4. Does there exist a convex polygon with 2015 diagonals?

#
# Solution.

The number of diagonals in a convex $n$-gon is

$$
\frac{n(n-1)}{2}-n=\frac{n^{2}-3 n}{2}
$$

Solve the equation

$$
\frac{n^{2}-3 n}{2}=2015 \quad \Longleftrightarrow \quad n^{2}-3 n-4030=0
$$

Its discriminant $D=9+4 \cdot 4030=16129$. Note that $120^{2}<16129<130^{2}$ and $D$ ends with the digit 9. If $D$ is the square of an integer $m$, then $m$ ends with the digit 3 or 7, i.e., only two cases are possible: $m=123$ or $m=127$. It is easy to check that $123^{2}=(120+3)^{2} \neq 16129$, while $127^{2}=(130-3)^{2}=$ $16900+9-2 \cdot 3 \cdot 130=16129$. Hence, $n=\frac{3 \pm 127}{2}$. Choosing the positive value, we get $n=65$. Therefore, the specified polygon exists, it is a 65-gon.

Answer. Yes. It is a 65-gon.