Mixing
Prove a case of Dirichlet's Theorem: that there are infinity many primes of the form $8k+1$
$begingroup$
Prove a case of Dirichlet's Theorem: that there are infinity many primes of the form $8k+1$ using these steps.
(Dirichlet’s Theorem). Let a, b be two positive integers. If gcd(a, b) = 1,
then there exists an infinite number of primes of the form ak + b.
The aim of this exercise is to prove Dirichlet's Theorem when $a = 8$ and $b = 1$.
Let $x$ be an even integer and $p$ be a prime divisor of $x^4+1$.
Show that $left(frac{-1}{p}right) = 1$.
Prove that $x$ and $p$ are coprime and deduce that $x$ is invertible modulo $p$.
Show that $left(frac2pright) = 1$. Hint: You might find the following identity useful:
$$ x^4+1 = (x^2+1)^2-2x^2 $$
Show that $p equiv 1 bmod 8$.
Deduce that there are infinitely many primes $p$ congruent to $1$ modulo $8$.
So far I've got
$N = (2p_1p_2ldots p_r)^4+1$
Let $p$ be a prime divisor of $N$. If $p|N$ then
$$p| (2p_1p_2ldots p_r)^4+1$$
$$-1 = (2p_1p_2ldots p_r)^4 mod p$$
$$(-1/p) = 1$$
This is where I get stuck.
My lecturer has replied with:
'This is indeed the beginning of the correct answer. You have that N is of the form x^4+1, so you can use question 4) and deduce N=1 mod 8. Now, can it be equal to one of the p_i?'
I'm still unsure where to go from here?
abstract-algebra number-theory modular-arithmetic
asked Jan 11 at 16:15
Mayur ChauhanMayur Chauhan
63
$endgroup$
|
show 3 more comments
$begingroup$
Prove a case of Dirichlet's Theorem: that there are infinity many primes of the form $8k+1$ using these steps.
(Dirichlet’s Theorem). Let a, b be two positive integers. If gcd(a, b) = 1,
then there exists an infinite number of primes of the form ak + b.
The aim of this exercise is to prove Dirichlet's Theorem when $a = 8$ and $b = 1$.
Let $x$ be an even integer and $p$ be a prime divisor of $x^4+1$.
Show that $left(frac{-1}{p}right) = 1$.
Prove that $x$ and $p$ are coprime and deduce that $x$ is invertible modulo $p$.
Show that $left(frac2pright) = 1$. Hint: You might find the following identity useful:
$$ x^4+1 = (x^2+1)^2-2x^2 $$
Show that $p equiv 1 bmod 8$.
Deduce that there are infinitely many primes $p$ congruent to $1$ modulo $8$.
So far I've got
$N = (2p_1p_2ldots p_r)^4+1$
Let $p$ be a prime divisor of $N$. If $p|N$ then
$$p| (2p_1p_2ldots p_r)^4+1$$
$$-1 = (2p_1p_2ldots p_r)^4 mod p$$
$$(-1/p) = 1$$
This is where I get stuck.
My lecturer has replied with:
'This is indeed the beginning of the correct answer. You have that N is of the form x^4+1, so you can use question 4) and deduce N=1 mod 8. Now, can it be equal to one of the p_i?'
I'm still unsure where to go from here?
abstract-algebra number-theory modular-arithmetic
asked Jan 11 at 16:15
Mayur ChauhanMayur Chauhan
63
$endgroup$
$begingroup$
that a wonderful exercise. Have you tried anything or are you just asking us to do to everything for you? (in that case i m sorry to say it won't happen)
$endgroup$
– Alexis
Jan 11 at 16:18
1
$begingroup$
please type this in the question's body. It would help people to understand where you are stuck.
$endgroup$
– Alexis
Jan 11 at 16:23
1
$begingroup$
Edit your question to show your attempts, rather than just posting a picture. Also, write down the question rather than pointing to a picture as well. Then, state Dirichlet's theorem yourself to begin with.
$endgroup$
– Mefitico
Jan 11 at 16:23
$begingroup$
You might want to look at the statement of the referenced theorem.
