If prime $p$ such $p|F_{n}(nge 2)$, show that $pequiv pm 2 pmod 5$
$begingroup$
Let $p$ be a prime number and $$p|F_{n}, nge 2$$
where $F_{n}=2^{2^n}+1.$
Show that
$$pequiv pm 2pmod 5.$$
I have proved
$$F_{n}equiv 2pmod 5,$$
because $$F_{n}=2^{2^n}+1=(2^2)^{2^{n-1}}+1equiv (-1)^{2^{n-1}}+1=2pmod 5,$$
but I can't show $pequiv pm2pmod 5$. Thanks.
number-theory fermat-numbers
$endgroup$
add a comment |
$begingroup$
Let $p$ be a prime number and $$p|F_{n}, nge 2$$
where $F_{n}=2^{2^n}+1.$
Show that
$$pequiv pm 2pmod 5.$$
I have proved
$$F_{n}equiv 2pmod 5,$$
because $$F_{n}=2^{2^n}+1=(2^2)^{2^{n-1}}+1equiv (-1)^{2^{n-1}}+1=2pmod 5,$$
but I can't show $pequiv pm2pmod 5$. Thanks.
number-theory fermat-numbers
$endgroup$
add a comment |
$begingroup$
Let $p$ be a prime number and $$p|F_{n}, nge 2$$
where $F_{n}=2^{2^n}+1.$
Show that
$$pequiv pm 2pmod 5.$$
I have proved
$$F_{n}equiv 2pmod 5,$$
because $$F_{n}=2^{2^n}+1=(2^2)^{2^{n-1}}+1equiv (-1)^{2^{n-1}}+1=2pmod 5,$$
but I can't show $pequiv pm2pmod 5$. Thanks.
number-theory fermat-numbers
$endgroup$
Let $p$ be a prime number and $$p|F_{n}, nge 2$$
where $F_{n}=2^{2^n}+1.$
Show that
$$pequiv pm 2pmod 5.$$
I have proved
$$F_{n}equiv 2pmod 5,$$
because $$F_{n}=2^{2^n}+1=(2^2)^{2^{n-1}}+1equiv (-1)^{2^{n-1}}+1=2pmod 5,$$
but I can't show $pequiv pm2pmod 5$. Thanks.
number-theory fermat-numbers
number-theory fermat-numbers
edited Jan 20 at 3:08
J. W. Tanner
2,6121217
2,6121217
asked Jan 19 at 1:47
function sugfunction sug
2961438
2961438
add a comment |
add a comment |
1 Answer
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$begingroup$
No. $641$ is a prime divisor of $F_5$
other examples at https://en.wikipedia.org/wiki/Fermat_number#Factorization_of_Fermat_numbers
$endgroup$
1
$begingroup$
A quick look at OEIS sequence A023394 (Prime factors of Fermat numbers) shows quite a few more exceptions.
$endgroup$
– DanielWainfleet
Jan 19 at 2:56
add a comment |
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1 Answer
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$begingroup$
No. $641$ is a prime divisor of $F_5$
other examples at https://en.wikipedia.org/wiki/Fermat_number#Factorization_of_Fermat_numbers
$endgroup$
1
$begingroup$
A quick look at OEIS sequence A023394 (Prime factors of Fermat numbers) shows quite a few more exceptions.
$endgroup$
– DanielWainfleet
Jan 19 at 2:56
add a comment |
$begingroup$
No. $641$ is a prime divisor of $F_5$
other examples at https://en.wikipedia.org/wiki/Fermat_number#Factorization_of_Fermat_numbers
$endgroup$
1
$begingroup$
A quick look at OEIS sequence A023394 (Prime factors of Fermat numbers) shows quite a few more exceptions.
$endgroup$
– DanielWainfleet
Jan 19 at 2:56
add a comment |
$begingroup$
No. $641$ is a prime divisor of $F_5$
other examples at https://en.wikipedia.org/wiki/Fermat_number#Factorization_of_Fermat_numbers
$endgroup$
No. $641$ is a prime divisor of $F_5$
other examples at https://en.wikipedia.org/wiki/Fermat_number#Factorization_of_Fermat_numbers
answered Jan 19 at 2:45
Will JagyWill Jagy
103k5102200
103k5102200
1
$begingroup$
A quick look at OEIS sequence A023394 (Prime factors of Fermat numbers) shows quite a few more exceptions.
$endgroup$
– DanielWainfleet
Jan 19 at 2:56
add a comment |
1
$begingroup$
A quick look at OEIS sequence A023394 (Prime factors of Fermat numbers) shows quite a few more exceptions.
$endgroup$
– DanielWainfleet
Jan 19 at 2:56
1
1
$begingroup$
A quick look at OEIS sequence A023394 (Prime factors of Fermat numbers) shows quite a few more exceptions.
$endgroup$
– DanielWainfleet
Jan 19 at 2:56
$begingroup$
A quick look at OEIS sequence A023394 (Prime factors of Fermat numbers) shows quite a few more exceptions.
$endgroup$
– DanielWainfleet
Jan 19 at 2:56
add a comment |
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