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 $x\in \Bbb Z$? $x<0,x\neq-52$

Seems the problem has been edited - the way it looks now it has all the context needed. With the extra conditions, $x<0$ and $x \ne -52$, there is still at least one (almost) trivial affirmative answer, $x = -26$. — , Mar 26 '16 at 21:16

score 4 · Accepted Answer · answered Mar 26 '16 at 20:29

4

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, $144^2 + 52 \times 144 = 168^2$.

answered Mar 26 '16 at 20:29

Patrick Stevens

1 Answers1