"All vertices of a convex pentagon are lattice points, and its sides have integral length. Show that its perimeter is even."

Question

"All vertices of a convex pentagon are lattice points, and its sides have integral length. Show that its perimeter is even." - Problem Solving Strategies, Arthur Engel, pg. 27.

I have proven the theorem through two methods. I thought about a third method, but it appears faulty, since it shows that the theorem is false. Please indicate me where I have made the error.

Consider two vertices of the pentagon. Since the points are in a lattice, and the sides are integers, $\Delta x , \Delta y, d \in \mathbb{N} \ $ . It is also obvious that $ \Delta x^{2}+\Delta y^{2}=d^{2} $.

That means that those three integers form a Pythagorean triplet. $\therefore d$ is odd, which applies to all sides.

Since the perimeter is the sum of all sides,

$ \left. \begin{array}{l} p = \sum\limits_{i=1}^5 d_{i}\\ \forall d_{i}, 2\nmid d_{i} \end{array} \right\}\Rightarrow \text{$p$ is odd.} $

I can't seem to find the mistake. Help would be appreciated.

Answer 1

A Pythagorean triplet $a^2+b^2=c^2$ need have $c$ odd only if it's primitive i.e. if we ignore cases like $6^2+8^2=10^2$. Why should that be true here?

Answer 2

In the interest of sharing this, I've found a 3rd method that works for any lattice polygon. Place one vertex at $0$ and let the sides be represented by the vectors $a_1, \dots, a_n, -a_1-\dots-a_n.$ The perimeter is $\sum\limits_i ||a_i|| + ||\sum\limits_i a_i|| \equiv \sum\limits_i ||a_i||^2 + ||\sum\limits_i a_i||^2 = 2\sum\limits_i ||a_i||^2 + 2\sum\limits_{i < j} \langle a_i, a_j \rangle \equiv 0 \mod 2.$