Problem 4

Determine all pairs

(a,b)

of positive integers such that

ab^{2}+b+7

divides

a^{2}b+a+b

Step 5 of 6: Check the small cases

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In plain words

Finish by checking $b=1$ and $b=2$ .

b=1\Rightarrow a+8\mid57\Rightarrow a=11,49

Detailed analysis

For $b=1$ , $a+8\mid a^2+a+1$ , hence $a+8\mid7a-1$ and then $a+8\mid57$ . Since $a+8>8$ , this gives $a+8=19$ or $57$ , so $a=11$ or $49$ . For $b=2$ , $4a+9\mid2a^2+a+2$ implies $4a+9\mid79$ ; the only possible divisor greater than $9$ is $79$ , which would give $a=35/2$ , impossible.