Find all positive integers x such that x(x + 180) is a square.

Question

$x^2 + 180x = k^2$

$k^2 - x^2 = 180x$

$(k+x)(k-x) = 180x$

Now i wrote $180x$ as $1(180x) , (2)(90x) , \dots , (180)(x)$ and i compared these values with $(k+x)(k-x)$ and found $12,60$ to be the only 2 integers , are there more ? Also if you have a better method please share.

Answer 1

WLOG $k\ge0\implies x+90+k>x+90-k$

$\implies(x+90+k)^2>(x+90-k)(x+90+k)=8100$

$\implies(x+90+k)>\sqrt{8100}=90\ \ \ \ (1)$

As $x+90+k+x+90-k=2(x+90)$ is even, $x+90+k,x+90-k$ have the same parity

As $(x+90-k)(x+90+k)=8100$ is even

$$\dfrac{x+90-k}2\cdot\dfrac{x+90+k}2=2025$$

By $(1),\dfrac{x+90+k}2>45$ and must divide $2025$ so, must belong to $\in[75, 81, 135, 225, 405, 675, 2025]$

Answer 2

Hint: Rewrite it as $(x + 90 + k)(x + 90 - k) = 8100$. Now you can factorize $8100$.