4.32. Let $n$ be the number of sides of a convex polygon, and $d$ be the number of its diagonals. Specify all values of $n$ for which $n>d$.


4.32. The number of diagonals of an $n$-sided polygon is determined by the formula $d=\frac{(n-3) n}{2}$. We need to select only those values of $d$ that satisfy the condition $d \leq n$. Substituting the formula for $d$, we get $\frac{(n-3) n}{2} \leq n$. Simplifying, we have $(n-3) n \leq 2n \Rightarrow n^2 - 3n \leq 2n \Rightarrow n^2 - 5n \leq 0$, i.e., the number $n$ belongs to the interval $(0,5)$. Since $n$ is the number of sides of the polygon, $n \geq 3$, and finally, we get that the values of $n$ that satisfy the condition are $n=3$ and $n=4$.

Answer: $n=3$ and $n=4$.