Some years ago (during my master's thesis), I studied a problem known as Kotzig's conjecture (however not this famous one, but the lesser known one on $k$-paths in graphs).
Anyway, there was no way to solve the problem, but I was able to distill one specific instance of a sub-problem, that seemed easy enough. Still, I was not able to solve it. Here it is.
Question.
Take the $K_5$ (complete graph on $5$ vertices), and subdivide its edges by inserting new vertices (see the image below). There is only one condition: as you can see, I suggestively used different colors for the edges to divide the graph into an "outer cycle" and an "inner cycle". The condition now is, that both cycles must have the same number of vertices in it, i.e., both cycles must have the same length, say $k+1$.
Do there always exists two (not necessarily intersection-free) paths $P_1$ and $P_2$ of length $k$, that start in the same vertex, and end in the same vertex?

I made the experience, that given any specific instance of the problem, I had no trouble finding such a double-$k$-path. But I found it very hard (even impossible), to give a general construction of such one. I had a lot of heuristics, but the conditions for their applications where complicated.
The following image shows two 12-paths in the above subdivision with 13-cycles.

Label the vertices of the original $K_5$ as $u_i\;(i \in [1,5])$. Let $w_{i,j}\;(i,j \in [1,5])$ be one greater than the number of vertices inserted in the edge $u_i - u_j$. Then we can label those intermediate vertices $v_{i,j,a}$ numbered $u_i - v_{i,j,1} - v_{i,j,2} - \cdots - v_{i,j,w_{i,j}-1} - u_j$ with identification $v_{i,j,a} = v_{j,i,w_{i,j}-a}$.
The generation constraint is that $w_{1,2} + w_{2,3} + w_{3,4} + w_{4,5} + w_{1,5} = w_{1,3} + w_{3,5} + w_{2,5} + w_{2,4} + w_{1,4} = k + 1$.
Let $G_5$ denote the original $K_5$ with weights $w_{i,j}$, and $G_5 \setminus (u_i, u_j)$ denote the graph obtained by deleting edge $(u_i, u_j)$ from $G_5$.
Denote the shared endpoints of $P_1$ and $P_2$ as $s$ and $t$.
There are five cases:
Put together, we get a giant but finite disjunction of conjunctions of inequalities which we can negate and feed to a satisfiability solver such as Z3.
To generate the Z3 source I'm using Python. I'll probably come back to this and try to tidy it up a bit. It turned out not to be necessary to fully cover the final case. I realise that computer proofs aren't as satisfactory as non-computer proofs, but at least knowing that it's a theorem is a good starting point.