User:Sigmundur/wip/Mersenne primes

Theorems about Mersenne numbers

Theorem: If a and p are natural numbers such that a^p − 1 is prime, then a = 2 or p = 1.

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Proof: a ≡ 1 (mod a − 1). Then a^p ≡ 1 (mod a − 1), so a^p − 1 ≡ 0 (mod a − 1). Thus a − 1 | a^p − 1. However, a^p − 1 is prime, so a − 1 = a^p − 1 or a − 1 = ±1. In the former case, a = a^p, hence a = 0,1 (which is a contradiction, as neither 1 nor 0 is prime) or p = 1. In the latter case, a = 2 or a = 0. If a = 0, however, 0^p − 1 = 0 − 1 = −1 which is not prime. Therefore, a = 2.

Theorem: If 2^p − 1 is prime, then p is prime.

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Proof: suppose that p is composite, hence can be written p = a⋅b with a and b > 1. Then (2^a)^b − 1 is prime, but b > 1 and 2^a > 2, contradicting statement 1.

Theorem: If p is an odd prime, then any prime q that divides 2^p − 1 must be 1 plus a multiple of 2p. This holds even when 2^p − 1 is prime.

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Theorem: If p is an odd prime, then any prime q that divides $2^{p}-1$ must be congruent to $\pm 1{\pmod {8}}$.

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Proof: $2^{p+1}=2{\pmod {q}}$, so $2^{(p+1)/2}$ is a square root of 2 modulo $q$. By quadratic reciprocity, any prime modulo which 2 has a square root is congruent to $\pm 1{\pmod {8}}$.

Theorem: A Mersenne prime cannot be a Wieferich prime.

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Proof: We show if $p=2^{m}-1$ is a Mersenne prime, then the congruence $2^{p-1}\equiv 1{\pmod {p^{2}}}$ does not satisfy. By Fermat's Little theorem, $m|p-1$. Now write, $p-1=m\lambda$. If the given congruence satisfies, then $p^{2}|2^{m\lambda }-1$,therefore $0\equiv (2^{m\lambda }-1)/(2^{m}-1)=1+2^{m}+2^{2m}+...+2^{(\lambda -1)m}\equiv -\lambda {\pmod {2^{m}-1}}$. Hence $2^{m}-1|\lambda$,and therefore λ ≥ 2^m − 1. This leads to p − 1  ≥ m(2^m − 1), which is impossible since m ≥ 2.

Theorem: A prime number divides at most one prime-exponent Mersenne number