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### Prime Proofs

```Date: 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.

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/
```
Associated Topics:
College Number Theory
High School Number Theory

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