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Well, I have been trying to prove that:

$$x^2 + y^2 + z^2 = 2xyz \implies x = y = z = 0$$

and have made little progress. Till now, I have only been able to prove that if this is to happen then $x$, $y$ and $z$ must be even. I am explaining how I arrived to this result but don't know how to proceed further or even figuring this out would make sense.

If $x^2 + y^2 + z^2 = 2xyz$, then at least one of $x$, $y$ or $z$ must be a multiple of 2. Let that one number be x and now $2|x$. Also let $x = 2a$.Then, $4a^2 + y^2 + z^2 = 4ayz$.

Now assume that one of $y$ or $z$ is odd. So the other must be odd as well as odd and odd add upto even and we need an even number to make the equality true. But it can be proved that the sum of squares of two odd numbers is never a multiple of $4$. Then the LHS would no longer be a multiple of 4 ($4a^2$ is a multiple of $4$ but $y^2 + z^2$ isn't) while the RHS would be. This is a contradition. Hence, $y$ and $z$ both are even. That's what I have been able to prove till now and still don't know if that is useful.

Please help if you have an answer.


marked as duplicate by user236182, N. F. Taussig, Rory Daulton, user147263, Shailesh Feb 24 '16 at 0:07

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  • $\begingroup$ What about $x=y=z=(3/2)$? $\endgroup$ – Daron Feb 20 '16 at 12:17
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    $\begingroup$ I think he might mean integers. $\endgroup$ – SS_C4 Feb 20 '16 at 12:18
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    $\begingroup$ @Daron It's not stated explicitly in the question, but it's tagged under diophantine-equation, so I don't think that's a valid solution. $\endgroup$ – Arthur Feb 20 '16 at 12:18
  • $\begingroup$ Have you tried applying AM-GM in a clever way? $\endgroup$ – Derek Allums Feb 20 '16 at 12:25
  • $\begingroup$ @Daron Forget about the diophantine equation for a minute but how did you derive $3/2$? Was it some calculation or trial-and-error? $\endgroup$ – TheRandomGuy Feb 20 '16 at 12:50

Your proof that $x,y,z$ are all even is correct. You can use similar reasoning to continue.

Lemma $1$: If $x^2+y^2+z^2=2^n xyz$ for some $x,y,z\in\mathbb Z$, $n\in\mathbb Z^+$, then $x,y,z$ are all even.

The proof is analogous. $x,y,z$ cannot all be odd (because the LHS would then be odd and the RHS even), so WLOG let $x=2x_1$.

The equation becomes $4x_1^2+y^2+z^2=2^{n+1}x_1yz$, so $4\mid y^2+z^2$, so $y,z$ are both even (you know how to prove it).

Lemma $2$: If $x^2+y^2+z^2=2^nxyz$ for some $n\in\mathbb Z^+$, $x,y,z\in\mathbb Z$, $x=2x_1$, $y=2y_1$, $z=2z_1$, then $x_1^2+y_1^2+z_1^2=2^{n+1}x_1y_1z_1$.

The proof is trivial. Lemma $1$ and $2$ together imply that if $x^2+y^2+z^2=2^nxyz$ for some $n\in\mathbb Z^+$, $x,y,z\in\mathbb Z$, then $x,y,z$ are divisible by arbitrarily large powers of $2$.

I.e., for all $k\in\mathbb Z^+$ we have $2^k\mid x,y,z$, which can only be possible if $x=y=z=0$.

  • $\begingroup$ By Lemma 1, you simplified that idea that all must be even to a great extent. Thanks. $\endgroup$ – TheRandomGuy Feb 20 '16 at 12:46

You are close to an answer. Assuming $x = 2a, y = 2b, z = 2c$, we get $$ 4a^2 + 4b^2 + 4c^2 = 16abc\\ a^2 + b^2 + c^2 = 4abc $$ Now apply the exact same reasoning to see that $a, b, c$ must all be even as well. And this carries on forever. Therefore $x, y, z$ must be infinitely divisible by $2$, and the only such number is $0$.

  • $\begingroup$ That was cool! So you prove it as there is no number infìnitely divisible by 2. This was a great idea. I didn't know you could use recursion like this. $\endgroup$ – TheRandomGuy Feb 20 '16 at 12:43
  • $\begingroup$ @DhruvSomani It's more or less the same as the proof that $\sqrt 2$ is irrational. You show that if there is a fraction $\frac pq = \sqrt 2$, then $p$ and $q$ are both infinitely divisible by $2$. $\endgroup$ – Arthur Feb 20 '16 at 14:38

Now I will prove another claim: There is no triple of positive integers that satisfy the equation.

Suppose there is. Take $2^k$ as the largest power of $2$ that divide $x,y$ and $z$ so that $x=2^k\alpha$ $y=2^k\beta$ and $z=2^k\delta$. Then LHS can be written as $2^{2k}(\alpha^2+\beta^2+\delta^2)$. Now if you rewrite RHS as well and take out common factors

$$ \alpha^2+\beta^2+\delta^2=2^{k+1}\alpha\beta\delta $$

where at least one of $\alpha,\beta,\delta$ is odd. The RHS in this new equation is even then this implies all at least two of $\alpha,\beta,\delta$ is odd. You can pick one even member say $\delta$ bu then $\alpha^2=\beta^2=1(mod4)$ and $\delta^2=0(mod4)$ but then $LHS=2(mod4)$ and $RHS=0(mod4)$. So this is obviously a contradiction.

Given this then at least one of $x,y,z$ should be zero, say $z=0$, but then this implies both $x=y=0$.

  • $\begingroup$ LHS can be written as.. Why $2^{2k}$ not $2^k$? $\endgroup$ – TheRandomGuy Feb 20 '16 at 14:00
  • $\begingroup$ @DhruvSomani $\left(2^k\alpha\right)^2+\left(2^k\beta\right)^2+\left(2^k\delta\right)^2=2^{2k}\left(\alpha^2+\beta^2+\delta^2\right)$. What do you mean? $\endgroup$ – user236182 Feb 20 '16 at 19:41

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