China Mathematical Olympiad

It is easy to check that $(3,3,n)$ ($n=2,3,\dots$) satisfy both equations. Now let $(p,q,n)$ be another triple satisfying the condition. Then we must have $p\ne q$, $p\ne 3$, $q\ne 3$. We may assume that $q>p\ge 5$.

If $n=2$, then $q^2\mid p^4-3^4$, or $q^2\mid (p^2-3^2)(p^2+3^2)$. Then either $q^2\mid p^2-3^2$ or $q^2\mid p^2+3^2$, since $q$ cannot divide both $p^2-3^2$ and $p^2+3^2$. On the other hand, $0<p^2-3^2<q^2$, $\frac{1}{2}(p^2+3^2)<p^2<q^2$. This leads to a contradiction.

So $n\ge 3$. From $p^n\mid q^{n+2}-3^{n+2}$, $q^n\mid p^{n+2}-3^{n+2}$, we get $$p^n\mid p^{n+2}+q^{n+2}-3^{n+2},q^n\mid p^{n+2}+q^{n+2}-3^{n+2}.$$ Since $p<q$, and $p$, $q$ primes, we have $$p^n{q}^n\mid p^{n+2}+q^{n+2}-3^{n+2}.\stackrel{◯}{1}$$ Then $p^n{q}^n\le p^{n+2}+q^{n+2}-3^{n+2}<2q^{n+2}$. That means $p^n<2q^2$.

As $q^n\mid p^{n+2}-3^{n+2}$ and $p>3$, we have $q^n\le p^{n+2}-3^{n+2}<p^{n+2}$, and consequently $q<p^{1+\frac{2}{n}}$. Since $p^n<2q^2$, we have $p^n<2p^{2+\frac{4}{n}}<p^{3+\frac{4}{n}}$. So $n<3+\frac{4}{n}$, and we get $n=3$. Then $p^3\mid q^5-3^5$, $q^3\mid p^5-3^5$.

From $5^5-3^5=2\times 11\times 131$, we know $p>5$; from $p^3\mid q^5-3^5$ we know $p\mid q^5-3^5$. By Fermat's little theorem, we get $p\mid q^{p-1}-3^{p-1}$. Then $p\mid q^{(5,p-1)}-3^{(5,p-1)}$.

If $(5,p-1)=1$, then $p\mid q-3$. From $$\begin{array}{rl}\frac{q^5-3^5}{q-3}& =q^4+q^3\cdot 3+q^2\cdot 3^2+q\cdot 3^3+3^4\\ & \equiv 5\times 3^4(\mathrm{mod}p)\end{array}$$ and $p\ge 5$, we get $p\nmid \frac{q^5-3^5}{q-3}$. So $p^3\mid q-3$. From $q^3\mid p^5-3^5$, we get $q^3\le p^5-3^5<p^5=(p^3{)}^{\frac{5}{3}}<q^{\frac{5}{3}}$. This is a contradiction.

So we have $(5,p-1)\ne 1$, and that means $5\mid p-1$. In a similar way, we have $5\mid q-1$. As $(q,p-3)=1$ (since $q>p\ge 7$) and $q^3\mid p^5-3^5$, we know that $q^3\mid \frac{p^5-3^5}{p-3}$. Then $$q^3\le \frac{p^5-3^5}{p-3}=p^4+p^3\cdot 3+p^2\cdot 3^2+p\cdot 3^3+3^4.$$ From $5\mid p-1$ and $5\mid q-1$, we get $p\ge 11$ and $q\ge 31$. So $$q^3\le p^4\left(1+\frac{3}{p}+{\left(\frac{3}{p}\right)}^2+{\left(\frac{3}{p}\right)}^3+{\left(\frac{3}{p}\right)}^4\right)$$ $$<p^4\cdot \frac{1}{1-\frac{3}{p}}\le \frac{11}{8}{p}^4.$$ Then we have $p>{\left(\frac{8}{11}\right)}^{\frac{1}{4}}{q}^{\frac{3}{4}}$. Consequently, $$\frac{p^5+q^5-3^5}{p^3{q}^3}<\frac{p^2}{q^3}+\frac{q^2}{p^3}<\frac{1}{q}+{\left(\frac{11}{8}\right)}^{\frac{3}{4}}\frac{1}{31^{\frac{1}{4}}}<1.$$ But this contradicts ① which says $p^3{q}^3\mid p^5+q^5-3^5$.

So we reach the conclusion that $(3,3,n)$ ($n=2,3,\dots$) are all the triples that satisfy the conditions.