$endgroup$
– yberman
Jan 11 at 16:35
1
$begingroup$
What do you mean by "Prove using Dirichlet's Theorem"?
$endgroup$
– Barry Cipra
Jan 11 at 17:05
|
show 3 more comments
$begingroup$
Prove a case of Dirichlet's Theorem: that there are infinity many primes of the form $8k+1$ using these steps.
(Dirichlet’s Theorem). Let a, b be two positive integers. If gcd(a, b) = 1,
then there exists an infinite number of primes of the form ak + b.
The aim of this exercise is to prove Dirichlet's Theorem when $a = 8$ and $b = 1$.
Let $x$ be an even integer and $p$ be a prime divisor of $x^4+1$.
Show that $left(frac{-1}{p}right) = 1$.
Prove that $x$ and $p$ are coprime and deduce that $x$ is invertible modulo $p$.
Show that $left(frac2pright) = 1$. Hint: You might find the following identity useful:
$$ x^4+1 = (x^2+1)^2-2x^2 $$
Show that $p equiv 1 bmod 8$.
Deduce that there are infinitely many primes $p$ congruent to $1$ modulo $8$.
So far I've got
$N = (2p_1p_2ldots p_r)^4+1$
Let $p$ be a prime divisor of $N$. If $p|N$ then
$$p| (2p_1p_2ldots p_r)^4+1$$
$$-1 = (2p_1p_2ldots p_r)^4 mod p$$
$$(-1/p) = 1$$
This is where I get stuck.
My lecturer has replied with:
'This is indeed the beginning of the correct answer. You have that N is of the form x^4+1, so you can use question 4) and deduce N=1 mod 8. Now, can it be equal to one of the p_i?'
I'm still unsure where to go from here?
abstract-algebra number-theory modular-arithmetic
asked Jan 11 at 16:15
Mayur ChauhanMayur Chauhan
63
$endgroup$
Prove a case of Dirichlet's Theorem: that there are infinity many primes of the form $8k+1$ using these steps.
(Dirichlet’s Theorem). Let a, b be two positive integers. If gcd(a, b) = 1,
then there exists an infinite number of primes of the form ak + b.
The aim of this exercise is to prove Dirichlet's Theorem when $a = 8$ and $b = 1$.
Let $x$ be an even integer and $p$ be a prime divisor of $x^4+1$.
Show that $left(frac{-1}{p}right) = 1$.
Prove that $x$ and $p$ are coprime and deduce that $x$ is invertible modulo $p$.
Show that $left(frac2pright) = 1$. Hint: You might find the following identity useful:
$$ x^4+1 = (x^2+1)^2-2x^2 $$
Show that $p equiv 1 bmod 8$.
Deduce that there are infinitely many primes $p$ congruent to $1$ modulo $8$.
So far I've got
$N = (2p_1p_2ldots p_r)^4+1$
Let $p$ be a prime divisor of $N$. If $p|N$ then
$$p| (2p_1p_2ldots p_r)^4+1$$
$$-1 = (2p_1p_2ldots p_r)^4 mod p$$
$$(-1/p) = 1$$
This is where I get stuck.
My lecturer has replied with:
'This is indeed the beginning of the correct answer. You have that N is of the form x^4+1, so you can use question 4) and deduce N=1 mod 8. Now, can it be equal to one of the p_i?'
I'm still unsure where to go from here?
abstract-algebra number-theory modular-arithmetic
abstract-algebra number-theory modular-arithmetic
asked Jan 11 at 16:15
Mayur ChauhanMayur Chauhan
63
asked Jan 11 at 16:15
Mayur ChauhanMayur Chauhan
63
edited Jan 11 at 17:33
Mayur Chauhan
asked Jan 11 at 16:15
Mayur ChauhanMayur Chauhan
63
asked Jan 11 at 16:15
Mayur ChauhanMayur Chauhan
63
asked Jan 11 at 16:15
Mayur ChauhanMayur Chauhan
63
63
$begingroup$
that a wonderful exercise. Have you tried anything or are you just asking us to do to everything for you? (in that case i m sorry to say it won't happen)
$endgroup$
– Alexis
Jan 11 at 16:18
1
$begingroup$
please type this in the question's body. It would help people to understand where you are stuck.
