Sophie Germain primes that are Fermat primes
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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 $F_{32}$
number-theory prime-numbers
asked Jan 5 at 21:49
coDE_RPcoDE_RP
799
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add a comment |
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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 $F_{32}$
number-theory prime-numbers
asked Jan 5 at 21:49
coDE_RPcoDE_RP
799
$endgroup$
$begingroup$
Every Fermat number satisfies your equation. You probably mean the exponents $n$ for which $2F_n+1$ is prime.
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– Peter
Jan 6 at 8:36
add a comment |
$begingroup$
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
799
$endgroup$
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
799
asked Jan 5 at 21:49
coDE_RPcoDE_RP
799
edited Jan 8 at 16:53
coDE_RP
asked Jan 5 at 21:49
coDE_RPcoDE_RP
799
asked Jan 5 at 21:49
coDE_RPcoDE_RP
799
asked Jan 5 at 21:49
coDE_RPcoDE_RP
799
799
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
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
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
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 Jan 6 at 8:43
PeterPeter
48.9k1239136
$endgroup$
1
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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
add a comment |
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$begingroup$
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 Jan 6 at 8:43
PeterPeter
48.9k1239136
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
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 Jan 6 at 8:43
PeterPeter
48.9k1239136
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
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 Jan 6 at 8:43
PeterPeter
48.9k1239136
$endgroup$
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 Jan 6 at 8:43
PeterPeter
48.9k1239136
answered Jan 6 at 8:43
PeterPeter
48.9k1239136
answered Jan 6 at 8:43
PeterPeter
48.9k1239136
answered Jan 6 at 8:43
PeterPeter
48.9k1239136
48.9k1239136
1
$begingroup$
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
add a comment |
1
$begingroup$
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
1
1
$begingroup$
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
$begingroup$
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
add a comment |
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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