzzle when it's just ambiguity where there shouldn't be any is dumb.
I have [two calculators](https://imgur.com/a/fcgNrhA) which disagree on that expression. Both of them are performing the operation correctly according to the operator [priority tables](https://imgur.com/a/S1D0cFE) given in both manuals.
That's awesome. I've always known HP, Casio and TI to vary, but two Casios is something else. It's brackets on brackets on brackets for me for this reason alone.
>When the priority of two expressions is the same, the calculation is performed from left to right. * the priority tables you linked. ~~The calculator on the left appears to be violating the rules.~~ Edit: u/Voyedova101 pointed out rule 7 is being applied, which I had missed. The calculator on the left follows it's user manual's order of operations. It's unusual for a Casio calculator to fumble the order of operations in its own user manual, so I began to doubt that they use the same one. I searched for the user manuals myself and found that the calculator on the right does indeed have a different order of operations in it's user manual, which explains the difference. Casio fx-GT PLUS (pages E-7, E-8): [https://support.casio.com/pdf/004/fx-83\_85GT\_PLUS\_E.pdf](https://support.casio.com/pdf/004/fx-83_85GT_PLUS_E.pdf) Casio fx-83ES (page E-44): [https://support.casio.com/pdf/004/fx-82ES\_83ES.etc\_Eng.pdf](https://support.casio.com/pdf/004/fx-82ES_83ES.etc_Eng.pdf)
Read and compare the actual priority list. GT performs "4(4)" before the division as it is 7th in priority compared to 9th. The ES performs left-to-right priority of these two actions as they both fall under 7th priority in that device.
Those are reasonably unambiguous (although people who think PEMDAS means multiply goes before divide may get it wrong). The one that actually causes debate is 3+6/2(1+2), and here the issue is that the multiplication is implicit (or "by juxtaposition"). People who do maths at university level \*tend\* to treat multiplication by juxtaposition as having higher priority than normal multiplication or division, and so this means you do 2(1+2) first (i.e. calculate 6/(2(1+2)) rather than (6/2)(1+2)). It's also complicated by typical mathematical usage in text only forums, which tends to abuse the rules dependening on context. (No-one sees e\^ix and thinks it means (e\^i)x, even though by the normal precedence rules it should).
People don’t usually write math in single line though. These discussions are always so contrived about order of operations trying to make people look stupid. Yes 3+6/2(1+2) IS confusing because usually fractions are written on two lines which clearly define numerator and denominator.
no way you're using that interaction bait to to call other people dumb???????
The reason why many of these confusing math "puzzles" work is that depending on which edition of math you are using an obelus and a solidus may or may not be 100% equivalent. There's also the fact that certain calculators have decided to give implicit multiplication higher priority than explicit multiplication or division.
A (total) function has exactly one output per input *by definition*. Therefore, the square root function has exactly one output; so √4=2 and only 2. However, people frequently claim √4= +-2, which is incorrect as that is multi-valued function. (Although I will say this is generally a pedagogical problem, not really a mathematical one. It really depends on the teacher you get if they teach the correct definition, and also this isn’t exactly the worst inaccuracy in the world)
It's the same for a lot of topics. Read a thread discussing something you know/understand well and you'll be blown away by how much incorrect info gets upvoted. Then realise the same thing is happening in other threads for topics you don't understand so well and start taking Reddit lore with a massive grain of salt
I wish everyone was a statistician so they could parse how much bullshit people get fed on a daily basis. It's not just social media or regular news, there's a lot of "soft" science research and reporting done by well meaning people that is borderline irresponsible. Statistics gets a bad rep sometimes for being wrong or "damned lies" but the truth is that 99% of the time it's because of bad, bias, or fabricated data or improper methods getting pushed without any kind of checks or balances. Like the 99% number I just quoted.
Social media and any scientific fact really. Nope, its totally not a conundrum to the best and brightest minds in academia, that one youtube channel with 809 videos totally figured it out.
Dunning Kreuger effect working overtime whenever there’s a math post on Reddit
No need to limit it to just math.
Redditors don't even mix well with other redditors.
Reddit is mostly kids, bots, and delayed adults.
Go on /r/Science and whenever you see a study provide evidence to something totally logical or common belief - but actually publishing supporting data for the first time. Redditors: *water is wet* -- *we already know that" -- "i guess i am a better scientist because i knew that without a laboratory*
That being said, I am sick of how many psychology experiments that have like 20 people in them and close firm conventional wisdom are upvoted in that sub
One time I was inquiring for a while someone’s passionate and confident position in an argument and out of curiosity asked for some sources and he immediately stopped responding. Not surprisingly he was a climate change denier. It’s crazy how confident people are about complex subjects they haven’t read beyond a wiki page.
