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Consider the following Diophantine equation $$zx^2 +xy^2 +yz^2 =xyzt .$$

Is that true that all solutions of this equation are of the form $(x,y,z,t) =(a^2 b ,b^2 c ,c^2 a ,t)$ for some $a,b,c\in\mathbb{N}$?

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This need not be true. We have the solution $(x,y,z,t)=(7,7,7,3)$, and $7=a^2b=b^2c=c^2a$ gives first $a=1$ and $b=7$ from $a^2b=7$, but also $c=1$ and $a=7$ from $c^2a=7$, a contradiction.

Dietrich Burde
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  • Ok, but what if we assume that $x\neq y$? –  Sep 13 '14 at 19:58
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    Take $(x,y,z,t)=(3,6,12,5)$. Then $3=a^2b$ gives $a=1,b=3$, and $b^2c=6$ is impossible. See http://math.stackexchange.com/questions/916409/solutions-of-diophantine-equation. – Dietrich Burde Sep 14 '14 at 11:59