prove that there infinitely many primes of the form $8k-1$
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Using the fact that $$left ( frac2p right )=(-1)^fracp^2-18$$
for each prime $p>2$,prove that there infinitely many primes of the form $8k-1$.
I thought that we could I assume that there is a finite number of primes of the form $8k-1$: $p_1,p_2 dots ,p_k$
Could we maybe set $N=8p_1p_2 cdots p_k-1 >1$
Then $N$ has a prime divisor $p$.$p$ can be of the form $8n+1,8n+3,8n+5 text or 8n+7$..
How could I continue?? Also...how can I use this: $left ( frac2p right )=(-1)^fracp^2-18$ ?
number-theory
asked Jun 22 '14 at 22:30
evinda
4,04031643
add a comment |Â
up vote
2
down vote
favorite
Using the fact that $$left ( frac2p right )=(-1)^fracp^2-18$$
for each prime $p>2$,prove that there infinitely many primes of the form $8k-1$.
I thought that we could I assume that there is a finite number of primes of the form $8k-1$: $p_1,p_2 dots ,p_k$
Could we maybe set $N=8p_1p_2 cdots p_k-1 >1$
Then $N$ has a prime divisor $p$.$p$ can be of the form $8n+1,8n+3,8n+5 text or 8n+7$..
How could I continue?? Also...how can I use this: $left ( frac2p right )=(-1)^fracp^2-18$ ?
number-theory
asked Jun 22 '14 at 22:30
evinda
4,04031643
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Using the fact that $$left ( frac2p right )=(-1)^fracp^2-18$$
for each prime $p>2$,prove that there infinitely many primes of the form $8k-1$.
I thought that we could I assume that there is a finite number of primes of the form $8k-1$: $p_1,p_2 dots ,p_k$
Could we maybe set $N=8p_1p_2 cdots p_k-1 >1$
Then $N$ has a prime divisor $p$.$p$ can be of the form $8n+1,8n+3,8n+5 text or 8n+7$..
How could I continue?? Also...how can I use this: $left ( frac2p right )=(-1)^fracp^2-18$ ?
number-theory
asked Jun 22 '14 at 22:30
evinda
4,04031643
Using the fact that $$left ( frac2p right )=(-1)^fracp^2-18$$
for each prime $p>2$,prove that there infinitely many primes of the form $8k-1$.
I thought that we could I assume that there is a finite number of primes of the form $8k-1$: $p_1,p_2 dots ,p_k$
Could we maybe set $N=8p_1p_2 cdots p_k-1 >1$
Then $N$ has a prime divisor $p$.$p$ can be of the form $8n+1,8n+3,8n+5 text or 8n+7$..
How could I continue?? Also...how can I use this: $left ( frac2p right )=(-1)^fracp^2-18$ ?
number-theory
asked Jun 22 '14 at 22:30
evinda
4,04031643
asked Jun 22 '14 at 22:30
evinda
4,04031643
asked Jun 22 '14 at 22:30
evinda
4,04031643
asked Jun 22 '14 at 22:30
evinda
4,04031643
4,04031643
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
7
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Let $p_1,p_2, ldots, p_k$ be the list of ALL primes of the form $8s+7$. Let
$$N=(p_1p_2 dotsb p_k)^2-2.$$
Note that $N equiv 7 pmod8$ and is odd. If $p$ is a prime that divides $N$, then
$$(p_1p_2 dotsb p_k)^2 equiv 2 pmodp.$$ Thus
$$left(frac2pright)=1.$$
Thus $p equiv pm 1 pmod8$.
So all primes that divide $N$ must be of the from $8s+1$ or $8s+7$. But not all of them can be of the form $8s+1$ (ask why???)
