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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 at 8:36
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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 at 22:30

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