Re: prime~.
From: Michael Lockhart (ml1000_at_bellsouth.net)
Date: 10/04/04
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Date: Sun, 3 Oct 2004 22:01:44 -0400
"mina_world" <mina_world@hanmail.net> wrote in message
news:cjq9ds$a7o$1@news.hananet.net...
> hello......doctor~
>
> p is a prime.
> n is positive integer.
>
> show that
> p | (a^n) - 1 for some a in Z
> => (p^2) | (a^n) - 1
That's not true.
For example, let n = 1, and let a = p + 1. Then p | (p + 1) - 1 = p, but
p^2 does not divide p. That shows it isn't true for n = 1.
Now let n = 2, a = 2. Then 3 | (2^2 - 1) = 3, but 9 doesn't divide 3.
Now let n = 2, a = 5. Then we have p | 24--2 and 3 work, but 9 doesn't
divide 24.
So n = 2 doesn't work either.
Shall I try one more? Let n = 3, a = 2. Then we have p | 7, so p = 7, but
49 doesn't divide 7. Nope.
Let n = 3, a = 3. Then we have p | 26, so p is either 13 or 2, but neither
one's square divides 26. Nope.
Perhaps you wrote it down wrong?
> example)
> if p=3,
> 3 | (4^3)-1 (namely, 3 | 63)
> => (3^2) | 63
> ---------------------------------------------
> i need your advice to solve it.
>
> thank you very much for your advice.
Sure.
Michael
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