1
$\begingroup$

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 :")

$\endgroup$
0
$\begingroup$

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 $$1+2^m+2^{2m}+...+2^{(\lambda-1)m} \equiv 1+1+1+...+1 \equiv \lambda \pmod{2^m-1}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.