Infinitely many prime $n$ for which $n^2 = p + 8$ for some prime $p$. How to prove that there exist an infinite number of prime $n$ for which $n^2=p+8$ for some prime $p$?
Verification of the form $n^2=p+8$ where $n$ and $p$ are some $p$.
$$\begin{array}{|c|c|} \hline
       n     &         n^2 = 8 + p       \\
\hline
      11     &         121 = 8 + 113      \\
      23     &         529 = 8 + 521     \\
      31     &         961 = 8 + 953      \\
      37     &         1369= 8 + 1361    \\
\hline
\end{array}
$$
 A: No proof is available. There are no proofs that any polynomial in a single integer variable, of degree at least two, such as your $n^2 - 8,$ take on an infinite number of prime values. In comparison, $n^2 - 8 m^2$ certainly does, but this is a function of two integer variables.  
The best known is $n^2 + 1$ . People suspect there are an infinite number of primes of this form, but no proof. One of the most efficient such polynomials is $n^2 - n + 41,$ the value is prime for $1 \leq n \leq 40,$ then it seems to give relatively frequent primes thereafter as long as $n \neq 0,1 \pmod {41},$ but we cannot be sure about infinite sets of primes with this one either. See http://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions 
Alright, needs emphasis; people have suspicions about one-variable polynomials and primes, http://en.wikipedia.org/wiki/Bunyakovsky_conjecture  but no proofs are yet available for any such problem. No proofs are expected for the foreseeable future. This is not the same as saying something is unprovable, I know of no conjectures in day-to-day number theory that have been shown to be unprovable. There is a tiny collection of statements in set theory, the one I remember being The Continuum Hypothesis, where there can be no proof because the statement is independent of the other axioms. Well, the guy got the Fields Medal for that, it does not happen every day. I suppose it is valid to say that The Parallel Postulate is independent of the other axioms of plane geometry and unprovable, as there is the hyperbolic/non-Euclidean plane. 
A: Your question is equivalent to there's infinitely many $m$ that:  
$2n^2 + 2(n - 2) = m,\ $where $2n + 1,\ 2m+1$ is prime.    
Such as: $n=3, \ 2n^2 + 2(n - 2) = 20,(2 \cdot 3+1)^2=7^2=49=2 \cdot 20+1+8=41+8,\ $ where $7,41$ is prime. It seems this is a elementary question.
