Bulgarian National Olympiad

We show that $n=7$. Note that $$\cos \pi =-1,\cos \frac{\pi }{2}=0,\cos \frac{\pi }{3}=\frac{1}{2},\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2},\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}.$$ Further, it follows from $0=\cos \frac{3\pi }{5}+\cos \frac{2\pi }{5}$ that $x_5=\cos \frac{\pi }{5}$ is a root of the equation $0=4x^3-3x+2x^2-1=(x+1)(4x^2-2x-1)$, i.e. $x_5=\frac{1+\sqrt{5}}{4}$. It remains to prove that $x_7=\cos \frac{\pi }{7}$ can not be expressed in the form $p+\sqrt{q}+\sqrt[3]{r}$, where $p,q$ and $r$ are rational numbers. Since $0=\cos \frac{4\pi }{7}+\cos \frac{3\pi }{7}$ we have that $x_7$ is a root of $$0=2(2x^2-1{)}^2-1+4x^3-3x=(x+1)(8x^3-4x^2-4x+1),$$ i.e. $x_7$ is a zero of $P(x)=8x^3-4x^2-4x+1$. Suppose that $x_7=p+\sqrt{q}+\sqrt[3]{r}$, where $q\ge 0,p,r\in \mathbb{Q}$. Now $x_7$ is a zero of $Q(x)=(x-p-\sqrt{q}{)}^3-r$ with coefficients of the form $a+b\sqrt{q}$, $a,b\in \mathbb{Q}$, i.e. from $\mathbb{Q}[\sqrt{q}]$. Since the coefficient of $x$ is nonnegative we have that $P\ne 8Q$. Thus $P=8Q+R$, where $R$ is a

polynomial of degree 1 or 2 with coefficients from $\mathbb{Q}[\sqrt{q}]$ and $R(x_7)=0$. If $\mathrm{deg}R=1$, then $x_7\in \mathbb{Q}[\sqrt{q}]$. If $\mathrm{deg}R=2$ and $R$ does not divide $P$, then $x_7$ is a zero of the remainder of $P$ divided by $R$, i.e. again $x_7\in \mathbb{Q}[q]$. If $\mathrm{deg}R=2$ and $R$ divides $P$ then the zero of $\frac{P}{R}$, which is from $\mathbb{Q}[q]$ is also zero of $P$. Thus, $P$ has a zero from $\mathbb{Q}[q]$. It is a zero of a polynomial of degree 1 or 2 with rational coefficients. It follows as above that $P$ has rational zero. Direct verification shows that none of the numbers $\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{8}$ (which are all possible rational zeroes of $P$) is a zero of $P$, a contradiction.