Does there exist a perfect square in the form: $|x^2+52x|$, where $x\in \Bbb Z$?

Question

Does there exist a perfect square in the form: $|x^2+52x|$, where $x\in \Bbb Z$? $x<0,x\neq-52$

Answer 1

Yes. Let $x=-52$; then $(-52)^2 + 52 \times (-52) = 0$, which is square. Or let $x = 0$, to get $0^2 + 0 = 0$.

For a sensible answer, $144^2 + 52 \times 144 = 168^2$.