# Math Help - What about b?

1. ## What about b?

If (2^a -1) divides (2^{a(b-1)} + 2^{a(b-2)} + ... + 2^2a + 2^a + 1)
for a, b 2^a - 1 and 2^b -1 being odd primes
What does it tell us about b?

2. Originally Posted by helgamauer
If (2^a -1) divides (2^{a(b-1)} + 2^{a(b-2)} + ... + 2^2a + 2^a + 1)
for a, b 2^a - 1 and 2^b -1 being odd primes
What does it tell us about b?
It would imply that $b=2^a-1$.

Note that $(2^a)^k\equiv1\pmod{2^a-1}$ for any positive integer $k$. Therefore, $(2^a)^{b-1}+(2^a)^{b-2}+\cdots+(2^a)^2+(2^a)+1\equiv1+1+\cdots+1=b\equi v0\pmod{2^a-1}$.

Thus $2^a-1(>1)$ divides $b$, and since $b$ is prime we must have $b=2^a-1$.

By the way, $2^{a(b-1)}+2^{a(b-2)}+\cdots+2^{2a}+2^a+1=(2^a)^{b-1}+(2^a)^{b-2}+\cdots+(2^a)^2+(2^a)+1$.