So there must be one of the form $q=8s+7$. Now see if you can proceed from here.
answered Jun 22 '14 at 22:40
Anurag A
22.4k12243
Anurag A: Why $N equiv 7 pmod 8$? We know that $p_1, dots , p_k equiv 7 pmod 8$,but why also the difference $(p_1 cdots p_k)^2-2 equiv 7 pmod 8$ ?? $$$$ If $N$ has a prime divisor $p$ of the form $8s-1$,$p$ must be of of $p_1, dots, p_k$,so $p mid p_1 cdot p_2 cdots p_k Rightarrow p mid (p_1 cdots p_k)^2 text and as p mid N Rightarrow p mid 2 text that is a contradiction.$
– evinda
Jun 22 '14 at 23:06
I understood it now...thank you very much!!!!
– evinda
Jun 22 '14 at 23:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Let $p_1,p_2, ldots, p_k$ be the list of ALL primes of the form $8s+7$. Let
$$N=(p_1p_2 dotsb p_k)^2-2.$$
Note that $N equiv 7 pmod8$ and is odd. If $p$ is a prime that divides $N$, then
$$(p_1p_2 dotsb p_k)^2 equiv 2 pmodp.$$ Thus
$$left(frac2pright)=1.$$
Thus $p equiv pm 1 pmod8$.
So all primes that divide $N$ must be of the from $8s+1$ or $8s+7$. But not all of them can be of the form $8s+1$ (ask why???)
So there must be one of the form $q=8s+7$. Now see if you can proceed from here.
answered Jun 22 '14 at 22:40
Anurag A
22.4k12243
Anurag A: Why $N equiv 7 pmod 8$? We know that $p_1, dots , p_k equiv 7 pmod 8$,but why also the difference $(p_1 cdots p_k)^2-2 equiv 7 pmod 8$ ?? $$$$ If $N$ has a prime divisor $p$ of the form $8s-1$,$p$ must be of of $p_1, dots, p_k$,so $p mid p_1 cdot p_2 cdots p_k Rightarrow p mid (p_1 cdots p_k)^2 text and as p mid N Rightarrow p mid 2 text that is a contradiction.$
– evinda
Jun 22 '14 at 23:06
I understood it now...thank you very much!!!!
– evinda
Jun 22 '14 at 23:27
add a comment |Â
up vote
7
down vote
accepted
Let $p_1,p_2, ldots, p_k$ be the list of ALL primes of the form $8s+7$. Let
$$N=(p_1p_2 dotsb p_k)^2-2.$$
Note that $N equiv 7 pmod8$ and is odd. If $p$ is a prime that divides $N$, then
$$(p_1p_2 dotsb p_k)^2 equiv 2 pmodp.$$ Thus
$$left(frac2pright)=1.$$
Thus $p equiv pm 1 pmod8$.
So all primes that divide $N$ must be of the from $8s+1$ or $8s+7$. But not all of them can be of the form $8s+1$ (ask why???)
So there must be one of the form $q=8s+7$. Now see if you can proceed from here.
answered Jun 22 '14 at 22:40
Anurag A
22.4k12243
Anurag A: Why $N equiv 7 pmod 8$? We know that $p_1, dots , p_k equiv 7 pmod 8$,but why also the difference $(p_1 cdots p_k)^2-2 equiv 7 pmod 8$ ?? $$$$ If $N$ has a prime divisor $p$ of the form $8s-1$,$p$ must be of of $p_1, dots, p_k$,so $p mid p_1 cdot p_2 cdots p_k Rightarrow p mid (p_1 cdots p_k)^2 text and as p mid N Rightarrow p mid 2 text that is a contradiction.$
– evinda
Jun 22 '14 at 23:06
I understood it now...thank you very much!!!!
– evinda
Jun 22 '14 at 23:27
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Let $p_1,p_2, ldots, p_k$ be the list of ALL primes of the form $8s+7$. Let
$$N=(p_1p_2 dotsb p_k)^2-2.$$
Note that $N equiv 7 pmod8$ and is odd. If $p$ is a prime that divides $N$, then
$$(p_1p_2 dotsb p_k)^2 equiv 2 pmodp.$$ Thus
$$left(frac2pright)=1.$$
Thus $p equiv pm 1 pmod8$.
So all primes that divide $N$ must be of the from $8s+1$ or $8s+7$. But not all of them can be of the form $8s+1$ (ask why???)
