Prime ProofsDate: 10/08/2002 at 19:19:23 From: Julie Subject: If a^(n) - 1 is prime, show that a=2 and that n is prime. Please help. If a^(n) - 1 is prime, show that a=2 and that n is a prime. If a^(n) + 1 is a prime, show that a is even and that n is a power of 2. I'm sure these questions are very similar, but I can't get my mind around them. Date: 10/10/2002 at 20:32:27 From: Doctor Paul Subject: Re: If a^(n) - 1 is prime, show that a=2 and that n is prime. >(a) if a^n -1 is prime, show that a=2 and that n is a prime. a^n - 1 can always be factored as: (a - 1) * (a^(n-1) + a^(n-2) + ... + a + 1) If a^n - 1 is prime, it must be the case that one of the factors above is equal to one. Can you argue that the second factor cannot be equal to one? If so, then you know that a - 1 = 1, which implies that a = 2. So to show that n has to be prime, suppose that n is not prime and then exhibit a nontrivial factorization of a^n - 1. This nontrivial factorization contradicts the supposed primality of a^n - 1 and so the conclusion we must make is that our assumption that n is not prime must in fact be false. There's more here: If 2^n-1 is prime, then so is n. - Prime Pages Proofs http://www.utm.edu/research/primes/notes/proofs/Theorem2.html >(b) if a^n + 1 is prime, show that a is even and that n is a power > of 2. If n is not a power of two, then it has an odd prime factor m. So we can write n = m*r where we know for sure that m is odd. Since m is odd, we can write m = 2*k + 1 for some integer k. Thus n = 2*k*r + r Now exhibit a nontrivial factorization of a^n + 1: a^n + 1 = a^(2*k*r + r) + 1 = (a^r + 1) * (a^(2*k*r) - a^(2*k*r - r) + a^(2*k*r - 2*r) - ... + a^(2*r) - a^r + 1) To see that a has to be even note that if a were odd, then we could write: a = 1 mod 2. We know that n has to be a power of two, so n is even. So we can write n = 2*q for some integer q. Then a^n = a^(2*q) = 1^(2*q) = 1 mod 2. So if a is odd and n is even, a^n will always be odd. Then a^n + 1 will always be even and hence cannot be prime. So if a^n + 1 is to be prime, we cannot have a odd. I hope this helps. Please write back if you'd like to talk about this some more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/