I am reading Birkhoff and Vandiver (1904) Integral divisors of $a^n-b^n$ and I am a bit confused.
From Fermat we know that $$x\equiv y \mod \phi(n) \Rightarrow a^x\equiv a^y \mod n$$ or $$n\equiv 0 \mod (p-1) \Rightarrow m^n\equiv 1 \mod p$$
but on page 2, what I read seems to be something like
...$ \mathit{m^n\equiv1 \mod p}$. It now follows from the definition of p, that m appertains to n, modulo p, so that $\mathit{p - 1\equiv 0\mod n}$...
which I interpret (in thoose particular conditions) as $$ m^n\equiv 1 \mod p\Rightarrow (p-1)\equiv 0 \mod n$$
What am I missing? Because it would mean $$n\equiv 0 \mod (p-1) \Rightarrow (p-1)\equiv 0 \mod n$$ or $n=(p-1)$ which is not the case.
Thanks
It isn't necessary that if $m^n \equiv 1 \pmod{p} \implies n \equiv 0 \pmod{p-1}$. For example, we can take $$2^3 \equiv 1 \pmod{7}$$ The smallest value $n$ can take, say $d$ is the order of $m$ modulo $p$. Here, we say that $3$ is the order of $2$ modulo $7$. It is not necessary that the order takes the value $p-1$. However, we can see that by definition of minimality of $d$, if $p-1=dq+r$ by division algorithm, then: $$m^{p-1-dq} \equiv 1 \pmod{p} \implies m^r \equiv 1 \pmod{p} \implies r=0$$ as $0<r<d$ would contradict the minimality of $d$. That is why it follows that $p-1 \equiv 0 \pmod{d}$. In the above context, I guess that $n$ was referring to the order.