Measure the central angles, say, from one of the given points. We have points exactly at angles $360^\circ\cdot \displaystyle\frac kN$ with $k=0,...,N-1$.
If $N$ is not divisible by $3$, then try to prove that the difference of any two angles cannot be $120^\circ$ which would be required for the equilateral triangle.
If $3|N$, then look at the vertex which is on the first third, i.e. with angle in $[0^\circ,\,120^\circ)$. An equilateral triangle must have exactly one vertex on each third of the circle, and one vertex already determines the other two.
Now, the vertex on the first third can be any out of those $N/3$.