Find the sum of all primes $p$ such that $p+2$ and $p^2+2p-8$ are also primes.

The answer is 3.

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For $p^{2} + 2p - 8 = (p + 4)(p - 2)$ to be prime, one of $p + 4$ or $p - 2$ must equal $1$ . As $p$ must be prime the only option is that $p - 2 = 1 \Longrightarrow p = 3$ , which is prime, as is $p + 2 = 5$ . Thus the only such prime, and hence the desired sum, is $\boxed{3}$ .