
2pseudoprime!
A number m is called a 2pseudoprime 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 2pseudoprime.
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?

Re: 2pseudoprime!
First note that $\displaystyle m = 2^n  1$ implies that $\displaystyle 2^n \equiv 1 (\bmod. m)$ , thus in fact $\displaystyle 2^r \equiv 2^{ r \bmod n} (\bmod. m)$ for any $\displaystyle r\in \mathbb{Z}$ (because adding  or substracting $\displaystyle n$'s to the exponent won't change the result)
But now, by hypothesis $\displaystyle 2^{n1} \equiv 1 (\bmod. n)$ thus $\displaystyle 2^n \equiv 2 (\bmod. n)$ ... (Wink)