[CmKaN]

Could you please see what's wrong with my solution again
1.Let $n=2k,k \in \mathbb{N}$
$ (2k)^{2}\times 2^{3(2k)^{2}}+1=4k^{2}\times2^{12k^{2}}+1=4k^{2}\times(2^{4})^{3k^{2}}+1\equiv 4k^{2}+1 \equiv 0(mod17)$
$4k^{2} \equiv 16(mod17)$
$k^{2} \equiv 4(mod17)$
$\therefore k^{2}=17t+4,n \in \mathbb{N}$
$t=\frac{(k+2)(k-2)}{17}$
$\because t \in \mathbb{N},17 \in \mathbb{P} $
we get two case
CASE1
$k+2=17m,m\in \mathbb{N}$
$k=17m-2$
$\therefore n=34m-4$
All n is$\left\lceil\ \frac{1004}{34} \right\rceil =29 $
CASE2
$k-2=17m,m \in \mathbb{N}$
$n=34m+4$
All n is $\left\lceil\ \frac{996}{34} \right\rceil =29 $
from case1,2 $\therefore$ all n is $58$