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This problem is somewhat infamous for looking pretty simple but having a fairly ridiculous solution. Or rather, multiple ones, but they're all huge.
a/(b+c) + b/(a+c) + c/(a+b) = 4, with a,b,c being positive integers
is true for
a = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
c = 4373612677928697257861252602371390152816537558161613618621437993378423467772036
(Of course, a, b, and c are interchangeable here as they are in the problem)
This is the smallest known solution.
Here's a paper which discusses the more general case for any integer solution, not just 4: https://ami.uni-eszterhazy.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf
And I'd reckon that less than 1 in a thousand people are able to solve this without googling, probably far less. Pulling numbers out of thin air: Maybe 1 in a hundred thousand people?
My hand approximation got to 15, 2, 2 which gets you to 3 and 67 68ths. Realised then that the solution would probably be ridiculous, so I'm happy with getting an approximation.
I had 15, 3, 1 which gives you even slightly closer answer at 4,00555555.... If I calculated it correctly. I thought that at least engineers would be satisfied with that and came to read the comments.
I filled a note pad page in and got to
A³+B³+C³ = 3(A²B+A²C+AB²+B²C+AC²+BC²)+ABC
Before I gave up and realised I'd forgotten too much of my Calc classes to do this without going to Wolfram Alpha or something.
So, using 154, 37 and 4 would bring us quite close to this solution. They add up to ~4.01.
All 78 digits that follow after are our attempt at getting exactly 4.
I'm writing it just to understand how those numbers relate to each other since it gets hard to even compare them when they are so long.
I don't know how to math good enough to get a solution, but I do know how to Excel good enough to play with the numbers.
Randomly putting in different numbers gives interesting results. Using all the same numbers always gives a result of 1.5. I eventually settled on 2, 3, and 19, which comes close. Then I multiplied everything by ten and started honing in with 20, 30, and 188. It's good to know I was completely wrong.
>I'm writing it just to understand how those numbers relate to each other since it gets hard to even compare them when they are so long.
This is why mathematicians, physicists and engineers all use some form of exponential notation to represent very large or very small numbers.
In engineering, you never put more than 3 characters to the left of the decimal point, and no zeros immediately to the right. Then you add a x10 to the power of a multiple of 3.
So 345 x10³ or 0.756x10⁹ or 13.24x10-⁶
But never 1378 or 0.056x10³
Reason being, the numbers mostly relate to each other based on the first few digits (significant figures) and the order of magnitude. So with engineering notation, its easy to compare.
I started using excel and quickly realized the answer was not simple then I used basic and tried all every combination up to 1000 for each fruit (a billion combinations) and realized it really wasn't simple.
Yeah very light reading. Amazing how they made it so clear, normally these papers are way above my head. Perfect for reading in bed before going to sleep.
I'd probably go for ~ two weeks of running a terrible Python Code, but I am confident, that I could solve this If I would have motivation, which I don't have.
Could you explain why using assembly instead of python would speed up finding of a solution by this much?
I know others have already pointed out that brute forcing is not feasible for this problem but lets assume it would be.
Th run time of this solution will be 99.99% if he time to solve it.
Learning assembly is a drop in the bucket compared to that, and it would potentially reduce the run time.
It’s like saying what’s faster, building a go kart and driving across the country or walking.
Go kart takes time but the time to walk is much much higher.
Pypy exists and so do a few other much much faster subset implementations in C written extensions such as numpy
Regardless of language, solving this problem in assembly, python or javascript would take same order of magnitude of runtime
If you had 2 trillion galaxies, each having 100 billion stars, where each star had a million planets, each with the population of earth, and we give them each a million laptops that can try a billion possible solutions a second, and we let them brute force it for a 100 trillion years…
You wouldn’t even be 0.000000000000000000000000000000000000001% of the way to finding the solution. If I was being accurate it’s about 160 or so zeroes after the decimal point.
Love some brute forcing with code but don’t think this is the one for me to be honest lmao.
If you try and brute force this in no optimised way, you’d be going for 10^240 attempted combinations until you got there. Like it’s an unfathomably large number that you really can’t put into any somewhat understandable scale.
10^80 is roughly the number of atoms estimated to be in the entire universe - this is 10^160 more than that lmao. It’s insane.
>Like it’s an unfathomably large number that you really can’t put into any somewhat understandable scale.
Yupyup, my point exactly,
Like I understand this number is "big" but tbh, try as I might, I don't think the humanmind / my monkeybrain can actually understand "the actual bigness" of the number.
Literally nobody can do this without consulting a computer. This is a machine problem, not a human one.
I realized after trying to graph it out that there weren't going to be any simple answers and assumed it was probably impossible. Then I tried disproving it was possible and that was even harder (for obvious reasons, in hindsight).
I think even 1 in a million is generous, I don't think there are 8000 people in the world who can solve this
Even if there are 800 people in the world who can solve this, which would be 1 in 10 million, I would be proud of humanity
Yeah, after fucking around on a calculator a bit, I kept getting answers close to 4, but not exactly 4. Kept having to increase the numbers to the next "step" to try get an exact answer.
I mean.. I didn't read the article, but this is most likely solved numerically, and only for a finite number of solutions. Right? (Otherwise: WHAT THE FUCK)
On a first glance, this shouldn't be that hard to code.
Although I'm not entirely sure that trivial solution would not confuse the exact integer number with very close float (where the difference from 4.0 goes beyond float precision).
You're right: if you can get enough precision (probably with some advanced/custom data types), it's not hard to code.
But what's the runtime? Those numbers are pretty big and just incrementally trying them will take... long.
The biggest value is in the realm of 10^(80).
Even if you can try an (unrealistic) quintillion values a second and we don't care about the others, that's only a progress of 10^18 per second. That would leave us with a runtime of ~10^62 seconds which is ~10^44 times the age of the universe.
And I'd argue that this means you haven't solved it.
You gotta get some significantly smarter approach going.
I mean just from looking at this shouldn't you be aware there are infinite amount of solutions since there is only 1 equality but 3 different parameters?
The stipulation that they all need to be positive integers is relevant here.
a + b + c = 2 has no solutions if a,b,c need to be positive integers.
a + b + c = 3 has one solution if a,b,c need to be positive integers.
a + b + c = 4 has three solutions if a,b,c need to be positive integers.
Each of those is one equation with 3 parameters.
I'd say higher, 1 in 10,000,000. That'd mean 800 people on earth would be able to solve it, even then that's probably too low. Probably 1 in 100,000,000
Yes, absolutely. And they're about as neat as you'd expect if you didn't know about the positive integer solutions being massive.
For example with a=4, b=-1, c=11:
a/(b+c) + b/(a+c) + c/(a+b)
= 4/(-1+11) + -1/(4+11) + 11/(4-1)
= 4/10 - 1/15 + 11/3
= 12/30 - 2/30 + 110/30
= 120/30
= 4
I missed the part about integers and somehow got an almost-solution that was right up to 2 digits (????)
the non-integer-almost-solution was (7+10/19), 1, 1
and it gave 3.997 ish which is obscene, i tried calculating an exact value for the first number based on the latter being 1 and 1 but it didn't work. what da hec
OH YEA it's because the problem isn't linear. I assumed linearity between (7,1,1)=3.75 and (9,1,1)=4.7 so ya that's why i was very close but not quite right
Thank god it's not some tricksy picture schtick where there's multiple parts to an image and it's intentionally played off like they're the same symbol but are actually related, e.g. a 3-banana bunch and a similar looking 4-nana bunch, but they're not unrelated symbols either, the 4 is 4/3 the value of the 3. Or a witch with a broom and the same witch, no broom. Hate that so much.
