<s>[INST] How many squares are of the form $x^3+7x^2-x$? 
How many squares are of the form $x^3+7x^2-x$, where $x$ is a positive integer, that are not multiples of $16$?

Since $x^3+7x^2-x = x(x^2+7x-1)$, we need both $x$ and $x^2+7x-1$ to be perfect squares and so let $x = k^2$, for some $k \in \mathbb{Z}^+$ . Then we need $(k^2)^2+7k^2-1 = k^4+7k^2-1$ to be a perfect square that is not a multiple of $16$. How do we do this?
 [/INST] @ovi Why both $x$ and $ x^2+7x-1$ must be a square? Because they are coprime.
From $x^2+7x-1=k^2$ we get $x^2+7x-1-k^2=0$ and the discriminant is $53+4k^2 = m^2 $. From here $53=(m-2k)(m +2k)$ and $53$ is a prime number. It follows $m+2k = 53, m-2k = 1$ and $m=27, k=13$. Therefore $x^2+7x-170=0$ and, from here, $x=10$ but $10$ is not a perfect square.
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