Re: Prime numbers problem

From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 12/13/04 

Date: Mon, 13 Dec 2004 17:54:10 +0000 (UTC)

In article <cpkj9s$d5a$1@news2.ipartners.pl>,
Waldek <waldek69@nospam.hotmail.com> wrote:
>
>"When I divide any prime number by 30, the remainder is always prime
>number."
>
>Can anybody help me to prove this?

Doubtful, since 31 is a prime but 1 is not.

Now, let us assume instead that you want to show that if p is a prime,
then the remainder of dividing p by 30 is either 1 or a prime.

Well, this is obviously true if p<30. For p>30, p must be odd, so it
must be congruent to an odd number modulo 30. In addition, it cannot
be congruent to 3 (because 3+ 30k is divisible by 3), 5 (5+30k is a
multiple of 5), 9 (9 + 30k is divisible by 3), 15 (divisible by 5), 21
(divisible by 3), 25 (divisible by 5), or 27 (divisible by 3).

That leaves as the only possibilities 1, 7, 11, 13, 17, 19, 23, and
29, which are all either prime or equal to 1.

Note that in addition, there are infinitely many primes of the form
30k+1, so 1 occurs infinitely often. After 31, it occurs at 61, 151,
181, 211, etc.

>Is there any helpful theorem for this issue?

The observation that 30 is divisible by 3 and 5, is about it.

>Can anybody point me to some links on Web with theorems for prime number
>theory?

You can start at the Prime Pages:

http://www.utm.edu/research/primes/ 

-- 
======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu

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