First note that implies that , thus in fact for any (because adding - or substracting- 's to the exponent won't change the result)
But now, by hypothesis thus ...
A number m is called a 2-pseudoprime if
(a) m is composite, and
(b) 2^(m -1) == 1 mod m (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 if n is 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?