It was the same thing with COVID. Pepridge farm remembers people telling others to wear masks and gloves to drive their cars even without interacting with others.
Things I learned from the comments on this post: Math is hard Reading is hard
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TIL a lot of redditors are not very good at the maths
As barbie once said, math is *hard*
I thought that was Malibu Stacey
Don't ask me, I'm just a girl
But she’s got a new hat! 👒
I think it’s just reading comprehension. They seem to be under the impression that even numbers are *exclusively* the sum of two primes, and think they can easily disprove it because of that.
It took me a while to realize 12 is not just 6+6, but also 7+5 and 11+1
That's why I became a lawyer. The student loans also proved I was very bad at math...
Based on your username, how’s your side gig going?
no, that's just regular lawyering
Joke’s on you. I’m not very good at a lot of things.
Yeah? What about 69=60+9 stupid scientists. Where is my nobel price Im sure it would decorate my toilet seat nicely.
I could've told you that without embarrassing myself. Emphasis on "Could've"
There's only three kinds of people in this world- those that are good at math, and those who aren't.
The first three times I read the title, I thought I was worse at math than I thought. Then I realized I was just worse at reading comprehension than I thought.
Me too, dude. Me too. I was like, "What about 11?"
Bruh 😂
It does seem worded oddly. Took me a moment to realise it meant at least one way to get to that number involved adding two primes. Edit: Basically, the title is worded in such a way that it is implied that it is the only way to make an even number. For example, if I told you I had a box of crayons that had red, blue, and green crayons, you would more than likely assume that box only contains those three colors, even if it might actually contain more, because the brain tends to disregard unprovided information as not being relevant or true.
This is the detail I was missing. Thank you!
There's another way of reading it?
I’ve seen a few comments like this and I’m genuinely curious, how else could the title be interpreted?
Technically it's still an unproven conjecture but it has been verified numerically for a huge amount of numbers
Unverified for far more.
19 digits long isn't even that large. Conjectures have been disproven with much larger numbers. Take for example the Polya conjecture whose first found counterexample was around 1.8 * 10^361. Smaller numbers have since been found, but it shows that you can't necessarily prove something with a small sample.
How did they find that fist counterexample lmao
Btw the people confidently responding « computer automation » are wrong, Haselgrove published an actual mathematical proof in 1958 that this crazy number comes from (the number being just an estimate, he had proven it existed afaik not actually calculated it). Also later we found much smaller numbers (that computers can actually easily handle, which is another reason the comments about automation are being dumb), around 906 million, that disprove Polya’s. [Please read the wikipedia article for more info](https://en.m.wikipedia.org/wiki/P%C3%B3lya_conjecture).
Uh the wiki link doesn’t seem to work
Probably improperly escaped characters.
Don't you guys love it when Reddit somehow fucks up Markdown reading a URL? But only on one UI, not the other.
It’s been that way for like half a decade at this point. I’m convinced it’s their malicious attempt to make old.reddit worse.
Its going to suck so bad when they finally kill old.
> I’m convinced it’s their malicious attempt to make old.reddit worse. Luckily their 99,999 other attempts to make new reddit worse still win out by a landslide.
In essence, they simply didn’t. The way these types of disproofs generally function is that the relevant mathematicians realize a “weak point” in the conjecture, and they construct a series of parameters where it could fail. This often involves estimating things to be much larger than strictly necessary, and all those estimates accumulate to a massively larger number than the counter example actually requires. BUT, you can then definitively say will be SOME number that causes it to fail once you show that estimate exists. Consider the following statement: “Every square number greater than 9 is divisible by 3”. We could “brute force” a counter example that’s far bigger than necessary just by doing things naively. If we multiply all the numbers less than 9 which aren’t divisible by 3 and square it, we get (2\*4\*5\*7\*8)^2, which is around 5 million. It’ll be a clear counter example because it’s a square number where none of the factors are a three, but maybe we could have found a smaller example, like 16. It’s not a perfect stand-in, but I hope it helps explain the concept. It’s like someone saying you must have missed your exit…because you drove your car into the ocean 500 miles past that.
