Does $N(mn)|N(m)N(n)$ for $\gcd(m,n)=1$? (Fibonacci Sequence)

Question

Consider Fibonacci Sequence Mod m, n and mn. And let N(m) be the period of Fibonacci Sequence mod m. For several $m,n$ I tried to compare $N(mn)$ with $N(m)N(n)$. It looks that there is no specific pattern, but for $\gcd(m,n)=1$ it looks $N(mn)|N(m)N(n)$. Could you write some hint, I don't know how to start proving if it is true?

Answer 1

Suppose $m$ and $n$ are coprime. Since $F_{j+N(m)N(n)} \equiv F_j \mod m$ and $F_{j+N(m)N(n)} \equiv F_j \mod n$ we have $F_{j+N(m)N(n)} \equiv F_j \mod mn$. Thus $N(mn) \mid N(m) N(n)$.

There's nothing special about Fibonacci numbers here: this would work for any sequence that is periodic mod $m$ and mod $n$.