2. Case1$P>5 \therefore P \equiv 1,-1(mod6)$
1.1$P \equiv 1(mod6)--->P=6k+1,k\in \mathbb{N} $
$P^{2}+11=(6k+1)^{2}+11$
$36k^{2}+12k+12=[(2)^{2}(3)](3k^{2}+k+1)$
$\because k>1$ $\therefore$ factor of $P^{2}+11>6$
1.2$P \equiv -1(mod6)--->P=6k-1,k\in \mathbb{N}$
$P^{2}+11=(6k-1)^{2}+11$
$36k^{2}-12k+12=[(2)^{2}(3)](3k^{2}-k+1)$
$\because k\in \mathbb{N}$ $\therefore$ factor of $P^{2}+11>6$
$\therefore$ no p for case1
Case2$P=2,3$
we put in the equation and get that only 3 is true
$\therefore P=3$
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