
NT 2. Find all pairs $(m, n)$ of positive integers such that

$$
125 \cdot 2^{n}-3^{m}=271
$$


Solution. Considering the equation mod5 we get

$$
3^{m} \equiv-1 \quad(\bmod 5)
$$

so $m=4 k+2$ for some positive integer $k$. Then, considering the equation $\bmod 7$ we get

$$
\begin{aligned}
& -2^{n}-9^{2 k+1} \equiv 5 \quad(\bmod 7) \Rightarrow \\
& 2^{n}+2^{2 k+1} \equiv 2 \quad(\bmod 7)
\end{aligned}
$$

Since $2^{s} \equiv 1,2,4(\bmod 7)$ for $s \equiv 0,1,2(\bmod 3)$, respectively, the only possibility is $2^{n} \equiv 2^{2 k+1} \equiv$ $1(\bmod 7)$, so $3 \mid n$ and $3 \mid 2 k+1$. From the last one we get $3 \mid m$, so we can write $n=3 x$ and $m=3 y$. Therefore, the given equation takes the form

$$
5^{3} \cdot 2^{3 x}-3^{3 y}=271
$$

or

$$
\left(5 \cdot 2^{x}-3^{y}\right)\left(25 \cdot 2^{2 x}+5 \cdot 2^{x} \cdot 3^{y}+3^{2 y}\right)=271
$$

It follows that $25 \cdot 2^{2 x}+5 \cdot 2^{x} \cdot 3^{y}+3^{2 y} \leq 271$, and so $25 \cdot 2^{2 x} \leq 271$, or $x<2$. We conclude that $x=1$ and from (2) we get $y=2$. Thus, the only solution is $(m, n)=(6,3)$.