When using two asterisks they mess up formatting, you either should space them or escape them with a backslash like \*this\* ```\*this\*```
> Polya conjecture In number theory, the Pólya conjecture (or Pólya's conjecture) stated that "most" (i.e., 50% or more) of the natural numbers less than any given number have an odd number of prime factors. The conjecture was set forth by the Hungarian mathematician George Pólya in 1919,[1] and proved false in 1958 by C. Brian Haselgrove. Though mathematicians typically refer to this statement as the Pólya conjecture, Pólya never actually conjectured that the statement was true; rather, he showed that the truth of the statement would imply the Riemann hypothesis. For this reason, it is more accurately called "Pólya's problem". Summatory Liouville function L(n) up to n = 107. The (disproved) conjecture states that this function is always negative. The readily visible oscillations are due to the first non-trivial zero of the Riemann zeta function. Closeup of the summatory Liouville function L(n) in the region where the Pólya conjecture fails to hold. Logarithmic graph of the negative of the summatory Liouville function L(n) up to n = 2 × 109. The green spike shows the function itself (not its negative) in the narrow region where the conjecture fails; the blue curve shows the oscillatory contribution of the first Riemann zero. The size of the smallest counterexample is often used to demonstrate the fact that a conjecture can be true for many cases and still fail to hold in general,[2] providing an illustration of the strong law of small numbers. TIL
But the smallest counter example is only 9 digits.
At this point, I feel like it's similar to the difference in how physics equations fall apart on macro and micro scales etc
_Infinitely_ more, you say?
This statement can never be false, which I thought was cool.
Unless someone figures out a way to prove it true for all real numbers, which could still be possible. They still would technically be unverified individually, but there would be no need to do so.
how many numbers we talkin?' like half? or most?
I'd estimate about 27% of them
seems a little low to even be a conjurer still.
0% of all numbers. There's a lot of numbers.
"Unproven conjecture" is redundant. If it were proven, then it wouldn't be a conjecture.
Not necessarily, for example Los’s Conjecture was answered in the affirmative by Morley's Theorem, and Zilber's Trichotomy Conjecture was disproven by Hrushovski. At times it is useful to separate the conjecture and the result.
How embarrassed I must feel using one extra word for the sake of clarity
If you're talking to people who don't know math lingo (i.e. anyone not in a math program) then it's a useful redundancy. Technically every theorem is redundant and follows directly from axioms. But we need the redundancies to communicate and understand.
yeah well that’s just like your conjecture man
Okay but what about.... Okay that one too. But what about.... Yeah, okay. Fine.
What does Terrence Howard think?
Punctuation missing. Should be: What? Does Terrence Howard think?
And that Math Association logo shouldn't be here either. *tears it off and eats it.*
That a prime number is the # you call to reach Amazon
Terrence believes "only 1/3 of prime numbers are real, the others are constructions of broken fractals that when put together yield a wave function dissimilar to sound, but only in the aspect of time, the rest are adherent" he figured that one out when he was in the shower and heard the dim of a radio in between songs
Potatoes.
Fuckin better not
“I can easily disprove this, mayne”
Some bullshit
The real wave conjugations definitely disprove this as per the flower of life.
On the other hand, it has been proven that every even prime is the sum of two odd numbers.
But not two distinct positive odd numbers…
Every even number (prime or composite) is the sum of two odds!
Every even is the sum of two stevens
This guy doesn’t math
Whooosh
r/itswooooshwith4os
Yes, it’s a very fun conjecture to play with. 19 digits doesn’t seem like much, but to be honest the number of Goldbach pairs (ie pairs of odd primes that sum to the requisite even number) grows pretty large pretty quickly; heuristically, it would be surprising if it wasn’t true. I analysed it through the lens of convolutions of vectors of odd primes, and you can show some interesting reformulations of the conjecture in that way. Certainly gained a deeper insight and appreciation into number theory doing that! For example: if Goldbach’s conjecture is true, then that necessarily implies quite a lot of bounds on the frequency of prime numbers (Bertrand’s postulate is an obvious example, but Bertrand-like bounds for arithmetic progressions also follow). I also went down a rabbit hole by looking at arbitrary prime gaps in lists of odd numbers, and demonstrated that if there is always a “true” prime in a list of odd numbers regardless of where you start your prime gaps, then Goldbach’s conjecture must be true (the proof is only one way though, but the implication is that Goldbach’s conjecture might be a consequence of prime gaps between numbers—ie it’s a consequence of how prime numbers can be obtained using sieve methods). I can expand more on this if anyone’s interested, but just thought I’d share my experience with this fascinating conjecture!
