# Difference between revisions of "1998 IMO Shortlist Problems/N1"

## 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| 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 $$(a,b)=(11,1),(49,1),(7k^2,7k).$$