45th Mongolian Mathematical Olympiad

If $z_1,z_2,\dots ,z_n$ are vertices of $P$ then $|z_1|=|z_2|=\cdots =|z_n|=1$. By the rotation we can assume that ${\prod }_{i=1}^n{z}_i=1$. We know that

$|w-z_1|\cdot |w-z_2|\cdots |w-z_n|=|(w-z_1)\cdots (w-z_n)|$.

Let $P(w)={\prod }_{i=1}^n(w-z_i)$ and $P(w)=w^n+Q(w)+1$, here $Q(w)=a_{n-1}{w}^{n-1}+\cdots +a_{1}w$.

Consider the $n$th roots of unity ${ϵ}_0,{ϵ}_1,\dots ,{ϵ}_{n-1}$. Therefore ${\sum }_{s=0}^{n-1}{ϵ}^s=0$.

Here $s\not\equiv 0(\mathrm{mod}n)$ we get $$\sum _{i=1}^{n-1}Q({ϵ}_i)=\sum _{j=1}^{n-1}\left(a_j\cdot \sum _{i=0}^{n-1}{ϵ}^j\right)=0$$

a) Assume to the contrary. If there exist $i$ such that $Q({ϵ}_i)=a_i+b_i$, $a_i>0$ then $$P({ϵ}_i)={ϵ}_i^n+1+Q({ϵ}_i)=2+a_i+b_i\cdot i,$$ thus we have $|P({ϵ}_i)|>2$. Hence $Q({ϵ}_i)=b_i\cdot i$ for arbitrary $i$. Because we have ${\sum }_{i=0}^{n-1}{a}_i=0$ and all of $a_i$ is less than or equal to zero, thus $a_i=0$ for arbitrary $i$. Now suppose that for arbitrary $i$: $b_i\ne 0$ then $$P({ϵ}_i)={ϵ}_i^n+1+Q({ϵ}_i)=2+b_i\cdot i.$$ So $|P({ϵ}_i)|>2$. This leads contradiction.

Now for arbitrary $i$, we have $Q({ϵ}_i)=0$, $i=\overline{0,n-1}$. But $\mathrm{deg}Q(w)=n-1$, hence $Q(w)$ is zero polynomial. Thus $P(w)=w^n+1$, we get $|P({ϵ}_i)|=|{ϵ}_i^n+1|=2$



b) From part a), the multiplication greater than or equal to $2$, it must be $P(w)=w^n+1$, $w^n+1=0$ equation's solutions are $$z_i=\cos \frac{2\pi }{2n}(2k+1)+i\sin \frac{2\pi }{2n}(2k+1),k=\overline{0,n-1}$$