How can I prove that the GCD is less or equal than the square root of the numbers' sum?

Question

The numbers $a$ and $b$ are natural numbers. $\frac{a+1}{b} + \frac{b+1}{a}$ is an integer number. How can I prove that $gcd(a, b) \le \sqrt{a+b}$?

Answer 1

After rearrangement, we can get $a^2 + a + b^2 + b = nab$ for some positive integer $n,$ and from there $(a + b)(a + b + 1) = (n+2)ab$. Now, $(a,b)^2|ab$, so $(a,b)^2 | (a+b)(a+b+1)$; but unless $(a, b) = 1$, we have $a + b + 1 = k(a,b) + 1$ coprime to $(a, b)$. Of course if $(a,b) = 1$ then the inequality is trivial, while if not, then we must have $(a,b)^2 | (a+b)$, which gives the inequality.