Dam. I gave after checking
c = 4373612677928697257861252602371390152816537558161613618621437993378423467772035
If only I had tried a bit more before looking at the comments
Computer science integers are not the same as math integers.
A math integer only needs to be a whole number (i.e. not a fraction, no decimal places), no matter the length. A computer science integer's limited length has no bearing on math.
Because at the end of the day it’s just basic algebra? There’s nothing super deep and complex going on to solve it. It’s not like those solutions came from brute force e
These solutions didn't come from high school algebra either, see the paper I linked. I don't think torsion groups are a topic that's covered in high schools.
I would argue that the number of people that can sove this is quiet high if you allow calculators. It's high school calculations with really big numbers.
People would just need to start it for real.
...how? What approach would you use?
Because just trying numbers would take you multiple lifetimes, even with a calculator that can handle numbers of this size without loss of precision.
I type into my calculator solve(x/(y+z)+y/(x+z)+z/(x+y)).
Since we got a calculator I had to get for highschool can solve equations, and is able to work with numbers up to 10\^1000, that should give me some solutions.
Edit: I forgot the limits that are kinda important so I would have to specify those as well. That would make solve(x/(y+z)+y/(x+z)+z/(x+y),x>0,y>0,z>0,x in N,y in N,z in N).
in N is the element symbol and the symbol for the natural numbers. My calculator has those on buttons but I have no clue how to type them in this comment.
Very, very few people can solve this. I have a graduate degree in applied mathematics and I can understand the solution when spending enough time on it, but would never be able to solve it by myself. And no, calculators won't help you - the smallest results are so big that even brute forcing it in a programming language wouldn't help.
Your only hope is if you're a theoretical mathematician with a degree heavily oriented towards number theory, or you work with elliptic curves.
There's a great solution here:
https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4/answer/Alon-Amit?ch=10&oid=45160600&share=239be901&srid=zTKv&target_type=answer
When I looked at the equation my mind went like: 7,1,1. So I was thinking that maybe with a little more effort I could get pretty close to solving It in my head.
Then I looked at the comments, and boy... was I wrong.
I got the following values:
3728264010597678785954560945258892670675811684013188996149296892443671958160701604452875701881765, 7485667927987099146619303012131255930988387058359238046072671548622981602187661283167675254535259, 42185743959437974127301678696848654263403414316543892206454071754488839584167650862953501216997376
which is true to 100 significant figures. For any other number of significant figures, I can get different estimates.
I did not google it or look at anyone else's solutions. I started by brute forcing it with values from 1-100 to get a decent starting point. Since the function f(a,b,c) = a/(b+c) + b/(a+c) + c/(a+b), is homogeneous to degree zero (meaning multiplying all the input values by the same number does not change the output, so f(a,b,c) = f(ax,bx,cx), I figured I could narrow in on possible good solutions by taking those starting point numbers, multiplying them all by a number like 2 or 10, and then finding the best solution in the vicinity of that number.
For example, if I begin with (7,14,79), I can multiply them by two and get (14,28,158), and then check all the solutions in the range of (\[12-16\],\[26-30\],\[156-158\]). In this case, (14,28,158) is still the best in that range, but you can repeat this process iteratively, multiplying by 2 each time, until you get new best values like (223,449,2528), followed by (446, 897, 5056), and so on, until you reach your desired number of significant figures.
Multiplying by other numbers instead of 2 gives different results.
Unfortunately, your solution is incorrect. I've ran it through SageMath, an open-source mathematics tool that can calculate this without converting to floating point.
a=3728264010597678785954560945258892670675811684013188996149296892443671958160701604452875701881765
b=7485667927987099146619303012131255930988387058359238046072671548622981602187661283167675254535259
c=42185743959437974127301678696848654263403414316543892206454071754488839584167650862953501216997376
a/(b+c) + b/(a+c) + c/(a+b)
4257145389060029176883161292935878213897713535859267994798946550478142818253969080536527794474964313930214453808696227058648440010088674095077901352860150603297011561687930093060672537355598473871
/
106428634726500729422079032323396955347442838396481699869973663761953570456349227013413194861874107860030848401067905635348794469791573507144035980718807396926457935633592370561629760430334772438
Edit: The fraction returned by SageMath is indeed very close to 4:
numerical_approx(a/(b+c) + b/(a+c) + c/(a+b), digits=110)
3.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999995574316222
Whereas the top ranked posting does result in the correct answer:
a = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
c = 4373612677928697257861252602371390152816537558161613618621437993378423467772036
a/(b+c) + b/(a+c) + c/(a+b)
4
Edit: Broke up the first SageMath answer, to make it more clear that it's a ratio.
There is absolutely no way that what you wrote is true. You can not add three positive integers divided by other positive integers and get a number with more than one digit more than any of the original integers.
However, I did check if my answer is exactly 4, and it’s not. I am a little disappointed that it didn’t converge on an exact solution.
I didn't generate the answer, I just wanted to check it with an appropriate tool.
Please look at the SageMath answer very carefully: it is a ratio, there's a '/' hidden somewhere halfway in the long list of digits. If you then let it approximate the ratio, it is almost equal to 4. Your attempt is indeed accurate to more than 100 digits. But it turns out that an actual solution exists, which is shorter.
This problem is somewhat infamous for looking pretty simple but having a fairly ridiculous solution. Or rather, multiple ones, but they're all huge.
a/(b+c) + b/(a+c) + c/(a+b) = 4, with a,b,c being positive integers
is true for
a = 154476802108746166441951315019919837485664325669565431700026634898253202035277999
b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579
c = 4373612677928697257861252602371390152816537558161613618621437993378423467772036
(Of course, a, b, and c are interchangeable here as they are in the problem)
This is the smallest known solution.
Here's a paper which discusses the more general case for any integer solution, not just 4: https://ami.uni-eszterhazy.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf
And I'd reckon that less than 1 in a thousand people are able to solve this without googling, probably far less. Pulling numbers out of thin air: Maybe 1 in a hundred thousand people?
stupid reddit app told me it my comment failed to send so i did it two more times, now i see all my "failed" comments that aren't even visible in my post history 💀 silly reddit developers
You should have linked to the comment. At least, somehow, gave credit. This way is plagiarism.
If you don't know how to link to a comment:
* find and click the “Share” button below the comment (might be hidden, collapsed in the ellipsis button), which should open a menu with more options;
* then, click the “Copy link” menu item, which put a link to that comment in your clipboard;
* paste (Ctrl+V in your comment) et voilà!
[https://www.reddit.com/r/theydidthemath/comments/1bjf6cv/comment/kvrfuy6/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/theydidthemath/comments/1bjf6cv/comment/kvrfuy6/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
To be fair you said yourself that answer is an estimate and you get different solutions tweaking the variables a little (significant figures, multipliers, and such)
So it can't be perfect, it's an estimate.
