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 5 of 7: Case a=b+1
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a=b+1:a2+bb2−a=1+4b+2b2−b−1.a=b+1:\quad\frac{a^2+b}{b^2-a}=1+\frac{4b+2}{b^2-b-1}.
Detailed analysis

For b=1 the denominator is negative and the pair (2,1) works. For b at least 6, the positive denominator b^2-b-1 is larger than the remainder 4b+2, so the fraction lies strictly between 1 and 2 and cannot be integral.