11. find all natural numbers $k$ such that $3^{k}+5^{k}$ is a power of a natural number with exponent $\geq 2$.

## 1st solution
Assume that $3^{k}+5^{k}=n^{t}$ with $t \geq 2$. If $k$ is even, then $3^{k}+5^{k} \equiv 1+1=2$ $(\bmod 4)$, so the left side is even but not divisible by 4, a contradiction to $t \geq 2$. If $k$ is odd, then the following applies

$$
3^{k}+5^{k}=(3+5)\left(3^{k-1}-3^{k-2} \cdot 5+\ldots-3 \cdot 5^{k-2}+5^{k-1}\right)
$$

The second factor on the right is odd, because it consists of an odd number of summands, which are all odd. Consequently, $3^{k}+5^{k}$ is divisible by 8 but not by 16, so $t=3$ :

$$
3^{k}+5^{k}=n^{3}
$$

Obviously $k=1$ is a solution, we show that it is the only one. For $k \geq 2$ we consider the equation modulo 9. Since $k$ is odd, a quick calculation yields

$$
3^{k}+5^{k} \equiv 5^{k} \equiv\left\{\begin{array}{lll}
5, & k \equiv 1 & (\bmod 6) \\
8, & k \equiv 3 & (\bmod 6) \\
2, & k \equiv 5 & (\bmod 6)
\end{array} \quad(\bmod 9)\right.
$$

For the right-hand side, however, $n^{3} \equiv 0,1,8(\bmod 9)$. Consequently, $k \equiv 3(\bmod 6)$ must apply. We now look at the equation modulo 7. Since $k$ is odd, a quick calculation shows

$$
3^{k}+5^{k} \equiv\left\{\begin{array}{lll}
1, & k \equiv 1.5 & (\bmod 6) \\
5, & k \equiv 3 & (\bmod 6)
\end{array} \quad(\bmod 7)\right.
$$

For the right-hand side, however, $n^{3} \equiv 0,1,6(\bmod 7)$. Therefore, $k \equiv 1,5(\bmod 6)$ must apply, in contradiction to $k \equiv 3(\bmod 6)$.

## 2nd solution

You can also finish the proof as follows: The calculation modulo 9 shows that $k$ is divisible by 3, $k=3 l$. The equation now follows

$$
\left(3^{l}\right)^{3}+\left(5^{l}\right)^{3}=n^{3}
$$

which, according to Fermat's theorem, has no solution, contradiction.