The goal of this short "footnote" is to prove the
following theorem used in the discussion of Mersenne
primes.
Theorem.
If for some positive integer n, 2n-1 is prime,
then so is n.
Proof.
Let r and s be positive integers, then the polynomial
xrs-1 is xs-1 times xs(r-1)
+ xs(r-2) + ... + xs
+ 1. So if n is composite (say r.s
with 1<s<n), then 2n-1 is also
composite (because it is divisible by 2s-1).
Notice that we can say more: suppose n>1. Since x-1 divides
xn-1, for the latter to be prime the former must be one.
This gives the following.
Corollary.
Let a and n be integers greater than one. If an-1
is prime, then a is 2 and n is prime.
Usually the first step in factoring numbers of the forms an-1
(where a and n are positive integers) is to factor the polynomial
xn-1. In this proof we just used the most basic
of such factorization rules, see [BLSTW88] for some
others.