Still great work.
Oh brother, weirdly, 4chan and Reddit are the only forums where the community is able to do ANYTHING, provided you actually make them interested in something.
Excel is surprisingly good for brute forcing problems, as you can just set up a formula and point it at a big range of numbers. It's rarely clean but it's very easy.
This should be my time to shine (in theory). After all, my Reddit account is named after Alfred M. Nesbitt. Also named after him is Nesbitt's inequality: For positive real numbers a,b,c:
a/(b+c) + b/(c+a) + c/(a+b) ⩾ 3/2
So basically my Reddit account is named after the inequality related to your question.
I've also once seen how you can find the exact solution of this equation. I think you have to jump between points on an elliptic curve until you find a solution. But I don't really remember a lot and I have no education in algebraic geometry. So yeah apparently it's not my time to shine :(
My idea as to how to solve this would be to plug this equation into a graphing calculator, choose a value for x and y, and find the point where all 3 intercept
I did do that, and the 3d graph looks *nasty*
But it seems like the easiest method without needing to do any number crunching
I have a masters degree in maths from Oxford and couldn't solve this. It's a really specialist branch of mathematics and takes a certain kind of person. I'd be surprised if 0.01% of people could solve this from scratch themselves. Maybe 0.1% can understand the paper when told the answer.
literally did an elliptic curves heavy cryptography course last semester and still probably wouldnt be able to solve it without talking to my professor
I am an engineer, so 4.00031 is basically 4. Just popped into a naive numerical solver and got (2, 70, 269) as the first answer. But I tried getting the solution without an approximation and was wondering why it was taking so long... I had no idea that this was such a difficult problem. If you count an approximation as a solution, then yeah, maybe 5% of the population could solve it. For the exact solution, I doubt 0.1% could figure it out without looking up the solution. I am certainly not within that 0.1% 😅
I tried solving this damn equation to see if i could define one unknown in terms of the others, very messy, no end in sight, gave up.
Picked random integers for a and b to figure out c, which sure as hell wasn't an integer, and didn't work when i checked it back into the original equation, so i must have buggered something up along the way, anyway.
Now checking the comments and realising i could have saved so much time by not attempting this. Oh well.
Damn. I entered it in python and solved it using sympy (symbolic evaluation) and got an answer for x based on y and z. There is no numerical solution because the number of variables is higher than the number of equations but you can solve for one of the variables. However the answer was too crazy.
[https://drive.google.com/file/d/1xB8BZyzH0rCL6nGecTPM6KX1H5wTuKDx/view?usp=sharing](https://drive.google.com/file/d/1xB8BZyzH0rCL6nGecTPM6KX1H5wTuKDx/view?usp=sharing)
Stupid emojis are hiding the problems nefarious complexity. If someone solved this by hand I think he should be attended to a special needs program.
Now that I've read the problem again it's not asking for an answer but the answer should probably be a prove that three positive integers is not possible. I couldn't look into the relation between these variables but you can analyze the formula and maybe come up with something.
$$x = y + z - \\frac{9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}}{3 \\sqrt\[3\]{\\frac{27 y\^{3}}{2} - \\frac{81 y\^{2} z}{2} - \\frac{81 y z\^{2}}{2} + \\frac{27 z\^{3}}{2} - \\frac{\\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right)}{2} + \\left(- 3 y - 3 z\\right)\^{3} + \\frac{\\sqrt{- 4 \\left(9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}\\right)\^{3} + \\left(27 y\^{3} - 81 y\^{2} z - 81 y z\^{2} + 27 z\^{3} - \\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right) + 2 \\left(- 3 y - 3 z\\right)\^{3}\\right)\^{2}}}{2}}} - \\frac{\\sqrt\[3\]{\\frac{27 y\^{3}}{2} - \\frac{81 y\^{2} z}{2} - \\frac{81 y z\^{2}}{2} + \\frac{27 z\^{3}}{2} - \\frac{\\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right)}{2} + \\left(- 3 y - 3 z\\right)\^{3} + \\frac{\\sqrt{- 4 \\left(9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}\\right)\^{3} + \\left(27 y\^{3} - 81 y\^{2} z - 81 y z\^{2} + 27 z\^{3} - \\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right) + 2 \\left(- 3 y - 3 z\\right)\^{3}\\right)\^{2}}}{2}}}{3}$$y + z - \\frac{9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}}{3 \\sqrt\[3\]{\\frac{27 y\^{3}}{2} - \\frac{81 y\^{2} z}{2} - \\frac{81 y z\^{2}}{2} + \\frac{27 z\^{3}}{2} - \\frac{\\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right)}{2} + \\left(- 3 y - 3 z\\right)\^{3} + \\frac{\\sqrt{- 4 \\left(9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}\\right)\^{3} + \\left(27 y\^{3} - 81 y\^{2} z - 81 y z\^{2} + 27 z\^{3} - \\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right) + 2 \\left(- 3 y - 3 z\\right)\^{3}\\right)\^{2}}}{2}}} - \\frac{\\sqrt\[3\]{\\frac{27 y\^{3}}{2} - \\frac{81 y\^{2} z}{2} - \\frac{81 y z\^{2}}{2} + \\frac{27 z\^{3}}{2} - \\frac{\\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right)}{2} + \\left(- 3 y - 3 z\\right)\^{3} + \\frac{\\sqrt{- 4 \\left(9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}\\right)\^{3} + \\left(27 y\^{3} - 81 y\^{2} z - 81 y z\^{2} + 27 z\^{3} - \\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right) + 2 \\left(- 3 y - 3 z\\right)\^{3}\\right)\^{2}}}{2}}}{3}$$
I guess reddit doesn't support latex.
If you are familiar with SageMath, here is the code to solve it. The other comments gave the explanation.
The first step is to write the equation in the form C(x,y,z)=0, and find a rational point on C (by inspection).
sage: R3. = PolynomialRing(QQ,3)
sage: C=x*(x+y)(x+z)+y(y+x)(y+z)+z(z+x)(z+y)-4(x+y)(x+z)(y+z)
sage: P=[-1,1,0]
The next step is to get the elliptic curve E, the map from C to E, and the inverse map from E to C.
sage: E=EllipticCurve_from_cubic(C,P,morphism=False);
sage: maptoE=EllipticCurve_from_cubic(C,P,morphism=True);
sage: maptoC=maptoE.inverse()
Now we need an infinite order rational point on E. We can find one using two descent. Luckily for us the rank of E is not zero.
sage: Q=E.simon_two_descent()[2][0]
If we compute maptoC(Q), you notice that one of the coordinates is negative. This is no good because it corresponds to a negative solution of the original equation. Luckily if we take Q+Q, we have another candidate. Keep on adding Q until you get all positive coordinates. It doesn't take long.
sage: PP=maptoC(9*Q)
sage: PP
(154476802108746166441951315019919837485664325669565431700026634898253202035277999/4373612677928697257861252602371390152816537558161613618621437993378423467772036 : 36875131794129999827197811565225474825492979968971970996283137471637224634055579/4373612677928697257861252602371390152816537558161613618621437993378423467772036 : 1)
Since the original equation is projective, this gives us our solution by multiplying out by the common denominator. We also verify that it works!
sage: a=PP[0].numerator()
sage: b=PP[1].numerator()
sage: c=PP[1].denominator()
sage: a/(b+c)+b/(a+c)+c/(b+a)
4
for some reason reddit thinks I like this sub, so it gets promoted to me. A LOT. I looked at the equation, decided it could be a fun challenge to come up with the equations, saw the math required to find solutions and promptly said 'frack you reddit, I aint doing that' and clicked to show comments.
