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
$$p^n \mid p^{n+2} + q^{n+2} - 3^{n+2}, \quad q^n \mid p^{n+2} + q^{n+2} - 3^{n+2}.$$
Since $p < q$, and $p$, $q$ primes, we have
$$p^n q^n \mid p^{n+2} + q^{n+2} - 3^{n+2}. \qquad \textcircled{1}$$
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
$$\begin{aligned} \frac{q^5 - 3^5}{q - 3} &= q^4 + q^3 \cdot 3 + q^2 \cdot 3^2 + q \cdot 3^3 + 3^4 \\ &\equiv 5 \times 3^4 \pmod{p} \end{aligned}$$
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.