Prime $p$ with $p^2+8$ prime

Any number can be written as $6c,6c\pm1,6c\pm2=2(3c\pm1),6c+3=3(2c+1)$

Clearly, $6c,6c\pm2,6c+3$ can not be prime for $c\ge 1$

Any prime $>3$ can be written as $6a\pm 1$ where $a\ge 1$

So, $p^2+8=(6a\pm 1)^2+8=3(12a^2\pm4a+3)$.

Then , $p^2+8>3$ is divisible by 3,hence is not prime.

So, the only prime is $3$.

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Any number$(p)$ not divisible by $3,$ can be written as $3b\pm1$

Now, $(3b\pm1)^2+(3c-1)=3(3b^2\pm2b+c)$.

Then , $p^2+(3c-1)$ is divisible by 3

and $p^2+(3c-1)>3$ if $p>3$ and $c\ge1$,hence not prime.

The necessary condition for $p^2+(3c-1)$ to be prime is $3\mid p$

$\implies$ if $p^2+(3c-1)$ is prime, $3\mid p$.

If $p$ needs to be prime, $p=3$, here $c=3$

Suppose there exists a prime $p$ (not equal to 3) such that $p^2 + 8$ is prime. Since $p$ is indivisible by 3, therefore $p \equiv 1 \pmod 3$ or $p \equiv -1 \pmod 3$, therefore $p^2 \equiv 1 \pmod 3$. Thus, $p^2 + 8 \equiv 9 \equiv 0 \pmod 3$, therefore $p^2 + 8$ is a prime greater than 3 that is divisible by 3 (a contradiction).

Put $\rm\,q,n = 3\:$ in: $\, $ for $\rm\,p\ne q\:$ primes, Fermat $\rm\,\Rightarrow\, q\:|\: p^{q-1}\!-\!1+qn\ $ so it is prime iff it $\rm = q.$