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How many twin primes are there whose sum is a power of a prime? I have started with the form p+(2+p), which reduces to 2(p+1), which suggests the number must be even. Where could I go from here?

ROS
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    Well, that would suggest that the sum is a power of two, so $p = 2^n -1$ for some $n$. Then $p+2 = 2^n +1$ also needs to be prime – xxxxxxxxx Sep 17 '19 at 23:58
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    What do you know about Mersenne and Fermat primes? – xxxxxxxxx Sep 18 '19 at 00:02
  • I don't know anything about Mersenne and Fermat primes, but I will read about these, thank you. – ROS Sep 18 '19 at 00:06
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    Except for 3 and 5, all other twin primes are of the form $3k-1, 3k+1$, hence their sum is also divisible by 3, in addition to being even. No multiple of 6 is a power of a prime. – Catalin Zara Sep 18 '19 at 00:08
  • Thank you for your insights- I will have to think more about why no multiple of 6 is a power of a prime, and where the twin prime form of 3k-1, 3k+1 comes from. – ROS Sep 18 '19 at 00:26
  • from the fact that not prime greater than three can have 3 as a factor ... –  Sep 18 '19 at 12:34
  • Whoops, late night brain must have taken over. No multiple of 6 is a power of a prime because any such number contains both 2 and 3 in its prime factorization, which makes the number composite. Thanks Roddy. – ROS Oct 02 '19 at 00:45

2 Answers2

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As @Catalin has noted. Any twin primes bigger than 3 and 5 are congruent to $\pm1$ modulo 6. Thus their sum is divisible by 6 an not a prime power. So 3 and 5 are the only twin primes adding up to a prime power.

Cade Reinberger
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If $(p,p+2)$ are a pair of twin primes, their sum is even, so if it equals to the power of a prime then that prime must be $2$. (Can you see why?) Furthermore, if $p>3$ then modulo $3$ we have $p\equiv -1$ and $p+2\equiv 1$, since in any other case $3$ would divide either $p$ or $p+2$, which is absurd. This means that their sum $p+(p+2)\equiv(-1)+1\equiv 0$ modulo $3$. But $p+(p+2)$ is supposed to be a power of $2$, so how can it possibly be divisible by $3$?

Thus, the only answer is $(3,5)$.

YiFan Tey
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