A positive integer $n$ is Danish if a regular hexagon can be partitioned into $n$ congruent polygons. Prove that there are infinitely many positive integers $n$ such that both $n$ and $2^n+n$ are Danish.
First notice that $n=3k$ is danish: we can cut a regular hexagon into 3 rhombii, then slice each rhombus into $k$ equal pieces (slice parallel to the sides, say).
Now we claim that $n=2\times 4^m$ is also Danish: cut a hexagon into two trapezoids along a diameter, and now realize that we can dissect a 120-120-60-60 trapezoid into four congruent copies of itself (use the three short sides as the new longer bases). We can iterate this process to cut each trapezoid into 4 smaller trapezoids.
The problem is solved by noticing that $2^{2\times4^m}+2\times4^m \equiv 2+1\equiv 0 \pmod{3}$.