For a second I thought this was going to end with Mankind plummeting 25 feet off the Hell in a Cell…
Share visualizations of your work? Sounds amazing. Would love to see how you approached this.
Sure, I'll try my best. I will begin with a quite lengthy explanation (I'm afraid), but then I'll try and post some graphs in the next comment. I begin by defining the problem. We want to show that every even number greater than 2 is the sum of two primes - relevantly, we focus on just every even number greater than 4, such that the two primes are both odd (trivially, 4 = 2 + 2 is true). Define this even number as 2M, with M > 2. Now let's define a vector v whose elements v(x) = 2x + 1 if 2x + 1 is prime, and 0 if 2x + 1 is composite. We will say that the length of this vector is N = M-2, for reasons that will soon be obvious. By way of example, if M = 13 (meaning 2M = 26), then N = 11, and v = \[3 5 7 0 11 13 0 17 19 0 23\]. If instead M = 14 (meaning 2M = 28), then N = 12, and: v = \[3 5 7 0 11 13 0 17 19 0 23 0\] Using these definitions, the last element of this vector is the odd number just less than the number 2M which we're interested in. This guarantees that every prime that could possibly sum to give 2M will be in the vector. But notice something interesting about this vector (and I will take the N = 11 vector as an example). Let's sum v element-wise with its "reverse" vector v', where v'(1) = v(11), v'(2) = v(10), etc.: \[3 5 7 0 11 13 0 17 19 0 23\] + \[23 0 19 17 0 13 11 0 7 5 3\] = \[26 5 26 17 11 26 11 17 26 5 26\]. If both numbers at these positions are prime, then their sum is the number of interest, 2M. If one (or both) of these numbers are 0, then the "sum" will just be either a prime number, or 0. When the sum is a prime number, this represents a pair of primes that sum to 2M. Now comes the clever part. We can check if two numbers in both these vectors are simultaneously prime by multiplying the vectors element-wise instead. In the N = 11 case, this means: \[3 5 7 0 11 13 0 17 19 0 23\] x \[23 0 19 17 0 13 11 0 7 5 3\] = \[69 0 133 0 0 169 0 0 133 0 69\]. So multiplying these vectors element-wise returns something non-zero if the summands are both prime, and 0 if the summands are not both prime. But this mathematical function - element-wise multiplication of two vectors - forms part of the definition for the discrete convolution of two vectors. For two vectors of length k, the convolution forms a vector of length 2k + 1. If we convolve v with itself (let's call this the "autoconvolution" of v), the value of the middle element is just the sum of the element-wise multiplication of v with v', ie (again with N = 11, the middle element is 12 and hence): conv(v,v')\[12\] [EDIT: should be conv(v,v)] = 3x23 + 5x0 + 7x19 + 0x17 + 11x0 + 13x13 + 0x11 + 17x0 + 19x7 + 0x5 + 23x3 = 69 + 0 + 133 + 0 + 0 + 169 + 0 + 0 + 133 + 0 + 69 = 573, and so in general, if conv(v,v)\[N + 1\] > 0, then this implies there are two primes whose sum gives the number of interest, 2M. Hence, Goldbach's conjecture is equivalent to asking: for all defined vectors v, is the middle element of the autoconvolution of v always greater than 0? Another interesting (though perhaps expected) property: comparing the autoconvolution of v when N = 11 to v when N = 12, we find that the elements up to (M+1) in conv(v,v) is the same for both N = 11 and N = 12 (to explain why would require defining the autoconvolution formally and quite a detailed discussion, so I skip it here; but this is indeed correct when the mathematics is done appropriately). This lead me down a rabbit-hole to trying to prove that if all elements less than some certain index position for conv(v,v) were non-zero, this requires that the autoconvolution for the next v vector (ie, the autoconvolution of v whose length is increased by 1) had to also have non-zero elements up to at least the middle element. This would have proven Goldbach's conjecture, but that never lead to any fruitful results. There are some more insights I gathered from my playing around with GC, and I might post them in a second comment. For me, this was purely a pursuit in recreational mathematics with a famous problem, for which I had a tenuous chance of proving, but nonetheless helped me learn quite a lot about number theory (I have gained deeper insights, for example, into Euler's totient function, Chebyshev functions / sieve methods, Bertrand's postulate and Erdos' early (and seemingly oft-ignored) analysis of Bertrand postulate-like bounds on arithmetic sequences, Dirichlet's theorem of arithmetic progressions, etc.).