$endgroup$
– Alexis
Jan 11 at 16:23
1
$begingroup$
Edit your question to show your attempts, rather than just posting a picture. Also, write down the question rather than pointing to a picture as well. Then, state Dirichlet's theorem yourself to begin with.
$endgroup$
– Mefitico
Jan 11 at 16:23
$begingroup$
You might want to look at the statement of the referenced theorem.
$endgroup$
– yberman
Jan 11 at 16:35
1
$begingroup$
What do you mean by "Prove using Dirichlet's Theorem"?
$endgroup$
– Barry Cipra
Jan 11 at 17:05
|
show 3 more comments
$begingroup$
that a wonderful exercise. Have you tried anything or are you just asking us to do to everything for you? (in that case i m sorry to say it won't happen)
$endgroup$
– Alexis
Jan 11 at 16:18
1
$begingroup$
please type this in the question's body. It would help people to understand where you are stuck.
$endgroup$
– Alexis
Jan 11 at 16:23
1
$begingroup$
Edit your question to show your attempts, rather than just posting a picture. Also, write down the question rather than pointing to a picture as well. Then, state Dirichlet's theorem yourself to begin with.
$endgroup$
– Mefitico
Jan 11 at 16:23
$begingroup$
You might want to look at the statement of the referenced theorem.
$endgroup$
– yberman
Jan 11 at 16:35
1
$begingroup$
What do you mean by "Prove using Dirichlet's Theorem"?
$endgroup$
– Barry Cipra
Jan 11 at 17:05
$begingroup$
that a wonderful exercise. Have you tried anything or are you just asking us to do to everything for you? (in that case i m sorry to say it won't happen)
$endgroup$
– Alexis
Jan 11 at 16:18
$begingroup$
that a wonderful exercise. Have you tried anything or are you just asking us to do to everything for you? (in that case i m sorry to say it won't happen)
$endgroup$
– Alexis
Jan 11 at 16:18
1
1
$begingroup$
please type this in the question's body. It would help people to understand where you are stuck.
$endgroup$
– Alexis
Jan 11 at 16:23
$begingroup$
please type this in the question's body. It would help people to understand where you are stuck.
$endgroup$
– Alexis
Jan 11 at 16:23
1
1
$begingroup$
Edit your question to show your attempts, rather than just posting a picture. Also, write down the question rather than pointing to a picture as well. Then, state Dirichlet's theorem yourself to begin with.
$endgroup$
– Mefitico
Jan 11 at 16:23
$begingroup$
Edit your question to show your attempts, rather than just posting a picture. Also, write down the question rather than pointing to a picture as well. Then, state Dirichlet's theorem yourself to begin with.
$endgroup$
– Mefitico
Jan 11 at 16:23
$begingroup$
You might want to look at the statement of the referenced theorem.
$endgroup$
– yberman
Jan 11 at 16:35
$begingroup$
You might want to look at the statement of the referenced theorem.
$endgroup$
– yberman
Jan 11 at 16:35
1
1
$begingroup$
What do you mean by "Prove using Dirichlet's Theorem"?
$endgroup$
– Barry Cipra
Jan 11 at 17:05
$begingroup$
What do you mean by "Prove using Dirichlet's Theorem"?
$endgroup$
– Barry Cipra
Jan 11 at 17:05
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Someone has already answered the question
Suppose that there are only finitely many such primes p1,…,pk≡1(mod8). Then consider the following number
(2p1⋯pk)4+1,
which is coprime with each pi, and has remainder 1 modulo 8. Since it is odd and greater than 1, it has to be divisible by an odd prime p. Then
ordp(2p1⋯pk)=8
which divides φ(p)=p−1 by Fermat's theorem. Therefore p is another prime ≡1(mod8).
answered Jan 11 at 16:52
Alvin R. FergusonAlvin R. Ferguson
111
$endgroup$
$begingroup$
How do you get $pequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:03
$begingroup$
This is one of the sub questions to my original problem. Not sure how he got that???
