, 2008

Solution:
If $n=1$ then $2^{1}+3^{1}=5$ is not a perfect cube.

Perfect cubes give residues $-1, 0$ and $1$ modulo $9$. If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible by $3$ (consider the congruence $2^{n}+3^{n}=x^{3}$ modulo $9$).

If $n=3k$ then $2^{3k}+3^{3k} > (3^{k})^{3}$. Also, $(3^{k}+1)^{3} = 3^{3k} + 3 \cdot 3^{2k} + 3 \cdot 3^{k} + 1 > 3^{3k} + 3^{2k} = 3^{3k} + 9^{k} > 3^{3k} + 8^{k} = 3^{3k} + 2^{3k}$.

But, $3^{k}$ and $3^{k}+1$ are two consecutive integers so $2^{3k}+3^{3k}$ is not a perfect cube.