*) Với $n=2k$, ta có:
$A=2^{3^n}+1=2^{3^{2k}}+1=2^{9^k}+1$
Ta có: $9^k\equiv1$ (mod $8$) $\Rightarrow 9^k=8n+1$
$\Rightarrow A=2^{8n+1}+1=2.256^n+1\equiv 3$ (mod $17$) vì $256\equiv 1$ (mod $17$).
*) Với $n=2k+1$, ta có:
$A=2^{3^n}+1=2^{3^{2k+1}}+1=2^{3.9^k}+1$
Ta có: $3.9^k\equiv3$ (mod $8$) $\Rightarrow 3.9^k=8m+3$
$\Rightarrow A=2^{8m+3}+1=8.256^m+1\equiv 9$ (mod $17$) vì $256\equiv 1$ (mod $17$).
Vậy: $2^{3^n}+1$ không chia hết cho $17$ với mọi $n\in\mathbb{N}$