$endgroup$
– Mayur Chauhan
Jan 11 at 17:06
add a comment |
$begingroup$
First of all, note that all odd and prime divisors of n4+1 have the form 8k+1. Suppose that there are just finite prime numbers of form 8k+1. Construct the number a=(2p1…pn)4+1, where {pi}ni=1are all primes of form 8k+1. This number is odd, and has at least one prime divisor q. Then q has the form 8l+1. But then q=pj for some j. But it is not possible since q divides a and q divides (2p1…pn)4, i.e, q divides 1.
answered Jan 11 at 17:00
Teri J. HaaseTeri J. Haase
111
$endgroup$
1
$begingroup$
How do you know that $qequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:04
add a comment |
$begingroup$
My proof goes like this:
Assume by way of contradiction that there are only finitely many of such prime numbers, say p1,p2,…,pr. Consider p=16p41p42⋯p4r+1=(2p1p2⋯pr)4+1. Recall that the congruence x4≡−1modp is solvable if and only if p≡1mod8. p cannot be a prime of the form 8n+1, but p≡1mod8, contradiction. Thus, there must be infinitely many prime numbers of the form 8n+1.
answered Jan 11 at 17:54
Philip C. McMillianPhilip C. McMillian
111
$endgroup$
$begingroup$
Good answer, I saw this just after I posted mine
$endgroup$
– Mike
Jan 11 at 18:09
add a comment |
$begingroup$
Maybe this will help fill in the details. Let us write $x = 2p_1ldots p_r$, for some finite $r$, where $p_1 ldots p_r$ are in Teri's answer, all the primes that are 1 mod 8. Then as Teri already noted there is a prime $p$ that divides $x^4+1$, and $p not in {p_1,ldots, p_r}$. We claim that $p$ is of the form $8k+1$ for some integer $k$ next. As already noted by Teri, this will imply that $p_1,ldots, p_r$ are not all the primes that are 1 mod 8, which will give you what you want.
Note that
$$x^4+1 equiv_p 0 Rightarrow x^4 equiv_p -1 Rightarrow x^8 equiv_p 1;$$
That $x^4 not equiv_p 1$ and $x^8 equiv_p 1$, together imply that $x$ has order precisely 8 in the group $left(mathbb{F}_pright)^{times}$, which implies that $|left(mathbb{F}_pright)^{times}| = p-1$ is divisible by 8, which implies that $p$ is of the form $8k+1$ for some integer $k$.
answered Jan 11 at 18:07
MikeMike
3,871412
$endgroup$
$begingroup$
So how would I be able to use this to go through each exercise in the original question?
$endgroup$
– Mayur Chauhan
Jan 11 at 18:15
$begingroup$
I'm not sure. This will establish the main result though, which I would imagine should be fine with your instructor
$endgroup$
– Mike
Jan 11 at 18:16
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Someone has already answered the question
Suppose that there are only finitely many such primes p1,…,pk≡1(mod8). Then consider the following number
(2p1⋯pk)4+1,
which is coprime with each pi, and has remainder 1 modulo 8. Since it is odd and greater than 1, it has to be divisible by an odd prime p. Then
ordp(2p1⋯pk)=8
which divides φ(p)=p−1 by Fermat's theorem. Therefore p is another prime ≡1(mod8).
answered Jan 11 at 16:52
Alvin R. FergusonAlvin R. Ferguson
111
$endgroup$
$begingroup$
How do you get $pequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:03
$begingroup$
This is one of the sub questions to my original problem. Not sure how he got that???