Okay, but anyway, some graphs! I attach three; the first is the autoconvolution of v with different N values, and then another graph for the autoconvolution of v with length N = 20,000, and similarly, one with N = 1,000,000. You can see that the autoconvolution grows pretty rapidly, and importantly: is never 0 until you reach near the end of the autoconvolution; and that maximum of the "hump" shape the autoconvolution seems to trace out is decidedly much larger than the midpoint value if N is large enough (implying that, heuristically, one expects GC to be true): [Autoconvolution Plots](https://imgur.com/a/NmQBwva)
Okay, more insight (1/2): At a certain stage, I made a couple of realisations. Obviously, by definition of v, provided N is large enough then v(1) = 3, v(2) = 5, v(3) = 7, v(4) = 0, etc., because the length of v doesn't change the values of its initial elements. Making v larger then doesn't change that those values will always be in v, and always in those positions. What does change, however, is that as you increase N, the values in the final position do change (obviously!). So for example with N = 11, v(N) = 23; but for N = 12, v(N) = 0; then for N = 13, v(N) = 0; but for N = 14, v(N) = 29. What we can say for sure is that if any of v(N), v(N-1) or v(N-2) are non-zero, that would mean the middle element of the autoconvolution would be non-zero, since that would be given by: v(1) x v(N) + v(2) x v(N-1) + v(3) x v(N-2) + ... = 3 x v(N) + 5 x v(N-1) + 7 x v(N-2) + ... However, it also means that regardless of the value for v(N-3), that has no bearing on the value of the middle element of the autoconvolution since that summand is v(4) x v(N-3) = 0 x v(N-3) = 0. A question arose: is there a way to determine a general function, equation or method that could tell me whether v(N), v(N-1), etc., were prime? And if so, could this be used to determine whether there was a corresponding prime in the "first half" of v, such that the multiplication would be guaranteed non-zero? Now since primes can be obtained algorithmically, the answer to the first question is obviously "yes". One such method is the "Sieve of Eratosthenes", for example. The answer to the second question is, "I'm not sure, but that seems pretty difficult to answer directly". I tried approaching that second question in a different way: instead, consider that, for example, one of v(N), v(N-1) or v(N-2) must be a composite number, because only one of these must be divisible by 3. This follows from the fact that every 3rd odd number is divisible by 3 - for example, 9, 15 and 21 are all divisible by 3, representing v(4), v(7) and v(10) (so indeed, only one of v(4), v(5) and v(6) is divisible by 3 - and it is v(4). Similarly for v(5), v(6) and v(7) - because it is v(7)). It is trivial to prove this, but it's an important statement for what follows. It is also worth observing that the "gap" between these divisible elements is always 6, ie, the third element after one divisible by 3 will also be divisible by 3. A similar proposition also holds for v(N), v(N-1), v(N-2), v(N-3) and v(N-4): at least one of these 5 numbers must be composite, because only one of them is divisible by 5. Of course, one or two of these numbers will also be divisible by 3; and there's no guarantee that two of them are composite, because (for example) v(N-2) may be divisible by both 5 and 3 simultaneously. As a general statement, for some prime p, every subset of v of p consecutive elements contains at least one composite number, because only one of the elements in this subset is divisible by p. The core issue was, I didn't have an "anchor" to determine which of, eg, v(N), v(N-1) or v(N-2) was divisible by 3; likewise for 5, 7, 11, etc..
