MathLabs

Problem 4

Determine all pairs (a,b)(a,b) of positive integers such that ab2+b+7 ab^{2}+b+7 divides a2b+a+b a^{2}b+a+b .
Step 4 of 6: Bound the second variable
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In plain words

The positive remainder is too small when b b is large.

D≤7a−b2⇒b≤2D\le7a-b^2\Rightarrow b\le2
Detailed analysis

For 7a>b27a>b^2, divisibility gives D≤7a−b2 D\le7a-b^2. If b≥3 b\ge3, then D=ab2+b+7>7a−b2 D=ab^2+b+7>7a-b^2, a contradiction. Hence b=1 b=1 or b=2 b=2.