So, I was looking at this problem which asks to show that $x^2 + y^2 + z^2 = 2xyz $ where $ x,y,z \in \mathbf{N}$ has no solution. Viewing the equation as quadratic in $x$ and solving for $x$ gives $$ x = yz \pm \sqrt{y^2z^2-y^2-z^2} $$ From which we can conclude $x$ has integral solutions $\iff y^2z^2-y^2-z^2$ is a perfect square. How do I proceed from here to show that $y^2z^2-y^2-z^2$ can never be a perfect square?
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@Desperado This is not a duplicate. None of the answers in that post is using this approach. OP is asking how to proceed using this specific approach in this post. The post you linked demonstrates several other methods of solving this problem however, and is therefore helpful. – Martund May 12 '21 at 08:15
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1https://math.stackexchange.com/questions/1534522/find-all-integer-solutions-to-x2y2z2-2xyz?noredirect=1&lq=1 this link has solution using the approach I needed. – Aravind Nujella May 12 '21 at 08:24
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Looking at $y^2z^2-y^2-z^2$ modulo $4$, we see that both $y$ and $z$ must be even if that expression is a perfect square. Setting $y=2y_1, z=2z_1$, we get $$ y^2z^2-y^2-z^2=16y_1^2z_1^2-4y_1^2-4z_1^2\\ =4(4y_1^2z_1^2-y_1^2-z_1^2) $$ If this is a perfect square, then $4y_1^2z_1^2-y_1^2-z_1^2$ must also be a perfect square. And again, looking at it modulo $4$, we see that $y_1$ and $z_1$ must both be even.
This continues indefinitely. Which cannot be done with positive integers. So there is no solution (unless $0$ is allowed).
Arthur
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@KeithBackman I don't think you can take step 2 in modulo $3$. You get the $9y_1^2z_1^2$ term mucking things up, as that's always divisible by $3$, and then they could both $y_1$ and $z_1$ not be divisible by $3$ and you get $1\bmod 3$ and no contradiction. In the modulo $4$ case, you would get 2 instead if they were both odd, letting you keep assuming they are even. – Arthur May 12 '21 at 17:02
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