proof that $(b^n-1)/(b-1)$ is not prime if n is a pseudoprime not prime of the base $b$.

Question

The question of my exercise says:

Proof that, if $n$ is a pseudoprime not prime of the base $b$

(i.e. $b^{n-1}\equiv 1 (\mod n)$)

then $N=(b^n-1)/(b-1)$ is also a pseudoprime not prime.

I have proven that $N$ is a pseudoprime, as seen in the answer I posted here.

But I'm having trouble proving that $N$ is not prime.

Any help?

Answer 1

Let $d\mid n$ be any divisor, then $$ \frac{(b^n-1)/(b-1)}{(b^d-1)/(b-1)}=\frac{(b^d)^{n/d}-1}{b^d-1}=\sum_{i=0}^{n/d-1}(b^d)^i $$ is an integer, so $$ \frac{b^d-1}{b-1}\mid \frac{b^n-1}{b-1}. $$ Since $n$ is composite, we can pick $d$ such that $1< (b^d-1)/(b-1)<(b^n-1)/(b-1)$. Done.