From: Robin Chapman
Subject: Re: Primes of the form 8n+5
Date: Fri, 10 Dec 1999 12:27:22 GMT
Newsgroups: sci.math
Keywords: infinitude of primes in the congruence classes mod 8
In article
,
Amanda Pack wrote:
> how should i go about proving the infinitude of primes of the above form?
> i am currently studying quadratic reciprocity.
>
> would the case for primes of the form 8n+7 be similar.
I'm sure both of these are in Nagell's book on number theory, but
I don't have my copy to hand. Suppose we have an integer N which
is (a) the sum of two mutually coprime squares and (b) congruent
to 5 (mod 8). Then by (a) no prime factor of N is congruent to 3
mod 4, and as by (b) N is odd then each prime factor of N is
congruent to 1 mod 4. Thus each prime factor of N is congruent
to 1 or 5 modulo 8. But they can't all be congruent to 1 modulo 8
by (b). So N has a prime factor q congruent to 5 (mod 8).
Let p_1, ... , p_k be a bunch of odd primes. Then
(2 p_1 p_2 ... p_k)^2 + 1 satisfies (a) and (b). It has a prime factor
q = 5 (mod 8) and q can't equal any of the p_j.
One can modify this for the cases of primes congruent to 3 or
congruent to 7 (mod 8). There are also modulo 12 versions.
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"`Well, I'd already done a PhD in X-Files Theory at UCLA, ...'"
Greg Egan, _Teranesia_
Sent via Deja.com http://www.deja.com/
Before you buy.
==============================================================================
From: erick@sfu.ca (Erick Bryce Wong)
Subject: Re: Primes of the form 8n+5
Date: 10 Dec 1999 07:33:10 GMT
Newsgroups: sci.math
Amanda Pack wrote:
>how should i go about proving the infinitude of primes of the above form?
>i am currently studying quadratic reciprocity.
>
>would the case for primes of the form 8n+7 be similar.
In both cases, you typically start with a finite set of such primes, multiply
them together to form N, then show that some polynomial in N is divisible by
at least one prime of the desired form, which cannot be in the original set.
For 8n+5, you might try 4N^2+1. Can any prime factors of this number divide
into N? Can any prime factors be of the form 8k+7? What else can you say
about the prime factors of 4N^2+1? Can /all/ of them be of the form 8k+1?
(Note that N, being the product of odd primes, is necessarily odd.)
For 8n+7, try to find another function of N that lets you argue similarly.
(Hint: you should be able to use N^2 - c, provided you pick c carefully.)
-- Erick
==============================================================================
From: pete@bignode.southern.co.nz (Pete Moore)
Subject: Re: Primes of the form 8n+5
Date: 10 Dec 99 20:32:48 +1300
Newsgroups: sci.math
Amanda Pack (ap512@columbia.edu) wrote:
>how should i go about proving the infinitude of primes of the above form?
>i am currently studying quadratic reciprocity.
If p1,...,pn are primes of this form, consider N=(2*p1*...*pn)^2+1. Let p|N.
What does your knowledge of quadratic residues tell you about p?
>would the case for primes of the form 8n+7 be similar.
Very similar. In this case consider N=(p1*p2*...*pn)^2-2.
--
+---------------- pete@bignode.southern.co.nz ----------------+
The effort to understand the universe is one of the very few things
that lifts human life above the level of farce, and gives it some
of the grace of tragedy - Steven Weinberg