Finding at least two paths of same length in a directed graph
Consider a graph $G$, we want to know if there are two different paths from $A$ to $B$ of the same length. What to do? Simple: code two paths in one. Define graph $G'$ with vertices $V \times V \times \{0,1\}$. You make a step in $G'$ by making two independent steps in $G$. The additional bit tells you whether the two paths have already split from each other.

Formally, there is an edge $(i,j,e) \to (i',j',e')$ in $G'$ iff $i \to i'$, $j \to j'$ in $G$ and $e'=e ∨ (i,i')≠(j,j')$.

The algorithm checks if there is a path $(A,A,0)$ to $(B,B,1)$ in $G'$, which is $O(V^4)$, or something like $O((V+E)^2)$.

If you agree this algorithm is correct then, as a consequence, the path in $G'$ is of length at most $2n^2$, therefore potential "path collisions" must occur latest at that length. You can get a $O(V^{\omega} \log V)$ algorithm from this observation, where $\omega$ is matrix multiplication complexity (ask if you need a spoiler...).

I strongly feel there is a $O(V+E)$ algorithm, which uses more of the structure of the problem.