$endgroup$
– Mayur Chauhan
Jan 11 at 17:06
add a comment |
$begingroup$
Someone has already answered the question
Suppose that there are only finitely many such primes p1,…,pk≡1(mod8). Then consider the following number
(2p1⋯pk)4+1,
which is coprime with each pi, and has remainder 1 modulo 8. Since it is odd and greater than 1, it has to be divisible by an odd prime p. Then
ordp(2p1⋯pk)=8
which divides φ(p)=p−1 by Fermat's theorem. Therefore p is another prime ≡1(mod8).
answered Jan 11 at 16:52
Alvin R. FergusonAlvin R. Ferguson
111
$endgroup$
$begingroup$
How do you get $pequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:03
$begingroup$
This is one of the sub questions to my original problem. Not sure how he got that???
$endgroup$
– Mayur Chauhan
Jan 11 at 17:06
add a comment |
$begingroup$
Someone has already answered the question
Suppose that there are only finitely many such primes p1,…,pk≡1(mod8). Then consider the following number
(2p1⋯pk)4+1,
which is coprime with each pi, and has remainder 1 modulo 8. Since it is odd and greater than 1, it has to be divisible by an odd prime p. Then
ordp(2p1⋯pk)=8
which divides φ(p)=p−1 by Fermat's theorem. Therefore p is another prime ≡1(mod8).
answered Jan 11 at 16:52
Alvin R. FergusonAlvin R. Ferguson
111
$endgroup$
Someone has already answered the question
Suppose that there are only finitely many such primes p1,…,pk≡1(mod8). Then consider the following number
(2p1⋯pk)4+1,
which is coprime with each pi, and has remainder 1 modulo 8. Since it is odd and greater than 1, it has to be divisible by an odd prime p. Then
ordp(2p1⋯pk)=8
which divides φ(p)=p−1 by Fermat's theorem. Therefore p is another prime ≡1(mod8).
answered Jan 11 at 16:52
Alvin R. FergusonAlvin R. Ferguson
111
answered Jan 11 at 16:52
Alvin R. FergusonAlvin R. Ferguson
111
answered Jan 11 at 16:52
Alvin R. FergusonAlvin R. Ferguson
111
answered Jan 11 at 16:52
Alvin R. FergusonAlvin R. Ferguson
111
111
$begingroup$
How do you get $pequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:03
$begingroup$
This is one of the sub questions to my original problem. Not sure how he got that???
$endgroup$
– Mayur Chauhan
Jan 11 at 17:06
add a comment |
$begingroup$
How do you get $pequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:03
$begingroup$
This is one of the sub questions to my original problem. Not sure how he got that???
$endgroup$
– Mayur Chauhan
Jan 11 at 17:06
$begingroup$
How do you get $pequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:03
$begingroup$
How do you get $pequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:03
$begingroup$
This is one of the sub questions to my original problem. Not sure how he got that???
$endgroup$
– Mayur Chauhan
Jan 11 at 17:06
$begingroup$
This is one of the sub questions to my original problem. Not sure how he got that???
$endgroup$
– Mayur Chauhan
Jan 11 at 17:06
add a comment |
$begingroup$
First of all, note that all odd and prime divisors of n4+1 have the form 8k+1. Suppose that there are just finite prime numbers of form 8k+1. Construct the number a=(2p1…pn)4+1, where {pi}ni=1are all primes of form 8k+1. This number is odd, and has at least one prime divisor q. Then q has the form 8l+1. But then q=pj for some j. But it is not possible since q divides a and q divides (2p1…pn)4, i.e, q divides 1.
answered Jan 11 at 17:00
Teri J. HaaseTeri J. Haase
111
$endgroup$
1
$begingroup$
How do you know that $qequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:04
add a comment |
$begingroup$
First of all, note that all odd and prime divisors of n4+1 have the form 8k+1. Suppose that there are just finite prime numbers of form 8k+1. Construct the number a=(2p1…pn)4+1, where {pi}ni=1are all primes of form 8k+1. This number is odd, and has at least one prime divisor q. Then q has the form 8l+1. But then q=pj for some j. But it is not possible since q divides a and q divides (2p1…pn)4, i.e, q divides 1.