So there must be one of the form $q=8s+7$. Now see if you can proceed from here.
answered Jun 22 '14 at 22:40
Anurag A
22.4k12243
Let $p_1,p_2, ldots, p_k$ be the list of ALL primes of the form $8s+7$. Let
$$N=(p_1p_2 dotsb p_k)^2-2.$$
Note that $N equiv 7 pmod8$ and is odd. If $p$ is a prime that divides $N$, then
$$(p_1p_2 dotsb p_k)^2 equiv 2 pmodp.$$ Thus
$$left(frac2pright)=1.$$
Thus $p equiv pm 1 pmod8$.
So all primes that divide $N$ must be of the from $8s+1$ or $8s+7$. But not all of them can be of the form $8s+1$ (ask why???)
So there must be one of the form $q=8s+7$. Now see if you can proceed from here.
answered Jun 22 '14 at 22:40
Anurag A
22.4k12243
answered Jun 22 '14 at 22:40
Anurag A
22.4k12243
answered Jun 22 '14 at 22:40
Anurag A
22.4k12243
answered Jun 22 '14 at 22:40
Anurag A
22.4k12243
22.4k12243
Anurag A: Why $N equiv 7 pmod 8$? We know that $p_1, dots , p_k equiv 7 pmod 8$,but why also the difference $(p_1 cdots p_k)^2-2 equiv 7 pmod 8$ ?? $$$$ If $N$ has a prime divisor $p$ of the form $8s-1$,$p$ must be of of $p_1, dots, p_k$,so $p mid p_1 cdot p_2 cdots p_k Rightarrow p mid (p_1 cdots p_k)^2 text and as p mid N Rightarrow p mid 2 text that is a contradiction.$
– evinda
Jun 22 '14 at 23:06
I understood it now...thank you very much!!!!
– evinda
Jun 22 '14 at 23:27
add a comment |Â
Anurag A: Why $N equiv 7 pmod 8$? We know that $p_1, dots , p_k equiv 7 pmod 8$,but why also the difference $(p_1 cdots p_k)^2-2 equiv 7 pmod 8$ ?? $$$$ If $N$ has a prime divisor $p$ of the form $8s-1$,$p$ must be of of $p_1, dots, p_k$,so $p mid p_1 cdot p_2 cdots p_k Rightarrow p mid (p_1 cdots p_k)^2 text and as p mid N Rightarrow p mid 2 text that is a contradiction.$
– evinda
Jun 22 '14 at 23:06
I understood it now...thank you very much!!!!
– evinda
Jun 22 '14 at 23:27
Anurag A: Why $N equiv 7 pmod 8$? We know that $p_1, dots , p_k equiv 7 pmod 8$,but why also the difference $(p_1 cdots p_k)^2-2 equiv 7 pmod 8$ ?? $$$$ If $N$ has a prime divisor $p$ of the form $8s-1$,$p$ must be of of $p_1, dots, p_k$,so $p mid p_1 cdot p_2 cdots p_k Rightarrow p mid (p_1 cdots p_k)^2 text and as p mid N Rightarrow p mid 2 text that is a contradiction.$
– evinda
Jun 22 '14 at 23:06
Anurag A: Why $N equiv 7 pmod 8$? We know that $p_1, dots , p_k equiv 7 pmod 8$,but why also the difference $(p_1 cdots p_k)^2-2 equiv 7 pmod 8$ ?? $$$$ If $N$ has a prime divisor $p$ of the form $8s-1$,$p$ must be of of $p_1, dots, p_k$,so $p mid p_1 cdot p_2 cdots p_k Rightarrow p mid (p_1 cdots p_k)^2 text and as p mid N Rightarrow p mid 2 text that is a contradiction.$
– evinda
Jun 22 '14 at 23:06
I understood it now...thank you very much!!!!
– evinda
Jun 22 '14 at 23:27
I understood it now...thank you very much!!!!
– evinda
Jun 22 '14 at 23:27
add a comment |Â
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