[ Factorization ]  $[$ Residue Arithmetic (miscellaneous) $]$

Find all natural $n$, for which $2^{n}+33$ is a perfect square.

#
If odd, then $2^{n}+33 \equiv 2(\bmod 3)$, so it cannot be a square.

Let $n=2 k$ and $2^{n}+33=m^{2}$. Then $\left(m-2^{k}\right)\left(m+2^{k}\right)=33$. Hence, $m-2^{k}=1, \mathrm{~m}+2^{k}=33$ or $m-2^{k}=3, m+$ $2^{k}=11$. In the first case, $2^{k}=16$, in the second $2^{k}=4$.

## Answer

$n=4$ or 8.