answered Jan 11 at 17:00
Teri J. HaaseTeri J. Haase
111
$endgroup$
1
$begingroup$
How do you know that $qequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:04
add a comment |
$begingroup$
First of all, note that all odd and prime divisors of n4+1 have the form 8k+1. Suppose that there are just finite prime numbers of form 8k+1. Construct the number a=(2p1…pn)4+1, where {pi}ni=1are all primes of form 8k+1. This number is odd, and has at least one prime divisor q. Then q has the form 8l+1. But then q=pj for some j. But it is not possible since q divides a and q divides (2p1…pn)4, i.e, q divides 1.
answered Jan 11 at 17:00
Teri J. HaaseTeri J. Haase
111
$endgroup$
First of all, note that all odd and prime divisors of n4+1 have the form 8k+1. Suppose that there are just finite prime numbers of form 8k+1. Construct the number a=(2p1…pn)4+1, where {pi}ni=1are all primes of form 8k+1. This number is odd, and has at least one prime divisor q. Then q has the form 8l+1. But then q=pj for some j. But it is not possible since q divides a and q divides (2p1…pn)4, i.e, q divides 1.
answered Jan 11 at 17:00
Teri J. HaaseTeri J. Haase
111
answered Jan 11 at 17:00
Teri J. HaaseTeri J. Haase
111
answered Jan 11 at 17:00
Teri J. HaaseTeri J. Haase
111
answered Jan 11 at 17:00
Teri J. HaaseTeri J. Haase
111
111
1
$begingroup$
How do you know that $qequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:04
add a comment |
1
$begingroup$
How do you know that $qequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:04
1
1
$begingroup$
How do you know that $qequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:04
$begingroup$
How do you know that $qequiv1pmod 8$?
$endgroup$
– Lord Shark the Unknown
Jan 11 at 17:04
add a comment |
$begingroup$
My proof goes like this:
Assume by way of contradiction that there are only finitely many of such prime numbers, say p1,p2,…,pr. Consider p=16p41p42⋯p4r+1=(2p1p2⋯pr)4+1. Recall that the congruence x4≡−1modp is solvable if and only if p≡1mod8. p cannot be a prime of the form 8n+1, but p≡1mod8, contradiction. Thus, there must be infinitely many prime numbers of the form 8n+1.
answered Jan 11 at 17:54
Philip C. McMillianPhilip C. McMillian
111
$endgroup$
$begingroup$
Good answer, I saw this just after I posted mine
$endgroup$
– Mike
Jan 11 at 18:09
add a comment |
$begingroup$
My proof goes like this:
Assume by way of contradiction that there are only finitely many of such prime numbers, say p1,p2,…,pr. Consider p=16p41p42⋯p4r+1=(2p1p2⋯pr)4+1. Recall that the congruence x4≡−1modp is solvable if and only if p≡1mod8. p cannot be a prime of the form 8n+1, but p≡1mod8, contradiction. Thus, there must be infinitely many prime numbers of the form 8n+1.
answered Jan 11 at 17:54
Philip C. McMillianPhilip C. McMillian
111
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Good answer, I saw this just after I posted mine
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– Mike
Jan 11 at 18:09
add a comment |
$begingroup$
My proof goes like this:
Assume by way of contradiction that there are only finitely many of such prime numbers, say p1,p2,…,pr. Consider p=16p41p42⋯p4r+1=(2p1p2⋯pr)4+1. Recall that the congruence x4≡−1modp is solvable if and only if p≡1mod8. p cannot be a prime of the form 8n+1, but p≡1mod8, contradiction. Thus, there must be infinitely many prime numbers of the form 8n+1.
answered Jan 11 at 17:54
Philip C. McMillianPhilip C. McMillian
111
$endgroup$
My proof goes like this:
Assume by way of contradiction that there are only finitely many of such prime numbers, say p1,p2,…,pr. Consider p=16p41p42⋯p4r+1=(2p1p2⋯pr)4+1. Recall that the congruence x4≡−1modp is solvable if and only if p≡1mod8. p cannot be a prime of the form 8n+1, but p≡1mod8, contradiction. Thus, there must be infinitely many prime numbers of the form 8n+1.
