easy to see that n is odd ,so n is greater or equal to 3 If n=3 then m =1 (mod 3) If n >= 5 we call p is a prime divisor of n. h =ord p(a) Then h|3^(n+1) => h=3^s (0<s <=n+1) . If s<=n then 3^s| 3^n hence p| m^(3^n)-1 => n|3 ( contradiction) So s=n+1 ,hence 3^(n+1)|p-1 (Fermat Little theorem) => 3^(n+1)|n-1 (Impossible since n-1 is smaller than 3^(n+1)) . QED Can someone Latex my solution ?

Quote: easy to see that $n$ is odd ,so $n$ is greater or equal to $3$ If $n=3$ then $m =1 (mod 3)$ If $n >= 5$ we call p is a prime divisor of $n$. h =ord p(a) Then $h|3^{n+1} => h=3^s$ and ($0<s <=n+1$) . If $s<=n$ then $3^s| 3^n$ hence $p| m^{(3^n)}-1 => n|3$ ( contradiction) So $s=n+1$ ,hence $3^{n+1}|p-1$ (Fermat Little theorem) => $3^{n+1}|n-1$ (Impossible since $n-1$ is smaller than $3^{n+1}$) . QED Can someone Latex my solution ?

this is an old problem from bulgaria

Essentially @ILOVEMYFAMILY’s solution, rewritten: We claim that the solutions are $(m, n) = (3t - 2, 3)$ for any positive integer $t$. Plugging these in, they indeed work. Note that $n$ must be odd. If $n = 3$, then $1 + m^{27} + m^{54}$ must be a multiple of $3$. Since $m^2 \equiv 1 \pmod{3}$, this implies that $m \equiv -1 \pmod{3}$, which works. Otherwise, assume that $n \ge 5$. Let $p$ be any prime divisor of $n$. Since $n \mid (1 + m^{3^n} + m^{2 \cdot 3^n})$, then $p \mid (m^{3^{n + 1}} - 1)$, so $m^{3^{n + 1}} \equiv 1 \pmod{p}$. Let $\operatorname{ord}_m(p) = r$. Then, $r \mid 3^{n + 1}$. Suppose that $r = 3^{n + 1}$. By Fermat's Little Theorem, $m^{p - 1} \equiv 1 \pmod{p}$. Then, $\operatorname{ord}_m(p) = 3^{n + 1} \mid (p - 1)$. This implies that $p - 1 \ge 3^{n + 1}$, a contradiction as $p$ is a divisor of $n$. Thus, $r \mid 3^n$. Then, $m^{3^n} \equiv 1 \pmod{p}$. Multiplying this for every prime factor of $n$, we get that $m^{3^n} \equiv 1 \pmod{n}$. Plugging this into the condition $n \mid (1 + m^{3^n} + m^{2 \cdot 3^n})$, we get that $n \mid 3$, contradiction. Thus, no solutions exist other than the ones stated earlier.