Which values for $x$ make $x(x + 180)$ a square? 
Which values for $x$ make $x(x + 180)$ a square?

$x = 12, 16, 60$ are some values. 
Then maybe solving $x(x + 180) = y^2 $ would give other values? 
 Tried using the general formula, but that would only give 2 values each time I find a suitable $y$, not all values. And I am stuck in the process:
$x(x + 180) = y^2 \Rightarrow x^2 + 180x - y^2 = 0.$
$x = \frac{-180 \pm \sqrt{180^2 +4y^2}}{2} = \frac{-180 \pm \sqrt{8100 +y^2}}{2}$ And then I beleive that I should find a sum of squares that is a square, and that gives a positive $x.$

Note that $$y^2=x(x+180)=(x+90-90)(x+90+90)=(x+90)^2-90^2$$
and then $$90^2=(x+90)^2-y^2=(x+90+y)(x+90-y) $$
As the factors on th right have the same parity (differ by $2y$) and the left is even, we are looking for factorizations of $90^2$ into even  factors. Any such factorization $90^2=(2u)(2v)$ (or simply $45^2=uv$) with wlog $u\ge v$ leads to valid solution $y=u-v$ and $x=u+v-90$. (Don't forget to allow negative $u,v$)
