<s>[INST] integer solution of $(x-y)(x+y)xy=z^2$ By wolfram alpha, integer solution of  $(x-y)(x+y)xy=z^2$ is $x=y=z=0$.
How to show that there are not another solutions with $z \neq 0$.
Thanks.
 [/INST] If $\gcd(a,b) = d$, and $(x,y,z) = (a,b,c)$ is a nontrivial solution, then $(\frac{a}{d},  \frac{b}{d},\frac{c}{d^2})$ is also a solution, where $\gcd(\frac{a}{d},\frac{b}{d}) = 1$.  So if there is no solution $(a,b,c)$ having $\gcd(a,b) = 1$, then there are no nontrivial solutions.
Assume $\gcd(a,b) = 1$.  We want to find $a$ and $b$ where the expression $(a+b)(a-b)ab$ would be a perfect square.  But $(a+b)$, $(a-b)$, and their product $(a^2-b^2)$ are all coprime to both $a$ and $b$.  So it is necessary that $a$ and $b$ are both squares, because any unsquared factors they may have will not be able to have their squares completed by the other terms.
$a = a'^2$ and $b = b'^2$.  $(a+b)(a-b) = a^2 - b^2 = a'^4 - b'^4$.  If there is a solution, then there must be some $c$ where $c^2 = a'^4 - b'^4$, or $c^2 + b'^4 = a'^4$.  Fermat proved this was impossible, so there cannot be any nontrivial solutions.
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