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.

  • 1
    $\begingroup$ Also, x-36 must be a divisor of 72. $\endgroup$ – 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

$$n^2-1\le72\implies2\le n\le8$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.