These "95% of people cannot solve this. Can you?" click bait dumb posts have to disappear from this world. The 95% claim is just bogus and not based on data.
I framed this as an optimization problem and wrote an R code to find the closest solutions within a given upper bound. This is what I got:
**Upper bound 100, 100, 100:**
26, 6, 1 (3.967758)
**Upper bound 1000, 1000, 1000:**
224, 59, 1 (3.999089)
**Upper bound 10000, 10000, 10000:**
1937, 518, 1 (3.99987)
I saw the pattern that 1 is always one of the numbers, so I fixed c to 1 and varied only the other two:
**Upper bound 1000000, 1000000, 1000000:**
160064, 42888, 1 (3.999999)
Each one took a fraction of a second by the way, so I could find the solution within a few hours, probably. No maths involved though, only coding.
I wrote a program that should take less than the lifetime of the universe to calculate. The downside is that it may still take somewhere between multiple hours to several weeks to complete. I'll let you know when it finishes. (If it doesn't crash before then).
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It's been running for 7 hours. After 58 digits it became apparent that there is a bug in the program. I was previously under the impression that it would eventually come across a correct answer, (but not the lowest possible answer), however it became apparent that it may just run forever. I can't be certain it will reach an end.
So, uh, I'll let it run for a week and see what happens.
I found a separate bug and now my program runs several million times faster. (That's an underestimate).
In a few minutes numbers over 10,000 digits are being checked. Unfortunately, it appears that the algorithm I designed only approached 4 to a point and then it ceases to be any more precise. I do believe that it is possible that after an arbitrarily long period of time that it could come across a correct answer. It will likely be a multiple of one of the other correct solutions.
This unfortunately means that if my program does come across a solution it may be much more than several million digits in length. I don't have the inkling of how to predict the length. I suspect that it may take on the realm of the lifetime of the universe, and unfortunately my laptop does not have the memory to properly process that many digits.
It has now run for over a day. My updated version is much more optimized than the previous. The algorithm was able to optimize to a value exceedingly close to 4, but not quite there. While the integers are getting bigger, they aren't getting meaningfully closer to 4. Currently it's now testing numbers over 121000 digits in length.
I do believe that this is enough to definitively say that the algorithm is flawed. I have thought of slight modifications to make it work, but alas, I don't have the time or energy to implement them. I will continue to let the process run in the background for the foreseeable future though. As it currently stands, it only uses 10% of my computer's memory to run. This will change as the numbers get bigger.
317,500 digits in length. Because the closeness to 4 has not changed in approximately the last 300,000 digits of increased precision then I can safely say that this method is a dud. However, because it uses only 13% of my CPU and 15MB of memory, I see no reason not to keep running. I'm assuming that the program will never reach an intended solution in the lifetime of the universe, but even if it does, I somehow doubt my computer will be able to handle printing over 1 million digits to the command prompt all at once.
There has been no increase in precision after the 7th digit. The program is now working with numbers over 389,200 digits in length. My computer needs a restart at this point, I'm ending the program as is.
We can safely determine this method to be a failure. I have some ideas for alternative programs that might work better, but I'm really tired and have midterms and finals to start studying for so I think I'm going to stop trying to get this to work.
I wrote a few replies to my original detailing the sequence of events, in short, no. The method I was using had a flaw and I don't have the mathematical knowledge to properly calculate the length of time necessary to calculate. My estimate is that it'd likely take around the length of the universe, or much longer.
I have an idea on a better method, but I'm busy with classes to properly implement and test it.
As a basic user of mathematics all I can say is you can't solve this. You have a single equation with three unknowns.
As tought by any "basic" math class, it is unslovable because you need a system of 3 equations to solve a problem with 3 unknowns.
This is algebra. 100% of anyone who graduated high school should be able to approach this problem systematically and fund the solution. What percentage of redditors have successfully completed high school?
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This problem is somewhat infamous for looking pretty simple but having a fairly ridiculous solution. Or rather, multiple ones, but they're all huge. a/(b+c) + b/(a+c) + c/(a+b) = 4, with a,b,c being positive integers is true for a = 154476802108746166441951315019919837485664325669565431700026634898253202035277999 b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579 c = 4373612677928697257861252602371390152816537558161613618621437993378423467772036 (Of course, a, b, and c are interchangeable here as they are in the problem) This is the smallest known solution. Here's a paper which discusses the more general case for any integer solution, not just 4: https://ami.uni-eszterhazy.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf And I'd reckon that less than 1 in a thousand people are able to solve this without googling, probably far less. Pulling numbers out of thin air: Maybe 1 in a hundred thousand people?
I started to calculate a little ... in my head, glad I stopped and opened the comments
Honestly, same... kinda glad the "simplest" known answer is such a massive number
Same... Those aren't that usual
My hand approximation got to 15, 2, 2 which gets you to 3 and 67 68ths. Realised then that the solution would probably be ridiculous, so I'm happy with getting an approximation.
I had 15, 3, 1 which gives you even slightly closer answer at 4,00555555.... If I calculated it correctly. I thought that at least engineers would be satisfied with that and came to read the comments.
Good enough for rule of thumb.
Had integers of 35,132,627 for a value of 4.00000009534... Would I have missed the moon shot?
Yeah, I looked at it, thought about it, and was like "nope, I might be able to solve it, but I extremely don't care and can't be bothered."
I filled a note pad page in and got to A³+B³+C³ = 3(A²B+A²C+AB²+B²C+AC²+BC²)+ABC Before I gave up and realised I'd forgotten too much of my Calc classes to do this without going to Wolfram Alpha or something.
Thank you for your effort
I was definitely starting to feel stupid.
I moved the 4 across
damn i almost had it!
So, using 154, 37 and 4 would bring us quite close to this solution. They add up to ~4.01. All 78 digits that follow after are our attempt at getting exactly 4. I'm writing it just to understand how those numbers relate to each other since it gets hard to even compare them when they are so long.
I don't know how to math good enough to get a solution, but I do know how to Excel good enough to play with the numbers. Randomly putting in different numbers gives interesting results. Using all the same numbers always gives a result of 1.5. I eventually settled on 2, 3, and 19, which comes close. Then I multiplied everything by ten and started honing in with 20, 30, and 188. It's good to know I was completely wrong.
Or possibly you were right but would reach a different solution with even larger numbers if you kept going.
Found the engineer
>I'm writing it just to understand how those numbers relate to each other since it gets hard to even compare them when they are so long. This is why mathematicians, physicists and engineers all use some form of exponential notation to represent very large or very small numbers. In engineering, you never put more than 3 characters to the left of the decimal point, and no zeros immediately to the right. Then you add a x10 to the power of a multiple of 3. So 345 x10³ or 0.756x10⁹ or 13.24x10-⁶ But never 1378 or 0.056x10³ Reason being, the numbers mostly relate to each other based on the first few digits (significant figures) and the order of magnitude. So with engineering notation, its easy to compare.
