IMO-Auswahlklausur

Solution:
The pairs $(1,0)$ and $(2,1)$ are solutions. Let $(m, n)$ be a solution with $m, n \geq 2$. The solution consists of three steps:

1. We have $7^{n} \equiv -2 \bmod 9 \Longleftrightarrow n \equiv 1 \bmod 3$.
2. We have $3^{m} \equiv 2 \bmod 49 \Longleftrightarrow m \equiv 26 \bmod 42$.
3. The conditions from 1 and 2 yield a contradiction modulo 43.

Step 1: We have $7^{1} \equiv 7$, $7^{2} \equiv 4$ and $7^{3} \equiv 1 \bmod 9$, and from then on the residues are periodic with period 3. From the original equation it follows that $7^{n} \equiv 7 \bmod 9$, hence $n \equiv 1 \bmod 3$.

Step 2: First show $m \equiv 2 \bmod 6$ analogously to Step 1. By the Euler-Fermat theorem $3^{42} \equiv 1 \bmod 49$, so the period is at most 42. Using $3^{6} \equiv -6 \bmod 49$ one obtains:

Step 3: Write $m=42 m' + 26$ and $n=3 n' + 1$. Considering the original equation modulo 43, one obtains, using Fermat's little theorem and $7^{3} \equiv -1 \bmod 43$,

$$3^{42 m' + 26} - 7^{3 n' + 1} \equiv 3^{26} - (-1)^{n'} \cdot 7 \equiv 15 - (-1)^{n'} \cdot 7 \not\equiv 2 \quad \bmod 43$$

Hence there are no further solutions.