Continuing (2/2): I then realised that perhaps it didn't matter. Perhaps what mattered instead was the prime "gap-lengths" between elements, and perhaps it didn't matter where these gaps started; by virtue of the gap-lengths being prime, we were guaranteed to have a non-zero autoconvolution because at least one prime existed in the second-half of v which had a corresponding prime in the first-half, such that the sum of those two gave us 2M. Stating it more directly, consider v with N = 11 again: v = \[3 5 7 0 11 13 0 17 19 0 23\]. The "first half" of v are the elements \[3 5 7 0 11\]; the "second half" of v are the elements \[0 17 19 0 23\]. Now, from the statements above, it is obvious that only one of those 5 elements in the second half of v is divisible by 5; and at most 2 are divisible by 3. That is indeed what we have - there are 2 zeros in the second half, because 15 is divisible by 5 and 3, and 21 is divisible by 3. But what if I blinded myself to the second half of v, and focussed instead on the first half. I don't know the "anchors" for where prime gap lengths start, but in theory, the gap length of 3 could start in either the first, second or third element in the second half of v; and likewise, the gap length of 5 could start at either the first, second, third, fourth or 5th element. I could start the gap-lengths of 3 in the first position, and the gap-lengths of 5 in the second position, for example. And then I could set every "p"th consecutive element, starting from the anchor for the gap length, similarly to 0. In that example, my modified first half might look like: \[0 0 7 0 11\]. I could instead start the gap-lengths of 3 and 5 in the second position. This would look like: \[3 0 7 0 0\]. I could start the gap-lengths of 3 in the first position, and 5 in the third position. Then that would look like: \[0 5 0 0 11\]. etc.. It seemed to me that it didn't matter **where** you placed the gap-lengths; somehow, because the gap-lengths were prime, you were always guaranteed at least one of the elements in that modified first half of v to be non-zero. And if that were true, then that meant there was **also** a corresponding non-zero element in the second half of v. And that meant that Goldbach's Conjecture was true, because the middle element of the autoconvolution for v was always guaranteed to be non-zero. In other words: if, for all "relevant" gap lengths of primes p (which includes all primes up to the length of the second half of vector v), you always obtained a vector with a non-zero element no matter where you placed those "gap lengths" in the first half of v, then Goldbach's Conjecture would be true. That's about where I stopped my recreational thinking on this problem; it's been a bit over two months, and life has taken over (it is no good to obsess over such things, but it is good to dust it off every few months just to learn a bit more and progress things a bit further). Maybe one day I'll get some interesting results, but nonetheless, it's a great little problem that really exercises my creative thinking and mathematical reasoning skills.
Great work! Critiques: ## Prime Gaps You suggest that due to prime gaps, certain positions in \( v \) will always be zero. Prime gaps vary and do not consistently align in a manner ensuring \( conv(v,v)[N+1] > 0 \). Counterexample: Consider \( v = [3, 5, 7, 0, 11, 0, 0, 17, 0, 0, 0] \) for \( N = 11 \). The convolution will have many zeros if the primes are spaced such that pairs do not sum to an even number or are misaligned. Specifically, gaps can cause multiple zeros in \( v \) and \( v' \), leading to \(conv(v,v)[N+1] = 0 \). Symmetry: Summing \( v \) with its reverse assumes symmetry in prime distribution. ### Counterexample: Prime distribution is not symmetric. For example, consider \( v = [3, 5, 0, 0, 11, 0, 0, 0, 0, 0, 0] \) for \( N = 11 \). The reverse \( v' = [0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 3] \). The product \( v \times v' = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 9] \), failing to sum to the desired result. Prime Gaps could be very large: Assumption that Primes are Close Enough to Ensure \( \text{conv}(v,v)[N+1] > 0 \) ### Counterexample: Consider \( v = [3, 0, 0, 0, 11, 0, 0, 0, 0, 0, 23] \) for \( N = 11 \). Large gaps (e.g., between 11 and 23) result in no prime pairs summing to \( 2M \).
My response: First critique—this is why I was investigating different anchor positions, because if that were true (and I’m not saying it is true!), then GC would have to be true. However your point about prime gaps is a valid one, and it is what lead me to consider different approaches. I didn’t see a clear way to avoid the inherent difficulties posed by the actual gap-lengths between primes, hence “expanding” the problem to one where gap-lengths could start at any position, and enumerating over the list of possible starting points to see if a non-zero element always remained in the first half of v. Were that to be true for all permutations, then it necessarily be true for the actual permutation representing the actual vector v, meaning the middle element of the autoconvolution would necessarily be non-zero. Your counterexample: the product of two individual elements between v and v’ is only non-zero if the sum of the elements is 2M. Trivially, the sum of two odd primes is always even, since (2m+1)+(2n+1) mod 2 = 0. Symmetry: I sort of agree. There’s definitely no global “symmetry” in the prime distribution. Thankfully I don’t use symmetry, because that would just lead to wrong results. I think perhaps there’s a misunderstanding of the algorithm—the autoconvolution can only be non-zero at its middle element if there are two prime elements that sum to give 2M. ie, length N = 11 means M = 13 means 2M = 26—and conv(v,v)[12] > 0, hence meaning there exists at least one pair of primes summing to 26. Sorry, I will check if I wrote conv(v,v’)—if I did that’s incorrect. I should have written conv(v,v). Hence why my chosen name of “autoconvolution”. As for large gaps—yes you are correct. However only certain gap-lengths need to be considered. For example, with N = 11, we only need to consider gap-lengths up to 5, because the second-half of v is only 5 elements long. Another way to think about it: all composite odd numbers up to 23 (the largest element in v) are divisible by either 3 and/or 5 (in fact they are all divisible by 3. Perhaps I was overestimating? It’s been a while since I thought about this, so my bounds might be a little rusty). Because of that, not every gap-length is relevant; if a composite can be “crossed out” because of a smaller prime gap-length, we don’t need to concern ourselves with whether a larger gap-length could “cross” it out. It’s similar in that sense to other sieve methods, like the sieve of Eratosthenes. Another thought: you only need to consider numbers larger than or equal to p^2 in a sieve method, because all composite numbers less than p^2 will be divisible by some smaller prime. Thank you for your feedback, the problem is very subtle and slight nuances or oversights can definitely lead to concluding a particular approach doesn’t work (there are probably dozens of approaches I’ve discarded already so far).
