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I am just struggling, because I know chances are that I didn't find a solution to an unsolved problem in under 3 hours, but I can't find what's wrong with this. It's really bugging me.

https://www.scribd.com/document/332017929/Proof-that-a-perfect-cuboid-does-not-exist

Alex Youcis
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Tristen
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2 Answers2

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Your error is assuming that no solution to $A^2 + B^2 + C^2 = 2D^2$ exists because $D\sqrt{2}$ is irrational.

See, for example, $2 \cdot 29^2 = 28^2 + 27^2 + 13^2$

Michael Biro
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  • What integer when multiplied by the square root of 2 results in a rational number? – Tristen Nov 23 '16 at 04:58
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    @Tristen No integer has that property. You don't need an integer with that property for the equation to have a solution. That's the problem with your proof. – cardboard_box Nov 23 '16 at 05:00
  • Ugh, How do I use latex in the comments? Totally new, this is my first ever post, but I can show you exactly why you do need an integer with that property – Tristen Nov 23 '16 at 05:01
  • So, the reason it is necessary to have an integer with that property is because of the following system of equations and what must be true about them. I first prove that $ (X^2+ Y^2 + Z^2)=(A^2+Z^2)=(B^2+Y^2)=(X^2+C^2)=(D)^2$ In order for a perfect cuboid to exist, then all of these variables must be elements of the natural numbers. Since these are all equal to one another, I can say that $(X^2+ Y^2 + Z^2)+(A^2+Z^2)+(B^2+Y^2)+(X^2+C^2)=4(D)^2$ and since I know that $X^2 + Y^2 + Z^2 = D^2$ It is quite simple to show that $(A^2+Z^2)+(B^2+Y^2)+(X^2+C^2)-(X^2+ Y^2 + Z^2)$ must be equal to $2(D)^2$ – Tristen Nov 23 '16 at 05:10
  • @Tristen You have proven that $2D^2 $ is an integer. You have not proven that it is a perfect square. So there's no reason to think that $ D \sqrt{2} $ is also an integer. – cardboard_box Nov 23 '16 at 05:16
  • Therefore I have have an equation in which $(A^2+B^2+C^2)=2(D)^2$ and in order to satisfy the properties of the definition of a perfect cuboid, A, B, C, and D must be elements of the natural number. Therefore I end up with the following: $$\left{\begin{matrix}A\in \mathbb{N}\ B \in \mathbb{N} \ C \in \mathbb{N} \ A^2 \in \mathbb{N}\B^2 \in \mathbb{N}\C^2 \in \mathbb{N}\A^2+B^2+C^2 \in \mathbb{N} \end{matrix}\right}$$ and in order to find a solution, I don't need to know anything other than that $(A^2 + B^2 + C^2) = 2(D)^2$ and that $D$ must be an element of the natural numbers itself – Tristen Nov 23 '16 at 05:17
  • $D^2$ must be a perfect square. That's the whole question... $D$ is the length of the line passing through the cube. The question is if $D$ is a natural number. – Tristen Nov 23 '16 at 05:19
  • "I don't need anything other than $ A^2 + B^2 + C^2 = 2(D)^2 $" Then let $ A = 13, B=27, C=28 $. Then $ D = 29 $ is an integer. You would have noticed that if you actually read the above answer. – cardboard_box Nov 23 '16 at 05:24
  • What about $Z^2$ Did you read mine? What Pythagorean triple contains $13^2 + Z^2 = D^2$? and $Z^2$ – Tristen Nov 23 '16 at 05:26
  • The point is, your claim that "$A^2 + B^2 + C^2 = 2D^2$ cannot have solutions with $A$, $B$, $C$, $D$ integers" is false. – Alex Zorn Nov 23 '16 at 05:32
  • All I am asking, is for someone to actually read the paper, and then come at me with an explanation that is based on the proof. People are quickly throwing out reasons that it can't be right, which I understand, but none of those are even in the slightest bit relevant. Yes, they are correct, if the things you are saying were the only basis for the claims. You don't just have A, B, C, and D you have multiple other variables as well, It is a system of equations with 7 variables and properties of them are all defined. Don't say "Oh well here's how it works with only 4 variables" – Tristen Nov 23 '16 at 05:32
  • I didn't say only that. $A^2 + B^2 + C^2 = 2(X^2) + 2(Y^2) + 2(Z^2) = 2(D^2)$ Give me the integer answers for that and I will 100% give you 20 dollars, and also $A^2 = X^2 + Y^2$ and $X^2 + Z^2 = B^2$ and also $Y^2 + Z^2 = C^2$ Seriously, please actually read what the claims are, It's frustrating. Yes of course I understand why $A^2 + B^2 + C^2 = D^2$ can be all integers. I am not an idiot. – Tristen Nov 23 '16 at 05:34
