I am having some trouble in proving that the only solutions to $ -2^{m-1} \equiv m \pmod{7} $
are $m \equiv 3,5, 13 \pmod{42}$.
What I tried to use:
If $-2^{m-1} \equiv m \pmod{7}$, then $2^{m-1} \equiv 6m \pmod{7}$ and we know that $2^6 \equiv 1 \pmod{7}$.
But now I can't go on. Could someone give me a hint of what should I be doing next?