1998 IMO Shortlist Problems/N1

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Problem

Find all pairs of positive integers $(a,b)$ such that, \[ab^2+b+7|a^2b+a+b\]


Solution

We have the following divisibility relations, \[ab^2+b+7|a^2b+a+b|b(a^2b+a+b)=a^2b^2+ab+b^2\] \[ab^2+b+7|a(ab^2+b+7)=a^2b^2+ab+7a\] Subtracting, \[ab^2+b+7|   \;|b^2-7a|\]

If $b=1$, we may have \[a+8|7a-1\] Otherwise, we would have \[|b^2-7a|<ab^2+b+7\] In the first case, \[a+8|7a+56\] This gives, \[a+8|57\] Since, $a+8>3$ and $57=3\cdot19$, we must have $a+8=19$ or $57$ which yields the solutions $(a,b)=(11,1),(49,1)$.

In the second case, $b^2-7a=0$ or, \[b^2=7a\implies 7|b\] Say, $b=7k$. Then $a=7k^2$. Checking we find that it is indeed a solution. Thus, all solutions are \[(a,b)=(11,1),(49,1),(7k^2,7k).\]

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