A numbermis called a 2-pseudoprime if

(a)mis composite, and

(b) 2^(m -1) == 1 modm(where == is the congruent sign, the triple horizontal bars)

(it's pseudo because it looks like Fermat's little theorem.

I need to show that ifnis a pseudoprime then m = [2^n] - 1 is also a 2-pseudoprime.

I have shown that m is composite, now how do I go about showing the other condition? I tried playing around with the mod stuff, but I bring it down to something like this:

if (2^n - 2) / n is an integer then show that ([2^2^n] - 4) / (2^n - 1) is also an integer. But I'm stuck proving this too. Any help?