??? Please look that am I in the correct way
$p \equiv 1,-1 \pmod 6$
$p=6k+1,6m-1 ,k,m \in \mathbb{N} $
$(6k\pm 1)^{2}+2543$
$12(3k^{2} \pm k+ 212)$
$2^{2}3(3k^{2} \pm k+ 212)$
$\because 16=16\times 1=8 \times 2 =4\times 2 \times 2= 2\times 2\times 2\times 2$
((1,2,3,4,5,6,7,8,9,10,11,12,13,14,15))<<<will not happen
$\therefore (3k^{2} \pm k+ 212)=2^{5},2x ,x\in \mathbb{P}$
$3k^{2} \pm k+ 212=2^{5} \rightarrow 3k^{2} \pm k +180=0$<<< not gonna happen
$\therefore 3k^{2} \pm k+ 212=2x,x\in \mathbb{P}$
$k(3k \pm1) =2(x-106)$
$\because x-106 \in \mathbb{N} \therefore 2|k \vee 2|3k \pm 1$
$\because x$ is odd number $\therefore x-106$ is odd number
case 1
$2|k$
$k=2m,m\in \mathbb{N}$
$\therefore k=2$--->$p=13$
case 2
$2|3k \pm 1$
and then I don't know what to do