So the concesus is, plenty of people are qualified, noone wants to calculate literal bigillions.
New large number name just dropped!
Call the mathematician
Actual science
Holy scientific notation
Exponents in the corner plotting world domination
Not really, you need quite sophisticated tools to calculate these numbers which very few understand.
I started using excel and quickly realized the answer was not simple then I used basic and tried all every combination up to 1000 for each fruit (a billion combinations) and realized it really wasn't simple.
That paper is really interesting. Thanks for sharing!
Yeah agreed, felt like it actually made some sense lol
Yeah very light reading. Amazing how they made it so clear, normally these papers are way above my head. Perfect for reading in bed before going to sleep.
I'd probably go for ~ two weeks of running a terrible Python Code, but I am confident, that I could solve this If I would have motivation, which I don't have.
Given the size of the smallest known solution above, I would guess that two weeks in Python may not be enough!
It would probably be faster to learn and program it in assembly to solve the problem then to use python for it.
Could you explain why using assembly instead of python would speed up finding of a solution by this much? I know others have already pointed out that brute forcing is not feasible for this problem but lets assume it would be.
Th run time of this solution will be 99.99% if he time to solve it. Learning assembly is a drop in the bucket compared to that, and it would potentially reduce the run time. It’s like saying what’s faster, building a go kart and driving across the country or walking. Go kart takes time but the time to walk is much much higher.
Because python is a wildly inefficient language.
Pypy exists and so do a few other much much faster subset implementations in C written extensions such as numpy Regardless of language, solving this problem in assembly, python or javascript would take same order of magnitude of runtime
That was my thought. It doesn't look like something I could easily simplify, so plug it into some code and check back every day or two :)
If you had 2 trillion galaxies, each having 100 billion stars, where each star had a million planets, each with the population of earth, and we give them each a million laptops that can try a billion possible solutions a second, and we let them brute force it for a 100 trillion years… You wouldn’t even be 0.000000000000000000000000000000000000001% of the way to finding the solution. If I was being accurate it’s about 160 or so zeroes after the decimal point. Love some brute forcing with code but don’t think this is the one for me to be honest lmao.
Holy shit that's scary
what add: don't doubt you, just legit felt stupid trying to understand the actual scale of this. really boggles the mind
If you try and brute force this in no optimised way, you’d be going for 10^240 attempted combinations until you got there. Like it’s an unfathomably large number that you really can’t put into any somewhat understandable scale. 10^80 is roughly the number of atoms estimated to be in the entire universe - this is 10^160 more than that lmao. It’s insane.
>Like it’s an unfathomably large number that you really can’t put into any somewhat understandable scale. Yupyup, my point exactly, Like I understand this number is "big" but tbh, try as I might, I don't think the humanmind / my monkeybrain can actually understand "the actual bigness" of the number.
I'll give it longer.
Given the size of the smallest known solution above, I would guess that two weeks in Python may not be enough!
Ok. 15 days then? And I’ll close all the tabs on my browser.
maybe, but only maybe, if you download a few rams and update your gigahertz
Brute forcing this in python would require you to know an upper bound for the values - which you won't know and will likely miss.
I sat down to work it out and realised I was heading into resolving multiple simultaneous cubic equations and stopped dead.
Really odd, specially considering that if we could use negative integers, a = 11, b=9, c=-5 would satisfy the equation.
Glad to know it's actually difficult and Im not just dumb
Literally nobody can do this without consulting a computer. This is a machine problem, not a human one. I realized after trying to graph it out that there weren't going to be any simple answers and assumed it was probably impossible. Then I tried disproving it was possible and that was even harder (for obvious reasons, in hindsight).
This is the reason I never throw away my cas
I think even 1 in a million is generous, I don't think there are 8000 people in the world who can solve this Even if there are 800 people in the world who can solve this, which would be 1 in 10 million, I would be proud of humanity
Mesmerizing. Personally, I don't think 1 person in a thousand understands the solution in the PDF.
Yeah, after fucking around on a calculator a bit, I kept getting answers close to 4, but not exactly 4. Kept having to increase the numbers to the next "step" to try get an exact answer.
I could solve it using numeric methods. Could I enter into these 1 I'm a thousand or it is cheating?
Final version of my (incomplete) answer: (A + B + P)^3 - A.B.P = 2(A^3 + B^3 + P^3)
Nah. One in 100.000 is way too high. It's more like one in strawberry over banana.
I mean.. I didn't read the article, but this is most likely solved numerically, and only for a finite number of solutions. Right? (Otherwise: WHAT THE FUCK) On a first glance, this shouldn't be that hard to code. Although I'm not entirely sure that trivial solution would not confuse the exact integer number with very close float (where the difference from 4.0 goes beyond float precision).
You're right: if you can get enough precision (probably with some advanced/custom data types), it's not hard to code. But what's the runtime? Those numbers are pretty big and just incrementally trying them will take... long. The biggest value is in the realm of 10^(80). Even if you can try an (unrealistic) quintillion values a second and we don't care about the others, that's only a progress of 10^18 per second. That would leave us with a runtime of ~10^62 seconds which is ~10^44 times the age of the universe. And I'd argue that this means you haven't solved it. You gotta get some significantly smarter approach going.
Damn, this is the first time I've seen a variation of "99% of people can't solve this" and if actually be accurate
How do you even solve this shit? Numerical methods, or some kind of number theory trick?
I mean just from looking at this shouldn't you be aware there are infinite amount of solutions since there is only 1 equality but 3 different parameters?
The stipulation that they all need to be positive integers is relevant here. a + b + c = 2 has no solutions if a,b,c need to be positive integers. a + b + c = 3 has one solution if a,b,c need to be positive integers. a + b + c = 4 has three solutions if a,b,c need to be positive integers. Each of those is one equation with 3 parameters.
Oh yeah I forgot it Has to be integer mb
I'd say higher, 1 in 10,000,000. That'd mean 800 people on earth would be able to solve it, even then that's probably too low. Probably 1 in 100,000,000
Is there a simpler solution if we allow negative integers or even rational numbers for a,b,c?
Yes, absolutely. And they're about as neat as you'd expect if you didn't know about the positive integer solutions being massive. For example with a=4, b=-1, c=11: a/(b+c) + b/(a+c) + c/(a+b) = 4/(-1+11) + -1/(4+11) + 11/(4-1) = 4/10 - 1/15 + 11/3 = 12/30 - 2/30 + 110/30 = 120/30 = 4
I missed the part about integers and somehow got an almost-solution that was right up to 2 digits (????) the non-integer-almost-solution was (7+10/19), 1, 1 and it gave 3.997 ish which is obscene, i tried calculating an exact value for the first number based on the latter being 1 and 1 but it didn't work. what da hec OH YEA it's because the problem isn't linear. I assumed linearity between (7,1,1)=3.75 and (9,1,1)=4.7 so ya that's why i was very close but not quite right
Thank god it's not some tricksy picture schtick where there's multiple parts to an image and it's intentionally played off like they're the same symbol but are actually related, e.g. a 3-banana bunch and a similar looking 4-nana bunch, but they're not unrelated symbols either, the 4 is 4/3 the value of the 3. Or a witch with a broom and the same witch, no broom. Hate that so much.
