
Problem 11.3. Find all integer numbers $m, n \geq 2$ such that

$$
\frac{1+m^{3^{n}}+m^{2.3^{n}}}{n}
$$

is integer.


Solution: Let $m$ and $n$ satisfy the conditions of the problem. Since $n$ is an odd integer number, we get $(m, n)=1$ and $n>2$. When $n=3$, all $m \geq 4$ such that $m \equiv 1(\bmod 3)$ are solutions, because if $m \equiv-1(\bmod 3)$, then $1+m^{3^{n}}+m^{2 \cdot 3^{n}} \equiv 1-1+1 \equiv 1(\bmod 3)$. Let now $n>3$. It follows that $m^{3^{n}} \not \equiv 1(\bmod n)$, because otherwise $1+m^{3^{n}}+m^{2 \cdot 3^{n}} \equiv 3(\bmod n)$, i. e., $n / 3$. On the other hand $1+$ $m^{3^{n}}+m^{2 \cdot 3^{n}}=\frac{m^{3^{n+1}}-1}{m^{3^{n}}-1}$ and therefore $m^{3^{n+1}} \equiv 1(\bmod n)$. Let $k$ be the least natural number such that $m^{k} \equiv 1(\bmod n)$. Further, $k / 3^{n+1}$ and $k+3^{n}$, so $k=3^{n+1}$. Let $\varphi(n)$ be Euler's function. From $(m, n)=1$ it follows that $m^{\varphi(n)} \equiv 1(\bmod n)$, so $k \leq \varphi(n)$. Therefore $3^{n+1} \leq \varphi(n) \leq n-1$, which is impossible.

The required numbers are: $n=3$ and all $m \geq 4$ such that $m \equiv 1(\bmod 3)$.

## Winter mathematics competition-Pleven, $6-8$ February 1998

Dedicated to the One Hundredth Anniversary of the UBM

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