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 2 of 6: Handle the zero remainder
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In plain words

The zero remainder gives an infinite family.

7a=b2⇒(a,b)=(7t2,7t)7a=b^2\Rightarrow (a,b)=(7t^2,7t)
Detailed analysis

If 7a=b27a=b^2, then 7∣b7\mid b , say b=7t b=7t , and consequently a=7t2 a=7t^2. Conversely, for these values E=tD E=tD , so every (7t2,7t)(7t^2,7t) is a solution.