When $q$ is prime and a Wieferich prime $p$ divides $\ M=2^q-1\ $ . Why can we conclude $\ p^2\mid M\ $?

Question

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 $p^2\mid M_q$.

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

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 ?

Answer 1

Obviously, $q \mid p-1$. Write $p-1=q\cdot a$.
Then $p^2 \mid 2^{q\cdot a}-1=(2^q-1)(2^{q(a-1)}+2^{q(a-2)}+\ldots +1)$.
Observe now that the second parentheses is $\equiv a\pmod{p}$ since each summand is $\equiv 1\pmod{p}$.
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: $p-1=q\cdot a\ge a$. This means that $a\le p-1<p$ and hence $a<p$.

Answer 2

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

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

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 $A_d, d\mid p-1$, 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 $1\pmod{p}$. This is the 'casting-out nines rule', eg $1111 = 4 \pmod{9}$.

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