regarding pseudo-prime numbers.

Question

If $W$ is an odd composite number and $-1+2^{W-1}$ is divisible by $W$ yet not by $W^2$, then $W^2$ does not divide $-1+ 2^{W(W-1)}$. Is this true? (forgive my use of symbols,I have no good math editing program that's easy to use.)

Answer 1

Actually the converse is true:
If $p^k$ is a prime power dividing $W$, then $2^{W-1}\equiv1\pmod{p^k}$ implies $2^{p^k(W-1)}\equiv1\pmod{p^{2k}}$ (you can verify this with binomial expansion, but it's a well known result), hence $W^2\mid2^{W(W-1)}-1$.

Analoguous results hold for higher exponents: $W\mid2^{W-1}-1\implies W^n\mid2^{W^{n-1}(W-1)}-1$, i.e., $W$ 'behaves' as a prime, as if $\varphi(W^n)=W^{n-1}(W-1)$.