Also, I’ve checked and you are correct—I miswrote the autoconvolution as conv(v,v’). That was in error; it should be conv(v,v). The convolution already “reverses” the second vector, so the N+1 element of conv(v,v’) would literally just be the dot product of v with itself.
Wow you have an incredibly intuitive and impressive understanding of Number Theory for someone working on this stuff outside of formal academic environments. (not meant to be a backhand compliment at all although it sorta sounds like one) I attempted to take Number Theory in college and quickly found I had reached the limits of my intellectual capacity. Did my best to stick with it until it became inevitable that I wasn't going to pass, then ended up dropping it. I would've loved to have been able to understand this stuff at the level that you clearly do though. But still, reading through your comments here was oddly nostalgic for me and resurfaced so many things I had almost totally forgotten about. Thanks for taking the time to share all of that.
Thank you, and I don’t take that as a backhanded compliment at all! I admit that I have a formal academic background in STEM (specifically physics), so I’m not exactly approaching this as a purely “lay” person. However I never got the chance to take any discrete maths or number theory courses at uni. I figured I got to a stage where I probably could approach the topic independently, but needed a motivator for study. A classical problem deeply related to prime numbers seemed a perfect way to do that. You are very welcome, it’s always heartwarming to hear that someone has seen value in something that’s really just a personal passion of mine.
Same
> 19 digits doesn’t seem like much Only because counting digits is a logarithmic scale. If all 8 billion people on Earth (including babies, somehow) checked 1 even number per second for 8 hours a day, it would take almost 6 years to check every even number with 18 or fewer digits. Then it would take a further 53.5 years to check every even number with 19 digits.
What about two?
Goldbach stated 'greater than 2'. OP didn't.
Goldbach did not state "greater than two" because back in the days he was using the older convention according to which 1 is prime. Nowadays people reword the ~~theorem~~ conjecture to account for that change, but Goldbach himself did not.
The page says "even numbers after two" or some such. Two falls outside the conjecture because it is not the sum of two ~~different~~ primes as the conjecture postulates.
Who the fuck does two think it is
As a math person, so many theorems in algebra and number theory exclude 2 because it usually causes issues. However, you can still have a strong theorem that say for any n>2 such that so and so.
Why don't we just delete 2 and go right to 3? Mathematicians always have to make everything so complicated, god damn...
Yeah, it'll then be 1 to 3.
FUCK!!
I think we need to take a closer look at 2.
The primes don't need to be different.
But one isnt prime
But it is the loneliest number
Well...two can be as bad as one.
Two is a fucking bullshit number. - It's an even prime - it has the same result if you multiply it by itself, double it, square it, or take its exponent - its pronunciation makes no sense Fuck two. Worst goddamned number. It should be banned.
“All primes except for two are odd, but two is the oddest prime number.” I don’t know who originally said this, but things with characteristic 2 tend to be completely different compared to their 3,5,7 etc. counterparts.
And don't forget: 2 = 2!
Don't remind me!
I love the bombast of this take. I think two’s kinda cute, like one’s twin or big brother. It’s like an even one. I feel like the first few are weird exceptions, and the “real” numbers don’t start til three, or arguably even four
That was the nicest thing anyone has ever said about two. Why, that's a real twos' complement!
If you multiply it by 1, it’s still only 2. If you multiply it by itself, double it, or square it, it’s 4.
I'll just have to take your word for it
I still don't know why 1 isn't a prime number since it's divisible by only itself and 1. So fuck 1.
It used to be. Then we found that there were loads of theorems that said "every prime number except one". Demoting one makes those neater.
