<s>[INST] Why is this not a perfect square? Let $x, y$ and $z$ be positive integers with $x<y$. It appears that the integer
$$(y^2z^3-x)(y^2z^3+3x)$$
is never a perfect square. Why? A proof? I'm not sure if it is easy.
 [/INST] Rewrite your expression as $(y^2z^3+x)^2-4x^2$. This is clearly less than $(y^2z^3+x)^2$. On the other hand, it is larger than $(y^2z^3+x-2)^2=(y^2z^3+x)^2-4(y^2z^3+x)+4$, since $4(y^2z^3+x)-4\geq 4y^2+4x-4\geq 4y^2>4x^2$. So if this expression is a square, it must be $(y^2z^3+x-1)^2=(y^2z^3+x)^2-2(y^2z^3+x)+1$, so $4x^2=2(y^2z^3+x)-1$. LHS is even, while RHS is odd, so this can't be.
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