How to show that a regular pentagon can't have all coordinates rational

Question

This is pretty straightforward if we're allowed to use trigonometry, so I guess my question is

Are there any nice (trigonometry-less) proofs of the fact that a regular pentagon in the plane must have some irrational coordinate?

Now, it's possible to translate a trigonometric proof into an algebraic proof since $\sin(18^\circ)$ can be written in terms of the golden ratio, so I should say that something along these lines would be less than ideal.

I've been thinking about a proof using the fact that the (orientation-preserving) symmetries of the dodecahedron can be identified with $A_5$ (the alternating group on 5 letters) and the 3 dimensional irreducible representations of $A_5$ can't be expressed over the rationals, but I'm not quite sure how this would go. Is proof along these lines possible?

Maybe a better (or worse?) way to say what I'm looking for is this: Is there a proof that is so specific to the pentagon that we couldn't hope to generalize it to other polygons? (Maybe with the exceptional embedding $S_5\hookrightarrow S_6$ lurking in the background...?)

Answer 1

We will show that there is no regular polygon with $n$ odd sides that is embedded in the rectangular lattice.

Proof by contradiction. Suppose not. Take the smallest such polygon. Consider the $n$ vectors that point from one vertex to the next. Let $v_i = a_ i \hat{i} + b_i \hat{j}$. Let $ l^2 = a_i ^2 + b_i^2 $ be the length of the sides. (Note that $l$ need not be an integer, though $l^2$ is an integer.)

We have $\sum a_i = 0 $ and $\sum b_i = 0$. Squaring and adding these equations, we get that

$0 = \sum a_i ^2 + \sum b_i ^2 + 2 \sum a_i a_j + b_i b_j $. Hence, $n l^2 = \sum a_i ^2 + b_i^2 $ is even. Since $n$ is odd, thus $l^2$ is even. This implies that $a_i, b_i$ are either both odd, or both even.

If $a_1, b_1$ are both even, then $l^2$ is a multiple of 4, which implies that all $a_i, b_i$ are even. This contradicts the minimality of the polygon, since we can just scale by $\frac{1}{2}$.

If $a_1, b_1$ are both odd, then $l^2$ is not a multiple of 4, which implies that all $a_i, b_i$ are odd. This contradicts the fact that $\sum a_i = 0$, since the LHS is odd.

Answer 2

Let $A$ be rotation by the angle $2\pi/5$ counterclockwise. If you could draw a pentagon with all coordinates rational, that would imply that $A$ had rational entries. We know $A^5 = 1$, so $A$ is a solution to the polynomial equation $x^5 - 1 = 0$. Since $A$ is not $1$, it is a solution to $x^4 + x^3 + x^2 + x + 1 = 0$, which is irreducible. The set of polynomials in $\mathbb{Q}[x]$ satisfying $p(A) = 0$ is an ideal, and therefore a principal ideal, and since it contains an irreducible polynomial it must equal the multiples of that polynomial.

However, $A$ satisfies $ch_{A}(A) = 0$, where $ch_{A}$ is the characteristic polynomial of $A$, which is a quadratic equation with rational coefficients. This is a contradiction.