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 3 of 6: Exclude a negative remainder
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In plain words

A nonzero divisible remainder cannot be smaller than the divisor.

7a<b2 is impossible7a<b^2\text{ is impossible}
Detailed analysis

If 7a<b27a<b^2, then 0<b2−7a<b2<D0<b^2-7a<b^2<D , contradicting D∣(7a−b2) D\mid(7a-b^2). Thus 7a>b27a>b^2 in every remaining case.