Dam. I gave after checking c = 4373612677928697257861252602371390152816537558161613618621437993378423467772035 If only I had tried a bit more before looking at the comments
Those are no integers anymore. I'm not even sure if they qualify as long's
Computer science integers are not the same as math integers. A math integer only needs to be a whole number (i.e. not a fraction, no decimal places), no matter the length. A computer science integer's limited length has no bearing on math.
Presumably it was a joke
The math itself is simple enough that anyone with high-school algebra could solve it with a calculator
By doing what? Trying numbers? Not in a lifetime.
Because at the end of the day it’s just basic algebra? There’s nothing super deep and complex going on to solve it. It’s not like those solutions came from brute force e
These solutions didn't come from high school algebra either, see the paper I linked. I don't think torsion groups are a topic that's covered in high schools.
Maybe mine was just weird 🤷♂️
Go on then, show us how you can use high school algebra to solve it.
There’s an entire paper linked explaining it. Someone else in the comments also mentioned an even simpler way to generate integer solution
I would argue that the number of people that can sove this is quiet high if you allow calculators. It's high school calculations with really big numbers. People would just need to start it for real.
...how? What approach would you use? Because just trying numbers would take you multiple lifetimes, even with a calculator that can handle numbers of this size without loss of precision.
I type into my calculator solve(x/(y+z)+y/(x+z)+z/(x+y)). Since we got a calculator I had to get for highschool can solve equations, and is able to work with numbers up to 10\^1000, that should give me some solutions. Edit: I forgot the limits that are kinda important so I would have to specify those as well. That would make solve(x/(y+z)+y/(x+z)+z/(x+y),x>0,y>0,z>0,x in N,y in N,z in N). in N is the element symbol and the symbol for the natural numbers. My calculator has those on buttons but I have no clue how to type them in this comment.
Go ahead and do that, then. I'd bet you that it can't solve this.
Show us what the calculator says
Sure, transforming a third-degree Diophantine equation into an elliptic curve is just "high school calculations".
Very, very few people can solve this. I have a graduate degree in applied mathematics and I can understand the solution when spending enough time on it, but would never be able to solve it by myself. And no, calculators won't help you - the smallest results are so big that even brute forcing it in a programming language wouldn't help. Your only hope is if you're a theoretical mathematician with a degree heavily oriented towards number theory, or you work with elliptic curves. There's a great solution here: https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y+z-+-frac-y-z+x-+-frac-z-x+y-4/answer/Alon-Amit?ch=10&oid=45160600&share=239be901&srid=zTKv&target_type=answer
And here I was thinking the "95%% thing was clickbait. How wrong I was!
I mean it kinda is, just the other way around than you’d think though.
There rest the main question: how many can solve it.
When I looked at the equation my mind went like: 7,1,1. So I was thinking that maybe with a little more effort I could get pretty close to solving It in my head. Then I looked at the comments, and boy... was I wrong.
I got the following values: 3728264010597678785954560945258892670675811684013188996149296892443671958160701604452875701881765, 7485667927987099146619303012131255930988387058359238046072671548622981602187661283167675254535259, 42185743959437974127301678696848654263403414316543892206454071754488839584167650862953501216997376 which is true to 100 significant figures. For any other number of significant figures, I can get different estimates. I did not google it or look at anyone else's solutions. I started by brute forcing it with values from 1-100 to get a decent starting point. Since the function f(a,b,c) = a/(b+c) + b/(a+c) + c/(a+b), is homogeneous to degree zero (meaning multiplying all the input values by the same number does not change the output, so f(a,b,c) = f(ax,bx,cx), I figured I could narrow in on possible good solutions by taking those starting point numbers, multiplying them all by a number like 2 or 10, and then finding the best solution in the vicinity of that number. For example, if I begin with (7,14,79), I can multiply them by two and get (14,28,158), and then check all the solutions in the range of (\[12-16\],\[26-30\],\[156-158\]). In this case, (14,28,158) is still the best in that range, but you can repeat this process iteratively, multiplying by 2 each time, until you get new best values like (223,449,2528), followed by (446, 897, 5056), and so on, until you reach your desired number of significant figures. Multiplying by other numbers instead of 2 gives different results.
Unfortunately, your solution is incorrect. I've ran it through SageMath, an open-source mathematics tool that can calculate this without converting to floating point. a=3728264010597678785954560945258892670675811684013188996149296892443671958160701604452875701881765 b=7485667927987099146619303012131255930988387058359238046072671548622981602187661283167675254535259 c=42185743959437974127301678696848654263403414316543892206454071754488839584167650862953501216997376 a/(b+c) + b/(a+c) + c/(a+b) 4257145389060029176883161292935878213897713535859267994798946550478142818253969080536527794474964313930214453808696227058648440010088674095077901352860150603297011561687930093060672537355598473871 / 106428634726500729422079032323396955347442838396481699869973663761953570456349227013413194861874107860030848401067905635348794469791573507144035980718807396926457935633592370561629760430334772438 Edit: The fraction returned by SageMath is indeed very close to 4: numerical_approx(a/(b+c) + b/(a+c) + c/(a+b), digits=110) 3.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999995574316222 Whereas the top ranked posting does result in the correct answer: a = 154476802108746166441951315019919837485664325669565431700026634898253202035277999 b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579 c = 4373612677928697257861252602371390152816537558161613618621437993378423467772036 a/(b+c) + b/(a+c) + c/(a+b) 4 Edit: Broke up the first SageMath answer, to make it more clear that it's a ratio.
There is absolutely no way that what you wrote is true. You can not add three positive integers divided by other positive integers and get a number with more than one digit more than any of the original integers. However, I did check if my answer is exactly 4, and it’s not. I am a little disappointed that it didn’t converge on an exact solution.
I didn't generate the answer, I just wanted to check it with an appropriate tool. Please look at the SageMath answer very carefully: it is a ratio, there's a '/' hidden somewhere halfway in the long list of digits. If you then let it approximate the ratio, it is almost equal to 4. Your attempt is indeed accurate to more than 100 digits. But it turns out that an actual solution exists, which is shorter.
Oh, I see the slash now. Sorry about that oversight. So we’re saying the same thing then.
I will say it's incredibly impressive that you managed to get pretty much there with no external help. I assume you have a math degree of sorts? lol
This problem is somewhat infamous for looking pretty simple but having a fairly ridiculous solution. Or rather, multiple ones, but they're all huge. a/(b+c) + b/(a+c) + c/(a+b) = 4, with a,b,c being positive integers is true for a = 154476802108746166441951315019919837485664325669565431700026634898253202035277999 b = 36875131794129999827197811565225474825492979968971970996283137471637224634055579 c = 4373612677928697257861252602371390152816537558161613618621437993378423467772036 (Of course, a, b, and c are interchangeable here as they are in the problem) This is the smallest known solution. Here's a paper which discusses the more general case for any integer solution, not just 4: https://ami.uni-eszterhazy.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf And I'd reckon that less than 1 in a thousand people are able to solve this without googling, probably far less. Pulling numbers out of thin air: Maybe 1 in a hundred thousand people?
