For $n\ge 3$, define the $n$-gonal billiards conjecture as the statement

All convex $n$-gons admit periodic billiard trajectories.

To the best of my knowledge this question remains open for all $n$. The case which has received the most attention is the "simplest" case $n=3$. But is this really the simplest case, in a strict sense? I.e., does the $3$-gonal billiards conjecture follow from the $m$-gonal billiards conjecture for any $m>3$? Furthermore, does the $n$-gonal billiards conjecture follow from the $m$-gonal billiards conjecture for any $n$ and $m>n$?

There is one case for which this is easily seen. If $T$ is any triangle then we can let $T'$ be the reflection of $T$ across its longest side and we see that $T\cup T'$ is a convex quadrilateral, thus the $3$-gonal conjecture follows from the $4$-gonal conjecture. However, if $T$ is obtuse then repeating this process gives us a $5$-gon which is not convex. This also breaks down for any $n>3$, as the result can be non-convex.

It is tempting to argue that by shrinking the length of some side to $0$ on an $m$-gon, assuming the $m$-gonal conjecture we obtain periodic trajectories on every $m+1$ gon. However, I see no reason why the length of any periodic trajectories would not diverge as the length of the side approaches $0$, which breaks this line of reasoning.