Two . Prime . Power. Sum . Again prime?

Let $E = p^q+q^p$ . First, note that if $(p,q)$ is a solution, then so is $(q,p)$ . Now, $p$ and $q$ can’t be both even or both odd, else E will be even. Without loss of generality, assume p = 2 and q some odd prime. So, $E = 2^q + q^2$ . There are two cases to consider.

Case 1: $q = 3$ .

This yields $E = 2^3 + 3^2 = 17$ , which is prime. So, (2,3) and, hence (3,2) are solutions.

Case 2: $q > 3$ .

There are two sub-cases to consider.

1) $q = 3k+1$ , where k is some even integer. Then, we have $E = 2^{3k+1} + (3k+1)^2 \equiv (-1)^k(-1) + 1 \equiv -1 + 1 \equiv 0 \pmod 3$ . Hence, $3 \mid E$ ; so, E can’t be prime.

2) $q = 3k+2$ , where $k$ is some odd integer. Then we have $E = 2^{3k+2} + (3k+2)^2 \equiv (-1)^k(1) + 1 \equiv -1 + 1 \equiv 0 \pmod 3$ . Hence, $3 \mid E$ ; so, again, $E$ can’t be prime.

As we have exhausted all possible cases, we conclude $(2,3)$ and $(3,2)$ are the only possible solutions.

The number of ordered pairs are 2 , which are (2,3)and (3,2).Therefore the sum is 2(2+3) which is equal to 10.

I plugged in ( $2,3)$ for ( $p,q)$ to test if $p^q+q^p$ would make a prime and it did! $2^3+3^2=17$ . Since you can reverse the numbers and they will yield the same result, the sum of all primes that fit that produce another prime is $2(p+q)=2(2+3)=10$