Prove that there does not exist a $n$-regular polygon $(n\ge 4)$, such that its sides and diagonals are all integers.

Question

Prove that there does not exist a $n$-regular polygon $(n≥4)$, such that its sides and diagonals are all integers. Maybe a famous problem, but I don't know how to solve that.

Answer 1

Suppose that the lengths of the sides and the diagonals of $n$-regular polygon $A_1A_2\cdots A_n$ with side lenghths $a$ are all integers.

Since $$\overline{A_iA_{i+2}}=\frac{a\sin\left(\frac{360}{n}\right)^\circ}{\sin\left(\frac{180}{n}\right)^\circ}=2a\cos\left(\frac{180}{n}\right)^\circ$$

is an integer, we have that $\cos\left(\frac{180}{n}\right)^\circ$ is rational.

By (the extension of) Niven's theorem, we have $$\cos\left(\frac{180}{n}\right)^\circ=0,\quad \pm\frac 12,\quad \pm 1$$ which contradicts that $n\ge 4$.

Answer 2

Let $\zeta\in\Bbb C$ be a primitive $n$th root of unity. The existence of such a polygon would imply that for some $r>0$ the numbers $r\left|\zeta^k-1\right|$ are integer for all $k\in\Bbb Z$. In particular, $$\tag1q:=\left|\zeta+1\right|=\frac{r\left|\zeta^2-1\right|}{r\left|\zeta^1-1\right|}\in\Bbb Q$$ so that $q^2=(\zeta+1)(\overline\zeta+1)$ and $\zeta,\overline\zeta$ are the roots of the quadratic polynomial $$\tag2X^2- (q^2-2)X+1\in\Bbb Q[X].$$ Hence $\phi(n)=[\Bbb Q[\zeta]:\Bbb Q]\le 2$ and from this we conclude that $n$ is not divisible by any prime $p>3$ (which would make $\phi(n)\ge p-1>2$), nor divisible by $3^2$ (which would make $\phi(n)\ge 6$), nor divisible by $2^3$ (which would make $\phi(n)\ge 4$), nor divisible by $2^23$ (which would make $\phi(n)\ge 4$). Thus we are left only with $n=1,2,3,4,6$. For $n=6$, the minimal polynomial of $\zeta$ is $X^2-X+1$ so that by comparison with $(2)$ we have $q^2=3$, which is absurd. For $n=4$, the minimal polynomial of $\zeta$ is $X^2+1$ so that by comparison with $(2)$ we have $q^2=2$, which is again absurd. We conclude $n\le 3$.

Remark: We used only $(1)$, i.e., only that the side lengths $A_iA_{i+1}$ and the shortest diagonals $A_iA_{i+2}$ are integral.