Ireland_2017

Let $d=\gcd (ab,a^2+p{b}^2)$ and suppose $d>1$. Let $q$ be a prime factor of $d$. Because $a^3=a(a^2+p{b}^2)-pb(ab)$ we see that $q\mid a^3$ and so $q\mid a$ as well. As $q\mid a^2+p{b}^2$, this implies that $q\mid p{b}^2$. Because $\gcd (a,b)=1$ this can only happen if $q=p$.

So far we have shown that $d>1$ implies $p\mid a$ and $d=p^{\alpha }$ for some $\alpha >0$. In particular, if $p\nmid a$, then $d=1$. Obviously, if $p\mid a$, then $p\mid d$. We continue to assume $d>1$ and need to show $d=p$. Write $a=\tilde{p}\tilde{a}$, so that $$d=\gcd (\tilde{p}\tilde{a}b,p^2{\tilde{a}}^2+p{b}^2)=p\cdot \gcd (\tilde{a}b,p{\tilde{a}}^2+b^2).$$ Because $\gcd (\tilde{a},b)=1$, the argument at the start of this solution shows that $\gcd (\tilde{a}b,p{\tilde{a}}^2+b^2)>1$ would imply that $p\mid b$, which contradicts $\gcd (a,b)=1$. Hence $\gcd (\tilde{a}b,p{\tilde{a}}^2+b^2)=1$ and so $d=p$.