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I am looking for a proof of the following statement about Fermat numbers and Sophie Germain primes:

The only Fermat primes that are also Sophie Germain primes are $3$ and $5$.

I suspect these are the only ones because many mathematicians believe the only Fermat primes are $F_0, F_1, F_2,F_3,F_4$, and it is very unlikely that not only are there more Fermat primes, but that some of them are Sophie Germain primes. It might also be interesting if someone could find a Fermat number $F_n$ such that $2F_n+1=2^{2^n+1}+3$ is prime. I have checked this up to $F_{32}$

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  • $\begingroup$ Every Fermat number satisfies your equation. You probably mean the exponents $n$ for which $2F_n+1$ is prime. $\endgroup$
    – Peter
    Jan 6, 2019 at 8:36
  • $\begingroup$ I see a purpose in Sophie Germain primes. I maybe also see a purpose in Fermat primes such as 65,537 for the purpose of its modulo multiplication group being of order a power of 2 and making it be a hard problem finding which number you can raise 3 to the power of to get that number modulo 65,537. I don't see a purpose in a number having both of those properties. $\endgroup$
    – Timothy
    May 3, 2020 at 18:35

1 Answer 1

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Claim : $$2^{2^n+1}+3$$ is divisble by $5$ for $n\ge 2$

Proof : Because of $2^4\equiv 1\mod 5$ we can reduce the exponent modulo $4$, which gives $2^1+3=5$ which is divisble by $5$. QED

Hence there is no further prime of the form $2F_n+1$ and therefore no further Fermat-Sophie-Germain-prime.

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    $\begingroup$ Thank you very much, this probably should have been fairly obvious to me, but I never thought to work modulo 5. $\endgroup$
    – coDE_RP
    Jan 7, 2019 at 22:30
  • $\begingroup$ I thought of this myself and was going to write something like it as an answer. This already answers the question. There is no need for anymore answers. I know that the first two Fermat numbers, 3 and 5, are Sophie-Germain primes. $\endgroup$
    – Timothy
    Mar 18, 2020 at 2:15

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