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?

(a^2+b^2)/(ab+1)^2 = a^2/(a^2b^2+2ab+1) + b^2/(a^2b^2+2ab+1)

a^2<a^2b^2, and b^2<a^2b^2, and since a and b are non-negative integers, the remaining terms simply increase the difference.

Let a=0 (or b=0); then e.g (0+b^2)/(0*b+1)^2 = b^2, true for all b.

Let a=1 (or b=1); then e.g. (b^2+1)/(1*b+1)^2 = 1, for b=0 (see above), but for b>0 is a fraction.

Let a=2 (or b=2), with the other greater than 0; then e.g. (b^2+2)/(2*b+1)^2 = (b^2+2)/(4b^2+4b+1), again a fraction, and so on.

Edited on June 18, 2018, 1:03 am

Posted by broll on 2018-06-18 01:00:21