answered Jan 11 at 17:54
Philip C. McMillianPhilip C. McMillian
111
answered Jan 11 at 17:54
Philip C. McMillianPhilip C. McMillian
111
answered Jan 11 at 17:54
Philip C. McMillianPhilip C. McMillian
111
answered Jan 11 at 17:54
Philip C. McMillianPhilip C. McMillian
111
111
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Good answer, I saw this just after I posted mine
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– Mike
Jan 11 at 18:09
add a comment |
$begingroup$
Good answer, I saw this just after I posted mine
$endgroup$
– Mike
Jan 11 at 18:09
$begingroup$
Good answer, I saw this just after I posted mine
$endgroup$
– Mike
Jan 11 at 18:09
$begingroup$
Good answer, I saw this just after I posted mine
$endgroup$
– Mike
Jan 11 at 18:09
add a comment |
$begingroup$
Maybe this will help fill in the details. Let us write $x = 2p_1ldots p_r$, for some finite $r$, where $p_1 ldots p_r$ are in Teri's answer, all the primes that are 1 mod 8. Then as Teri already noted there is a prime $p$ that divides $x^4+1$, and $p not in {p_1,ldots, p_r}$. We claim that $p$ is of the form $8k+1$ for some integer $k$ next. As already noted by Teri, this will imply that $p_1,ldots, p_r$ are not all the primes that are 1 mod 8, which will give you what you want.
Note that
$$x^4+1 equiv_p 0 Rightarrow x^4 equiv_p -1 Rightarrow x^8 equiv_p 1;$$
That $x^4 not equiv_p 1$ and $x^8 equiv_p 1$, together imply that $x$ has order precisely 8 in the group $left(mathbb{F}_pright)^{times}$, which implies that $|left(mathbb{F}_pright)^{times}| = p-1$ is divisible by 8, which implies that $p$ is of the form $8k+1$ for some integer $k$.
answered Jan 11 at 18:07
MikeMike
3,871412
$endgroup$
$begingroup$
So how would I be able to use this to go through each exercise in the original question?
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– Mayur Chauhan
Jan 11 at 18:15
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I'm not sure. This will establish the main result though, which I would imagine should be fine with your instructor
$endgroup$
– Mike
Jan 11 at 18:16
add a comment |
$begingroup$
Maybe this will help fill in the details. Let us write $x = 2p_1ldots p_r$, for some finite $r$, where $p_1 ldots p_r$ are in Teri's answer, all the primes that are 1 mod 8. Then as Teri already noted there is a prime $p$ that divides $x^4+1$, and $p not in {p_1,ldots, p_r}$. We claim that $p$ is of the form $8k+1$ for some integer $k$ next. As already noted by Teri, this will imply that $p_1,ldots, p_r$ are not all the primes that are 1 mod 8, which will give you what you want.
Note that
$$x^4+1 equiv_p 0 Rightarrow x^4 equiv_p -1 Rightarrow x^8 equiv_p 1;$$
That $x^4 not equiv_p 1$ and $x^8 equiv_p 1$, together imply that $x$ has order precisely 8 in the group $left(mathbb{F}_pright)^{times}$, which implies that $|left(mathbb{F}_pright)^{times}| = p-1$ is divisible by 8, which implies that $p$ is of the form $8k+1$ for some integer $k$.
answered Jan 11 at 18:07
MikeMike
3,871412
$endgroup$
$begingroup$
So how would I be able to use this to go through each exercise in the original question?
$endgroup$
– Mayur Chauhan
Jan 11 at 18:15
$begingroup$
I'm not sure. This will establish the main result though, which I would imagine should be fine with your instructor
$endgroup$
– Mike
Jan 11 at 18:16
add a comment |
$begingroup$
Maybe this will help fill in the details. Let us write $x = 2p_1ldots p_r$, for some finite $r$, where $p_1 ldots p_r$ are in Teri's answer, all the primes that are 1 mod 8. Then as Teri already noted there is a prime $p$ that divides $x^4+1$, and $p not in {p_1,ldots, p_r}$. We claim that $p$ is of the form $8k+1$ for some integer $k$ next. As already noted by Teri, this will imply that $p_1,ldots, p_r$ are not all the primes that are 1 mod 8, which will give you what you want.