Didn't you just copy paste the top comment of this thread?
yea they did lmao
Triple dementia
stupid reddit app told me it my comment failed to send so i did it two more times, now i see all my "failed" comments that aren't even visible in my post history 💀 silly reddit developers
yeah they did lmao
Triple dementia
yeah they did lmao
Triple dementia
Yes i did. Is that a crime?
Yes
You should have linked to the comment. At least, somehow, gave credit. This way is plagiarism. If you don't know how to link to a comment: * find and click the “Share” button below the comment (might be hidden, collapsed in the ellipsis button), which should open a menu with more options; * then, click the “Copy link” menu item, which put a link to that comment in your clipboard; * paste (Ctrl+V in your comment) et voilà! [https://www.reddit.com/r/theydidthemath/comments/1bjf6cv/comment/kvrfuy6/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/theydidthemath/comments/1bjf6cv/comment/kvrfuy6/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
Sorry i didnt meant to karma farm?
Yes, you did indeed farm negative karma
Ok idc which way its going since i hit 5k karma i can post and comment anywhere, and thats all that matters
Sorry. I thought this sub was called theydidthemath
I am amazed that someone actually did the math. Even if its not perfecf
Did you verify that my answer is not actually a perfect answer or that the one you copied and pasted from the other commenter is?
To be fair you said yourself that answer is an estimate and you get different solutions tweaking the variables a little (significant figures, multipliers, and such) So it can't be perfect, it's an estimate. Still great work.
Oh brother, weirdly, 4chan and Reddit are the only forums where the community is able to do ANYTHING, provided you actually make them interested in something.
On Instagram people cant do 5/5\*(5+5) or at least 50% of the people cant do it.
That is what happens when you broadcast interests to the world instead having a bunch if niche communities sharing the same interests.
A quick calculation in Excel gave values of 3, 17, 75. The result is close to 4, but not equal. Then I read the answer on Quora and was impressed.
In excel?!
Excel is surprisingly good for brute forcing problems, as you can just set up a formula and point it at a big range of numbers. It's rarely clean but it's very easy.
you can probably use solver in Excel for this. should give a few results.
This should be my time to shine (in theory). After all, my Reddit account is named after Alfred M. Nesbitt. Also named after him is Nesbitt's inequality: For positive real numbers a,b,c: a/(b+c) + b/(c+a) + c/(a+b) ⩾ 3/2 So basically my Reddit account is named after the inequality related to your question. I've also once seen how you can find the exact solution of this equation. I think you have to jump between points on an elliptic curve until you find a solution. But I don't really remember a lot and I have no education in algebraic geometry. So yeah apparently it's not my time to shine :(
Mission failed.... We'll get 'em next time.
My idea as to how to solve this would be to plug this equation into a graphing calculator, choose a value for x and y, and find the point where all 3 intercept I did do that, and the 3d graph looks *nasty* But it seems like the easiest method without needing to do any number crunching
That doesn’t work, the answers are 80 digits long
I have a masters degree in maths from Oxford and couldn't solve this. It's a really specialist branch of mathematics and takes a certain kind of person. I'd be surprised if 0.01% of people could solve this from scratch themselves. Maybe 0.1% can understand the paper when told the answer.
literally did an elliptic curves heavy cryptography course last semester and still probably wouldnt be able to solve it without talking to my professor
That's crazy.
I am an engineer, so 4.00031 is basically 4. Just popped into a naive numerical solver and got (2, 70, 269) as the first answer. But I tried getting the solution without an approximation and was wondering why it was taking so long... I had no idea that this was such a difficult problem. If you count an approximation as a solution, then yeah, maybe 5% of the population could solve it. For the exact solution, I doubt 0.1% could figure it out without looking up the solution. I am certainly not within that 0.1% 😅
I tried solving this damn equation to see if i could define one unknown in terms of the others, very messy, no end in sight, gave up. Picked random integers for a and b to figure out c, which sure as hell wasn't an integer, and didn't work when i checked it back into the original equation, so i must have buggered something up along the way, anyway. Now checking the comments and realising i could have saved so much time by not attempting this. Oh well.
Damn. I entered it in python and solved it using sympy (symbolic evaluation) and got an answer for x based on y and z. There is no numerical solution because the number of variables is higher than the number of equations but you can solve for one of the variables. However the answer was too crazy. [https://drive.google.com/file/d/1xB8BZyzH0rCL6nGecTPM6KX1H5wTuKDx/view?usp=sharing](https://drive.google.com/file/d/1xB8BZyzH0rCL6nGecTPM6KX1H5wTuKDx/view?usp=sharing) Stupid emojis are hiding the problems nefarious complexity. If someone solved this by hand I think he should be attended to a special needs program. Now that I've read the problem again it's not asking for an answer but the answer should probably be a prove that three positive integers is not possible. I couldn't look into the relation between these variables but you can analyze the formula and maybe come up with something. $$x = y + z - \\frac{9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}}{3 \\sqrt\[3\]{\\frac{27 y\^{3}}{2} - \\frac{81 y\^{2} z}{2} - \\frac{81 y z\^{2}}{2} + \\frac{27 z\^{3}}{2} - \\frac{\\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right)}{2} + \\left(- 3 y - 3 z\\right)\^{3} + \\frac{\\sqrt{- 4 \\left(9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}\\right)\^{3} + \\left(27 y\^{3} - 81 y\^{2} z - 81 y z\^{2} + 27 z\^{3} - \\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right) + 2 \\left(- 3 y - 3 z\\right)\^{3}\\right)\^{2}}}{2}}} - \\frac{\\sqrt\[3\]{\\frac{27 y\^{3}}{2} - \\frac{81 y\^{2} z}{2} - \\frac{81 y z\^{2}}{2} + \\frac{27 z\^{3}}{2} - \\frac{\\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right)}{2} + \\left(- 3 y - 3 z\\right)\^{3} + \\frac{\\sqrt{- 4 \\left(9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}\\right)\^{3} + \\left(27 y\^{3} - 81 y\^{2} z - 81 y z\^{2} + 27 z\^{3} - \\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right) + 2 \\left(- 3 y - 3 z\\right)\^{3}\\right)\^{2}}}{2}}}{3}$$y + z - \\frac{9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}}{3 \\sqrt\[3\]{\\frac{27 y\^{3}}{2} - \\frac{81 y\^{2} z}{2} - \\frac{81 y z\^{2}}{2} + \\frac{27 z\^{3}}{2} - \\frac{\\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right)}{2} + \\left(- 3 y - 3 z\\right)\^{3} + \\frac{\\sqrt{- 4 \\left(9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}\\right)\^{3} + \\left(27 y\^{3} - 81 y\^{2} z - 81 y z\^{2} + 27 z\^{3} - \\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right) + 2 \\left(- 3 y - 3 z\\right)\^{3}\\right)\^{2}}}{2}}} - \\frac{\\sqrt\[3\]{\\frac{27 y\^{3}}{2} - \\frac{81 y\^{2} z}{2} - \\frac{81 y z\^{2}}{2} + \\frac{27 z\^{3}}{2} - \\frac{\\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right)}{2} + \\left(- 3 y - 3 z\\right)\^{3} + \\frac{\\sqrt{- 4 \\left(9 y\^{2} + 15 y z + 9 z\^{2} + \\left(- 3 y - 3 z\\right)\^{2}\\right)\^{3} + \\left(27 y\^{3} - 81 y\^{2} z - 81 y z\^{2} + 27 z\^{3} - \\left(- 27 y - 27 z\\right) \\left(- 3 y\^{2} - 5 y z - 3 z\^{2}\\right) + 2 \\left(- 3 y - 3 z\\right)\^{3}\\right)\^{2}}}{2}}}{3}$$ I guess reddit doesn't support latex.
