Sophie Germain primes that are Fermat primes
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 $F_{32}$
number-theory prime-numbers
asked Jan 5 at 21:49
coDE_RPcoDE_RP
599
add a comment |
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 $F_{32}$
number-theory prime-numbers
asked Jan 5 at 21:49
coDE_RPcoDE_RP
599
Every Fermat number satisfies your equation. You probably mean the exponents $n$ for which $2F_n+1$ is prime.
– Peter
2 days ago
add a comment |
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 $F_{32}$
number-theory prime-numbers
asked Jan 5 at 21:49
coDE_RPcoDE_RP
599
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 $F_{32}$
number-theory prime-numbers
number-theory prime-numbers
asked Jan 5 at 21:49
coDE_RPcoDE_RP
599
asked Jan 5 at 21:49
coDE_RPcoDE_RP
599
edited 6 hours ago
coDE_RP
asked Jan 5 at 21:49
coDE_RPcoDE_RP
599
asked Jan 5 at 21:49
coDE_RPcoDE_RP
599
asked Jan 5 at 21:49
coDE_RPcoDE_RP
599
599
Every Fermat number satisfies your equation. You probably mean the exponents $n$ for which $2F_n+1$ is prime.
– Peter
2 days ago
add a comment |
Every Fermat number satisfies your equation. You probably mean the exponents $n$ for which $2F_n+1$ is prime.
– Peter
2 days ago
Every Fermat number satisfies your equation. You probably mean the exponents $n$ for which $2F_n+1$ is prime.
– Peter
2 days ago
Every Fermat number satisfies your equation. You probably mean the exponents $n$ for which $2F_n+1$ is prime.
– Peter
2 days ago
add a comment |
1 Answer
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Claim : $$2^{2^n+1}+3$$ is divisble by $5$ for $nge 2$
Proof : Because of $2^4equiv 1mod 5$ 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.
answered 2 days ago
PeterPeter
46.8k1039125
1
Thank you very much, this probably should have been fairly obvious to me, but I never thought to work modulo 5.
– coDE_RP
yesterday
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
Claim : $$2^{2^n+1}+3$$ is divisble by $5$ for $nge 2$
Proof : Because of $2^4equiv 1mod 5$ 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.
answered 2 days ago
PeterPeter
46.8k1039125
1
Thank you very much, this probably should have been fairly obvious to me, but I never thought to work modulo 5.
– coDE_RP
yesterday
add a comment |
Claim : $$2^{2^n+1}+3$$ is divisble by $5$ for $nge 2$
Proof : Because of $2^4equiv 1mod 5$ 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.
answered 2 days ago
PeterPeter
46.8k1039125
1
Thank you very much, this probably should have been fairly obvious to me, but I never thought to work modulo 5.
– coDE_RP
yesterday
add a comment |
Claim : $$2^{2^n+1}+3$$ is divisble by $5$ for $nge 2$
Proof : Because of $2^4equiv 1mod 5$ 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.
answered 2 days ago
PeterPeter
46.8k1039125
Claim : $$2^{2^n+1}+3$$ is divisble by $5$ for $nge 2$
Proof : Because of $2^4equiv 1mod 5$ 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.
answered 2 days ago
PeterPeter
46.8k1039125
answered 2 days ago
PeterPeter
46.8k1039125
answered 2 days ago
PeterPeter
46.8k1039125
answered 2 days ago
PeterPeter
46.8k1039125
46.8k1039125
1
Thank you very much, this probably should have been fairly obvious to me, but I never thought to work modulo 5.
– coDE_RP
yesterday
add a comment |
1
Thank you very much, this probably should have been fairly obvious to me, but I never thought to work modulo 5.
– coDE_RP
yesterday
1
1
Thank you very much, this probably should have been fairly obvious to me, but I never thought to work modulo 5.
– coDE_RP
yesterday
Thank you very much, this probably should have been fairly obvious to me, but I never thought to work modulo 5.
– coDE_RP
yesterday
add a comment |
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Every Fermat number satisfies your equation. You probably mean the exponents $n$ for which $2F_n+1$ is prime.
– Peter
2 days ago