Note that
$$x^4+1 equiv_p 0 Rightarrow x^4 equiv_p -1 Rightarrow x^8 equiv_p 1;$$
That $x^4 not equiv_p 1$ and $x^8 equiv_p 1$, together imply that $x$ has order precisely 8 in the group $left(mathbb{F}_pright)^{times}$, which implies that $|left(mathbb{F}_pright)^{times}| = p-1$ is divisible by 8, which implies that $p$ is of the form $8k+1$ for some integer $k$.
answered Jan 11 at 18:07
MikeMike
3,871412
$endgroup$
Maybe this will help fill in the details. Let us write $x = 2p_1ldots p_r$, for some finite $r$, where $p_1 ldots p_r$ are in Teri's answer, all the primes that are 1 mod 8. Then as Teri already noted there is a prime $p$ that divides $x^4+1$, and $p not in {p_1,ldots, p_r}$. We claim that $p$ is of the form $8k+1$ for some integer $k$ next. As already noted by Teri, this will imply that $p_1,ldots, p_r$ are not all the primes that are 1 mod 8, which will give you what you want.
Note that
$$x^4+1 equiv_p 0 Rightarrow x^4 equiv_p -1 Rightarrow x^8 equiv_p 1;$$
That $x^4 not equiv_p 1$ and $x^8 equiv_p 1$, together imply that $x$ has order precisely 8 in the group $left(mathbb{F}_pright)^{times}$, which implies that $|left(mathbb{F}_pright)^{times}| = p-1$ is divisible by 8, which implies that $p$ is of the form $8k+1$ for some integer $k$.
answered Jan 11 at 18:07
MikeMike
3,871412
edited Jan 11 at 18:14
answered Jan 11 at 18:07
MikeMike
3,871412
answered Jan 11 at 18:07
MikeMike
3,871412
answered Jan 11 at 18:07
MikeMike
3,871412
3,871412
$begingroup$
So how would I be able to use this to go through each exercise in the original question?
$endgroup$
– Mayur Chauhan
Jan 11 at 18:15
$begingroup$
I'm not sure. This will establish the main result though, which I would imagine should be fine with your instructor
$endgroup$
– Mike
Jan 11 at 18:16
add a comment |
$begingroup$
So how would I be able to use this to go through each exercise in the original question?
$endgroup$
– Mayur Chauhan
Jan 11 at 18:15
$begingroup$
I'm not sure. This will establish the main result though, which I would imagine should be fine with your instructor
$endgroup$
– Mike
Jan 11 at 18:16
$begingroup$
So how would I be able to use this to go through each exercise in the original question?
$endgroup$
– Mayur Chauhan
Jan 11 at 18:15
$begingroup$
So how would I be able to use this to go through each exercise in the original question?
$endgroup$
– Mayur Chauhan
Jan 11 at 18:15
$begingroup$
I'm not sure. This will establish the main result though, which I would imagine should be fine with your instructor
$endgroup$
– Mike
Jan 11 at 18:16
$begingroup$
I'm not sure. This will establish the main result though, which I would imagine should be fine with your instructor
$endgroup$
– Mike
Jan 11 at 18:16
add a comment |
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$begingroup$
that a wonderful exercise. Have you tried anything or are you just asking us to do to everything for you? (in that case i m sorry to say it won't happen)
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– Alexis
Jan 11 at 16:18
1
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please type this in the question's body. It would help people to understand where you are stuck.
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– Alexis
Jan 11 at 16:23
1
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Edit your question to show your attempts, rather than just posting a picture. Also, write down the question rather than pointing to a picture as well. Then, state Dirichlet's theorem yourself to begin with.
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– Mefitico
Jan 11 at 16:23
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You might want to look at the statement of the referenced theorem.
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– yberman
Jan 11 at 16:35
1
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What do you mean by "Prove using Dirichlet's Theorem"?
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– Barry Cipra
Jan 11 at 17:05