MathLabs

Problem 2

Find all positive integers a,ba,b such that a2+bb2−a\frac{a^2+b}{b^2-a} and b2+aa2−b\frac{b^2+a}{a^2-b} are both integers.
Step 3 of 7: Force two cases
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b2+aa2−b∈Z>0  ⟹  b2+a≥a2−b  ⟹  (a−b)(a+b)≤a+b  ⟹  a−b≤1.\frac{b^2+a}{a^2-b}\in\mathbb Z_{>0}\implies b^2+a\ge a^2-b\implies (a-b)(a+b)\le a+b\implies a-b\le1.
Detailed analysis

The second denominator is positive. Since its quotient is a positive integer, its numerator is at least its denominator. Factoring the resulting inequality and using a+b>0 gives a-b at most 1.