How to prove $2^{m-1}-1 \over 127m$ is integer Question. Prove that $2^{m-1}-1 \over 127m$ is integer when $k>1$, $k \in \Bbb{N}$, $p=6k+1$, $p \in \Bbb{P}$, $m=2^p-1$.
Approach. Let's prove these three propositions.
A. (127, m)=1
B. $127 \mid 2^{m-1}-1$
C. $m \mid 2^{m-1}-1$
I did 'A' by myself, but I can't do B and C. I think that I have to use FlT in C. How can I prove it?
 A: As you say you have shown A yourself, I will start with your B statement.  Note that
$$2^3 = 8 \equiv 1 \pmod 7 \tag{1}\label{eq1}$$
and
$$127 = 2^7 - 1 \tag{2}\label{eq2}$$
It's also given that
$$p = 6k + 1 \tag{3}\label{eq3}$$
and
$$m = 2^p - 1 \Rightarrow m - 1 = 2^p - 2 \tag{4}\label{eq4}$$
Thus, using \eqref{eq1}, \eqref{eq3} and \eqref{eq4}, we have
$$2^p - 2 = 2\left(2^{3 \times 2k}\right) - 2 \equiv 2\left(1\right)^{2k} - 2 \equiv 0 \pmod 7 \Rightarrow m - 1 = 7j \tag{5}\label{eq5}$$
for some positive integer $j$. From \eqref{eq2}, we have that $2^7 \equiv 1 \pmod{127}$, so from \eqref{eq4} and \eqref{eq5}, we have that
$$2^{m - 1} - 1 = 2^{7j} - 1 = \left(2^7\right)^j - 1 \equiv 1^j - 1 \equiv 0 \pmod{127} \tag{6}\label{eq6}$$
This confirms your B statement.
As for C, note that as $p$ is prime, Fermat's little theorem gives that
$$p \mid 2^p - 2 \tag{7}\label{eq7}$$
With this, then
$$m - 1 = 2^p - 2 = pi \tag{8}\label{eq8}$$
for some positive integer $i$. Thus,
$$2^{m - 1} - 1 = 2^{pi} - 1 = \left(2^p\right)^i - 1 \equiv 1^i - 1 \equiv 0 \pmod m \tag{9}\label{eq9}$$
thus proving your C statement.