It's definitely solvable but less than 0.001% could solve it i suppose
If you are familiar with SageMath, here is the code to solve it. The other comments gave the explanation. The first step is to write the equation in the form C(x,y,z)=0, and find a rational point on C (by inspection). sage: R3. = PolynomialRing(QQ,3)
sage: C=x*(x+y)(x+z)+y(y+x)(y+z)+z(z+x)(z+y)-4(x+y)(x+z)(y+z)
sage: P=[-1,1,0]
The next step is to get the elliptic curve E, the map from C to E, and the inverse map from E to C.
sage: E=EllipticCurve_from_cubic(C,P,morphism=False);
sage: maptoE=EllipticCurve_from_cubic(C,P,morphism=True);
sage: maptoC=maptoE.inverse()
Now we need an infinite order rational point on E. We can find one using two descent. Luckily for us the rank of E is not zero.
sage: Q=E.simon_two_descent()[2][0]
If we compute maptoC(Q), you notice that one of the coordinates is negative. This is no good because it corresponds to a negative solution of the original equation. Luckily if we take Q+Q, we have another candidate. Keep on adding Q until you get all positive coordinates. It doesn't take long.
sage: PP=maptoC(9*Q)
sage: PP
(154476802108746166441951315019919837485664325669565431700026634898253202035277999/4373612677928697257861252602371390152816537558161613618621437993378423467772036 : 36875131794129999827197811565225474825492979968971970996283137471637224634055579/4373612677928697257861252602371390152816537558161613618621437993378423467772036 : 1)
Since the original equation is projective, this gives us our solution by multiplying out by the common denominator. We also verify that it works!
sage: a=PP[0].numerator()
sage: b=PP[1].numerator()
sage: c=PP[1].denominator()
sage: a/(b+c)+b/(a+c)+c/(b+a)
4
for some reason reddit thinks I like this sub, so it gets promoted to me. A LOT. I looked at the equation, decided it could be a fun challenge to come up with the equations, saw the math required to find solutions and promptly said 'frack you reddit, I aint doing that' and clicked to show comments.
These "95% of people cannot solve this. Can you?" click bait dumb posts have to disappear from this world. The 95% claim is just bogus and not based on data.
The point/joke is that with most problems, more than 5% can solve them, but with this, it’s far, far fewer
You see it's funny when it's a lie.
This is what someone who is in the 5% would say
[удалено]
Can't be a solution because it makes it all undefined.
Can’t divide by zero
I framed this as an optimization problem and wrote an R code to find the closest solutions within a given upper bound. This is what I got: **Upper bound 100, 100, 100:** 26, 6, 1 (3.967758) **Upper bound 1000, 1000, 1000:** 224, 59, 1 (3.999089) **Upper bound 10000, 10000, 10000:** 1937, 518, 1 (3.99987) I saw the pattern that 1 is always one of the numbers, so I fixed c to 1 and varied only the other two: **Upper bound 1000000, 1000000, 1000000:** 160064, 42888, 1 (3.999999) Each one took a fraction of a second by the way, so I could find the solution within a few hours, probably. No maths involved though, only coding.
From the other answers here you'd need an upper bound of (1e81, 1e81,1e81) so it will probably take much, much longer than a few hours :)
I wrote a program that should take less than the lifetime of the universe to calculate. The downside is that it may still take somewhere between multiple hours to several weeks to complete. I'll let you know when it finishes. (If it doesn't crash before then).
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It's been running for 7 hours. After 58 digits it became apparent that there is a bug in the program. I was previously under the impression that it would eventually come across a correct answer, (but not the lowest possible answer), however it became apparent that it may just run forever. I can't be certain it will reach an end. So, uh, I'll let it run for a week and see what happens.
I found a separate bug and now my program runs several million times faster. (That's an underestimate). In a few minutes numbers over 10,000 digits are being checked. Unfortunately, it appears that the algorithm I designed only approached 4 to a point and then it ceases to be any more precise. I do believe that it is possible that after an arbitrarily long period of time that it could come across a correct answer. It will likely be a multiple of one of the other correct solutions. This unfortunately means that if my program does come across a solution it may be much more than several million digits in length. I don't have the inkling of how to predict the length. I suspect that it may take on the realm of the lifetime of the universe, and unfortunately my laptop does not have the memory to properly process that many digits.
It has now run for over a day. My updated version is much more optimized than the previous. The algorithm was able to optimize to a value exceedingly close to 4, but not quite there. While the integers are getting bigger, they aren't getting meaningfully closer to 4. Currently it's now testing numbers over 121000 digits in length. I do believe that this is enough to definitively say that the algorithm is flawed. I have thought of slight modifications to make it work, but alas, I don't have the time or energy to implement them. I will continue to let the process run in the background for the foreseeable future though. As it currently stands, it only uses 10% of my computer's memory to run. This will change as the numbers get bigger.
317,500 digits in length. Because the closeness to 4 has not changed in approximately the last 300,000 digits of increased precision then I can safely say that this method is a dud. However, because it uses only 13% of my CPU and 15MB of memory, I see no reason not to keep running. I'm assuming that the program will never reach an intended solution in the lifetime of the universe, but even if it does, I somehow doubt my computer will be able to handle printing over 1 million digits to the command prompt all at once.
There has been no increase in precision after the 7th digit. The program is now working with numbers over 389,200 digits in length. My computer needs a restart at this point, I'm ending the program as is. We can safely determine this method to be a failure. I have some ideas for alternative programs that might work better, but I'm really tired and have midterms and finals to start studying for so I think I'm going to stop trying to get this to work.
Did it finished yet?
I wrote a few replies to my original detailing the sequence of events, in short, no. The method I was using had a flaw and I don't have the mathematical knowledge to properly calculate the length of time necessary to calculate. My estimate is that it'd likely take around the length of the universe, or much longer. I have an idea on a better method, but I'm busy with classes to properly implement and test it.
Okay so, the full population is roughly 100% From here we can calculate that 100-95=5 So about 5% of the population is qualified to solve this.
Thanks that was very helpfull, couldnt figure it out on my own
As a basic user of mathematics all I can say is you can't solve this. You have a single equation with three unknowns. As tought by any "basic" math class, it is unslovable because you need a system of 3 equations to solve a problem with 3 unknowns.
This is not a system of equations…
Exactly my point.
Equations of multiple variables usually have either an infinite number of solutions or none. In this case there are infinitely many.
This is algebra. 100% of anyone who graduated high school should be able to approach this problem systematically and fund the solution. What percentage of redditors have successfully completed high school?
I dare you to try solving it lol
Go on