Vtyjkyuk

Here : https://en.wikipedia.org/wiki/Wieferich_prime

I came across the following claim :

A prime divisor $p$ of $M_q$, where $q$ is prime, is a Wieferich prime if and only if $p^2mid M_q$.

The part, I cannot prove : If $M=2^q-1$ , $q$ prime and $pmid M$ is a prime factor of $M$ satisfying $2^p-1equiv 1mod p^2$ (in other words, $p$ is a Wieferich-prime) , then we also have $p^2mid M$ or alternatively formulated : $2^qequiv 1mod p^2$

I only could conclude that $ord_2(p^2)mid p-1$ , but not $ord_2(p^2)=q$ as claimed. What do I miss ? How can I prove this part ?

Obviously, $q mid p-1$. Write $p-1=qcdot a$.

Then $p^2 mid 2^qcdot a-1=(2^q-1)(2^q(a-1)+2^q(a-2)+ldots +1)$.

Observe now that the second parentheses is $equiv apmodp$ since each summand is $equiv 1pmodp$.

Observe also that $a<p$.

This shows that $p nmid(2^q(a-1)+2^q(a-2)+ldots +1)$ which means that the whole $p^2$ divides the first parentheses and so $p^2 mid 2^q-1$.

EDIT: $p-1=qcdot age a$. This means that $ale p-1<p$ and hence $a<p$.

We start with the observation that $a^m-b^m$ is divisible $a^n-b^n$ if $nmid m$.

From this, one deduces there is an algebraic factor $A_n$ that divides $a^m-b^m$ exactly when $nmid m$.

If prime $p$ divides some $a^q-b^q$, but for no value less than $q$, then $p mid A_q$, since this likewise does not divide any earlier $a^n-b^n$.

We now suppose that $q$ is a proper divisor of $p-1 = qr$ (say). Then $(a^p-1-b^p-1)$ is the product of all $A_d, dmid p-1$, and the same holds true for $q$. The result of the division of $(a^p-1-b^p-1)/(a^q-b^q)$ is the sum of $r$ terms, all having a value $1pmodp$. This is the 'casting-out nines rule', eg $1111 = 4 pmod9$.

So if $p mid A_qr$, then $p mid r$, and specifically, $r$ is $p^x$.

Since we can exclude that $p$ can not divide $p-1$, then if $p^2$ divides some $a^q-b^q$ and $p^2$ divides some $a^p-1-b^p-1$, and thus $p^2 mid A_q$.

QED