## Task 30/76

Determine all pairs $(n ; m)$ of natural numbers that satisfy the equation $4^{n}+65=9^{m}$.
We transform the equation equivalently so that it appears as an equation between products. From $4^{n}+65=9^{m}$ it follows that

$$
9^{m}-4^{n}=65=3^{2 m}-2^{2 n}=\left(3^{m}-2^{n}\right)\left(3^{m}+2^{n}\right)=5 \cdot 13
$$

Since $3^{m}+2^{n}>3^{m}-2^{n}$, it follows that $3^{m}-2^{n}=5 ; 3^{m}+2^{n}=13$.

By simple calculation, this yields $n=m=2$ as a solution. The verification confirms the correctness. The possibility of choosing $65=1 \cdot 65$ is ruled out, as it does not yield an integer $m$. Thus, the given solution is the only one.