Solution
= Solution
Both $n$ and $m$ divide $b^{2p}-1$, hence so does $N=nm$ after checking $\gcd(n,m)=1$ under the hypothesis. Also $n\equiv p\pmod{b-1}$ and the elementary factorization gives $N\equiv1\pmod{2p}$; part (a) with $k=2p$ applies. As odd primes $p$ grow, $N$ grows, yielding infinitely many examples.