Fraction and Its Representation

Since it is trivial to reduce $m/n$ to co-prime pairs, let's assume $m, n$ are co-prime. Also by definition of rational numbers we can say that "infinite -1" is not a valid output. If prime factors of $n$ are also prime factors for $b$ then the representation of $m/n$ is finite. Otherwise, there exists some prime factors of $n$ that doesn't divide $b$. It can be easily shown that $1/n$ has similar period length as $m/n$, so we mainly focus on $1/n$ in base $b$ for now. Let's further divide the problem in 2 cases as following:

$gcd(n, b) = 1$: $\frac{1}{n}_b = \frac{b^r}{nb^r}_b = \frac{qn+1}{nb^r}_b = \frac{q}{b^r}_b + \frac{1}{b^r}_b\frac{1}{n}_b$ where $b^r = q*n + 1 \Longleftrightarrow b^r \equiv 1 \mod n$. Then $\frac{1}{b^r}_b\frac{1}{n}_b = \frac{q}{b^{2r}}_b + \frac{1}{b^{2r}}_b\frac{1}{n}_b$. Thus the length of the repetition is $r$, this is also called the order of $b$ modulo $n$. And the output will be "infinite 0 r".

$gcd(n, b) \neq 1$: Let $n = st$ where $gcd(s, b) = 1$. Now $1/n = (1/s)(1/t)$ where $1/t$ has finite representation in base $b$ and $1/s$ is solved in previous case. One can verify that the repetition of $1/s$ gets shifted by the length of the representation of $1/t$. Hence output is "infinite l r" where $l$ is the length of $\frac{1}{t}_b$.

At this point there should be some algorithmic task to find $s,t$as defined above, which can be found by either prime factorizing $n, b$or repeatedly using gcd to seperate all common prime factors. And to find order one can utilze euler totient theorem and the fact if $x^u \equiv v \mod n$ then $x^{uy} \equiv v \mod n$, so take the divisors $d$of $\phi(n)$ and check if $b^d \equiv 1 \mod n$.