# Find all natural numbers x for which $\sqrt{\frac{x+36}{x-36}}$ is a natural number

I started with transforming the expression to $$\sqrt{1+\frac{72}{x-36}}$$ and I realised that $\frac{72}{x-36}$ must be one less than a square. Then I don't know what to do. Do I just try some number one less than a square and check if it's divisible by 72? Is there a smarter method I'm missing? I know there isn't a lot of possibilities, but I think there's a smarter method.

• Also, x-36 must be a divisor of 72. – steven gregory Dec 30 '17 at 9:47

The number in the square root is at most $73$, and a square. So it is one of $1,4,9,16,25,36,49,64$.
Also, the number in the square root is one more from a divisor of $72$, so it is one of $4,9,25$. All those values can be achieved.
Let $$\sqrt{\dfrac{x+36}{x-36}}=n\implies x=\dfrac{36(n^2-1+2)}{n^2-1}=36+\dfrac{72}{n^2-1}$$
So $n^2-1(\ge-1)$ must divide $72$ and
$$n^2-1\le72\implies2\le n\le8$$