I've been working on this for quite awhile, and am stumped after a little bit. I have some stuff written down, but I just don't know how to completely prove it. I don't have much done yet:
$(2^{p-1})(2^p -1)$
$(2^{p-1})(2^p -1)\equiv 4 \pmod 6$
I know $p$ must be prime, but I'm just having trouble going forward.
Let's forget about the case p=2. Then, p is prime implies: p is odd. It's easy to prove that when p is odd you have:
2^p ≡ 2 (mod 6).
It's so because 2*4 ≡ 2 (mod 6) so you can go to any odd power and keep the residue at 2.
The same way you prove that 2^(p-1) ≡ 4 (mod 6).
That's it.
(2^(p-1))(2^p - 1) ≡ (4)(2-1) ≡ 4 (mod 6).