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 2 of 7: Establish the denominator sign
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b2−a>0.b^2-a>0.
Detailed analysis

If b=1, the case b^2-a<0 can be checked directly; only a=2 survives. If b is at least 2 and a is at least b^2, then the second positive fraction is strictly between 0 and 1, impossible for an integer. Thus in the remaining cases b^2-a is positive.