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I am a complete beginner and I am studying the relation between mersenne and wieferich primes.

Wikipedia says

A prime divisor p of Mq, where q is prime, is a Wieferich prime if and only if $p^2$ divides Mq Thus, a Mersenne prime cannot also be a Wieferich prime.

why? I understand that any two mersenne primes are always coprimes, so a mersenne prime cannot divide another mersenne prime. But I have trouble making the logical leap from that statement to the above one.

Also, the proof on mersenne primes page here, I do not understand how

$1 + 2^m + 2^{2m} + ... + 2^{(λ − 1)m} ≡ −λ \mod (2^m − 1)$

I apologize if I this has been asked before. But everywhere I search, the statement is given without proof.

Thank you very much :")

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3 Answers 3

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Incomplete answer. This statement $$1+2^m+2^{2m}+...+2^{(\lambda-1)m} \equiv \lambda \pmod{2^m-1}$$ is easy deductible from $$1 \equiv 1 \pmod{2^m-1} $$ $$2^m\equiv 1 \pmod{2^m-1} \tag{1}$$ by raising $(1)$ at the power of $2$ $$(2^m)^2 \equiv 1^2 \pmod{2^m-1} \Rightarrow 2^{2m}\equiv 1 \pmod{2^m-1}$$ by raising $(1)$ at the power of $3$ $$(2^{m})^3 \equiv 1^3 \pmod{2^m-1} \Rightarrow 2^{3m}\equiv 1 \pmod{2^m-1}$$ $$...$$ by raising $(1)$ at the power of $\lambda -1$ $$(2^{m})^{\lambda -1} \equiv 1^{\lambda -1} \pmod{2^m-1} \Rightarrow 2^{(\lambda -1)m}\equiv 1 \pmod{2^m-1}$$ Now, we add altogether $$1+2^m+2^{2m}+...+2^{(\lambda-1)m} \equiv 1+1+1+...+1 \equiv \lambda \pmod{2^m-1}$$

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For any prime numbers p and q, p is a prime divisor of Mq when and only when 2^q is congruent to 1 modulo p. Since q is prime and $2^{p - 1}$ is congruent to 1 modulo p, q is a factor of p - 1. p is a Weiferich prime is and only if $2^{p - 1}$ is congruent to 1 modulo $p^2$. This also happens only when $2^q$ is congruent to 1 modulo $p^2$.

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rtybase has already answered the second half of the question. The first half is quite simple:

If $M_q$ is a Mersenne prime, then it is itself a prime divisor of $M_q$. Therefore (applying the first sentence quoted with $p = M_q$) it is a Wieferich prime iff $M_q^2$ divides $M_q$, which is clearly impossible.

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