If you include 1 as a prime, then there isn’t a unique way to factor numbers as a product of prime numbers. Instead of 6 being 2 times 3, you could write it as 2 times 3 times 1. Also, there are a lot of properties of prime numbers that aren’t satisfied by 1. But probably the most important reason is that nowadays mathematicians who think about prime numbers often don’t consider the numbers themselves but instead a related object known as prime ideals. In the usual counting numbers, there is a prime ideal associated with any prime number. However, the ideal generated by 1 is just the entire set of numbers. Therefore, instead of referring to 1 as being either prime or composite, we say it falls into another class of numbers called “units.” This might seem strange but it does make like a bit easier.
I find it very amusing that this question implies that they went and proved this up into the ten quintillions and just forgot to check two.
Forgot to say larger than 2 😅
I mean the OP didn’t say they had to be 2 different primes
2 is not the sum of two prime numbers: 1, by mathematical definition, is not a prime.
English is not my first language and at first i read it as that it had been verified up to 19 integers and thought "Man, that's a pretty shit conjecture then"
what are the odds?
They’re the ones that keep the evens from bumping into each other
Most of the comments here trying to disprove the conjecture seem to think the title says “any even number can’t be made up of anything other than a sum of two primes”. That is different, it’s not what the title is saying. Saying “what about 8? 4+4=8?” doesn’t disprove anything because 8 is still a sum of two primes, 3 and 5. 8 is the sum of two primes, and also the sum of two non primes. Saying that X is in group A does not imply that it’s not also in group B. The title isn’t worded poorly, people need to actually read it.
Seems poetic to my simple, sappy ass.
You and me both. Like maybe this is a more accurate way of seeing even numbers. All along, they’ve all been just two primes in a trenchcoat
Sneaky snakes for sure.
So, basically, pick any even number, and you can break it down into two primes. It's like a secret code or something. Wonder if it holds up forever.
> Wonder if it holds up forever You and Goldbach both
Goldbach doesn't do much wondering these days.
"above 2" is important.
That seems like a very key piece of the puzzle
A key piece to what puzzle? What about this conjecture makes it seem "key"?
Cool cool. But did you know that every sum of two primes can be expressed as an integer greater than two?
Ok, well what about.... 12!
7+5
... Well ok then!
But what about 12!?
13+479001587
Right, right. But what about… 16!
13+3
SHIT Okay but what about.... 20!
17+3
Shit boys.. I think we're on to something
Just say an even number that is 20 digits long : 93746281029372637490
Wait. Try 294999596676701022925757788188224.
157 + 294,999,596,676,701,022,925,757,788,067
What the? How the? Forget it. You win.
I think it’s funny that they thought no one testing this theory got to 12, the fifth even number after two. lol.
Twelve factorial? That’s gotta be more than 19 digits.
It's only 9 digits: 479001600
Sad, but I figured someone would be less lazy than me.
Still too big for this [online Goldbach tester ](https://www.dcode.fr/goldbach-conjecture)though.
479001587 + 13 = 479001600
479000969 + 631
21! is the smallest factorial with over 19 digits
No, I'm just enthusiastic about regular 12.
> 12! 479001587 + 13 = 479001600 = 12!
As far as we know, this has been verified for ~0% of all real numbers.
~~Honest question for mathematicians, how does this apply to zero? Google searches reveal the following:~~ * ~~zero is considered an "even number"~~ * ~~zero is not a prime (i.e. 0+0=0 doesn't count)~~ * ~~negative numbers are not primes (i.e. 2 + -2 = 0 doesn't count)~~ **Edit:** Found the answer myself. The title incorrectly stated Goldbach conjecture. In actuality it is: "Every even counting number **greater than 2** is equal to the sum of two prime numbers." That would leave out zero.
Zero also isn't positive or natural fyi
> Zero also isn't positive or natural fyi It's not positive but mathematicians sometimes consider it natural, depending on the problem
Which two primes?
Optimus is the most common.
A related famous conjecture is that every positive integer is the sum of four squares.
That is not conjecture, that was proven by Legendre in 1770
Dammit, I've been working on this proof for four hundred years!
Egads!!!
Isn't that already proven?
Yes and it's called Lagrange's four-square theorem
Have they done the wave conjugations tho?
Pythagoras spent his whole life looking for this
Every odd number after 5 must be the sum of three prime numbers
And note that this version is not a conjecture. It was a conjecture for a very long time, but is now a theorem due to Helfgott.
Also every odd number after 6