Let $\alpha(n)$ denote the measure of the interior angles of a regular $n$-gon. What is $n$ if $\alpha(n+3)-\alpha(n)=\alpha(n)-\alpha(n-2)$?
For a regular polygon with $n$ sides, the sum of the interior angles is $(n-2) \cdot 180^{\circ}$, so the measure of one angle is

$$
\begin{aligned}
\alpha(n) & =\frac{(n-2) \cdot 180^{\circ}}{n} ; \quad \text { and thus } \\
\alpha(n+3) & =\frac{(n+3-2) \cdot 180^{\circ}}{n+3} \\
\alpha(n-2) & =\frac{(n-2-2) \cdot 180^{\circ}}{n-2}
\end{aligned}
$$

Thus, the equation is:

$$
\frac{(n+1) \cdot 180^{\circ}}{n+3}-\frac{(n-2) \cdot 180^{\circ}}{n}=\frac{(n-2) \cdot 180^{\circ}}{n}-\frac{(n-4) \cdot 180^{\circ}}{n-2}
$$

We multiply both sides of the equation by $n(n-2)(n+3)$:

$$
\begin{gathered}
{[n(n-2)(n+1) \cdot 180]-\left[(n-2)^{2}(n+3) \cdot 180\right]=} \\
=\left[(n-2)^{2}(n+3) \cdot 180\right]-[(n+3)(n-4) \cdot n \cdot 180]
\end{gathered}
$$

From this, we get:

$$
\begin{aligned}
\left(n^{3}-n^{2}-2 n\right)-\left(n^{3}-n^{2}-8 n+12\right) & =\left(n^{3}-n^{2}-8 n+12\right)-\left(n^{3}-n^{2}-12 n\right) \\
6 n-12 & =4 n+12 \\
n & =12
\end{aligned}
$$

The sought regular polygon is the regular dodecagon.