  • It seems like, on the last page, you say "$A^2 + B^2 + C^2 = 2D^2$, and therefore $D\sqrt{2} \in \mathbb{N}$". The conclusion of that sentence does not follow from the equation. – Alex Zorn Nov 23 '16 at 05:39
  • And nobody will give you integer answers for $A,B,C,X,Y,Z,D$, because it's an unsolved problem. We're just trying to show you that your proof has flaws. – Alex Zorn Nov 23 '16 at 05:40
  • $(A^2 + B^2 + C^2) = 2D^2$, the $\sqrt{2D^2}$ = $D\sqrt{2}$ – Tristen Nov 23 '16 at 05:41
  • And so $D\sqrt{2} = \sqrt{A^2 + B^2 + C^2}$. But neither side has to be an integer. – Alex Zorn Nov 23 '16 at 05:42
  • Incorrect. $X^2 + Y^2 + Z^2 = D^2$ and both of those are integers, A,B,and C must be integers. A^2 + B^2 + C^2 must be integers, and D^2 must be an integer, and D must be an integer. – Tristen Nov 23 '16 at 05:44
  • X is in at least 3 pythagorean triples, and is not the largest value in any of those 3. Y is in at least 3 pythagorean triples, and is not the largest value in any of those 3. Z is in at least 3 pythagorean triples, and is not the largest value in any of those 3. A is in at least 2 pythagorean triples, and is the largest value in only one of those 2.
    B is in at least 2 pythagorean triples, and is the largest value in only one of those 2.
    D is in at least 3 pythagorean triples, and is the largest value in all of those 3.     – Tristen Nov 23 '16 at 05:46
  • $(A^2 + Z^2) + (B^2 + Y^2) + (X^2 + C^2)-(X^2+Y^2+Z^2)=2(D)^2=(A^2+B^2+C)^2$ – Tristen Nov 23 '16 at 05:48
  • @Tristen Realistically, as you pointed out yourself, this proof is wrong--it could be right, but it's highly unlikely. Given that, people are unwilling to sink the time in to reading your entire solution. So, when you make an incorrect claim that people can quickly locate, they just stop reading. While that might miss the forest through the trees, the onus is on you to fix it. You made the incorrect claim that was pointed out by Zorn, if that is not pivotal to your argument, then rewrite that portion of the note to reflect that. Maybe then someone will spend the time to read all of it. GL! – Alex Youcis Nov 23 '16 at 09:41
  • I won't hold my breath for that $20. :P – Michael Biro Nov 23 '16 at 13:03
  • Michael, How can I get you you're 20 dollars? I wish you had more clearly explained why that is, Because you are correct, I had made an inference that the reason did not need explaining, because it seemed simple and I didn't want the paper to drone on. I will be updating the paper with the explicit explanation of why, but I concede, You were correct. Had I threw in one last paragraph I wouldn't owe you 20 dollars, but alas, tis my mistake! XD Don't think I didn't mean that, I dropped out of college to spend 18-20 hours a day working on math problems. My biggest problem – Tristen Nov 23 '16 at 15:34
  • is that nobody will take the time of day to even look at my work. This can get so frustrating sometimes. I am an aspie, and already struggle to explain stuff to people, but it comes with it's perks including a spatial IQ of approx. 160, sadly my overall IQ is only around 130ish, but I just wish people would even read over my work sometimes and at least point me in the right direction. So, I apologize for getting frustrated. The problem is solved, I just kind of thought the rest of the solution was arbitrarily obvious. I will update the paper, add on the trivial stuff. But I do owe you twenty. – Tristen Nov 23 '16 at 15:41
  • I'm happy that you enjoy mathematics, and think you should keep working on it! However, you should be aware that producing correct proofs takes a lot of practice and starting with notorious open problems is probably not the best approach. Keep the $20 and good luck with fixing the proof. – Michael Biro Nov 23 '16 at 15:51
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You show that $A^2+B^2+C^2 = 2D^2 $.

But you have not shown that $A^2+B^2+C^2 $ is a perfect square, only that it is an integer.

Therefore, you have not shown that $2D^2$ is the square of an integer.

This is the error in your proof.

marty cohen
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  • $2D^2$ cannot possibly be the square of an integer (unless D were irrational), since every square can be uniquely represented as the product of prime factors raised to an even power, and 2 is definitely not a prime factor raised to an even power. – Mark May 05 '24 at 23:05