[ Factorization ]

 [ Arithmetic of residues (other) $]$ Solve the equation $3^{n}+55=m^{2}$ in natural numbers.
Let's consider two cases.

1) $n=2 k$. We can write the equation as $\left(m-3^{k}\right)\left(m+3^{k}\right)=55$. From this, $3^{k}<55$, that is, $k \leq 3$. By substituting the values $n=2,4,6$ into the original equation, we find the corresponding values of $m$.
2) $n=2 k+1$. We can write the equation as $3 \cdot 9^{k}+55=m^{2}$. The last digit of the left side is 2 or $8\left(9^{k}\right.$ ends in 9 or 1), and a square cannot end in such a digit.

## Answer

$(2,8),(6,28)$.

## Problem