This problem was inspired by the infamous problem 6 in the 1988 Math Olympiad:
When is a2+b2 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.
Proof (spoiler)
