This problem was inspired by the infamous problem 6 in the 1988 Math Olympiad:

When is a^{2}+b^{2} divisible by (ab+1)^{2} where a and b are non-negative integers?

Consider (a^2 + b^2) / (ab+1)^2 , which needs to be an integer.

The divisor divides the numerator only if a or > b are 0, in which case (ab + 1)^2 = 0. Clearly, this always works.

Otherwise, the divisor is bigger than the numerator and the fraction is less than 1

Case 1) a > 1, b > 1

(ab+1)^2 = (a^2)(b^2) plus some other positive terms

Clearly (a^2)(b^2) > (a^2 + b^2), so the divisor is bigger than the numerator

Case 2) a or b = 1

Without loss of generality, assume a = 1

Then (ab + 1)^2 = b^2 + 2b +1, which makes the divisor greater

than the numerator by 2b.

q.e.d.