Rioplatense Mathematical Olympiad

The answer is $n\ge 3$. We abbreviate PP = polygonal pythagorean.

First, assume that $n$ is PP. We will show then that $n+2$ is also PP. Let $a_1,\dots ,a_n$ be pairwise different positive integers such that $a_i^2+a_{i+1}^2$ is a perfect square for all $i=1,\dots ,n$, where $a_{n+1}=a_1$. Choose any Pythagorean triple $(x,y,z)$, that is, three positive integers such that $x^2+y^2=z^2$. We claim that $a_{1}x,a_{2}x,\dots ,a_{n}x,a_{n}y,a_{1}y$ satisfy the desired conditions for $n+2$. Indeed, $$\begin{array}{rl}(a_{i}x{)}^2+(a_{i+1}x{)}^2& =x^2(a_i^2+a_{i+1}^2),\\ (a_{n}x{)}^2+(a_{n}y{)}^2& =a_n^2(x^2+y^2),\\ (a_{n}y{)}^2+(a_{1}y{)}^2& =y^2(a_n^2+a_1^2),\text{and}\\ (a_{1}y{)}^2+(a_{1}x{)}^2& =a_1^2(y^2+x^2),\end{array}$$ which are all products of two perfect squares and therefore are perfect squares themselves.

However, it may happen that either $a_{n}y$ or $a_{1}y$ is equal to some $a_{i}x$. To make sure that this is not the case, take a prime number $p$ which does not divide any of the $a_i$'s, and use the Pythagorean triple $(x,y,z)=(p^2-1,2p,p^2+1)$. Since $y$ is divisible by $p$ and both $a_i$ and $x$ are not, no $a_{i}x$ can be equal to $a_{n}y$ or $a_{1}y$, and so the $n+2$ numbers $a_{1}x,a_{2}x,\dots ,a_{n}x,a_{n}y,a_{1}y$ are pairwise different, which proves our claim.

With the example given in the problem statement we are able to get solutions for all odd $n$. To solve the problem for even $n$, it is enough to find a solution for $n=4$. Considering the Pythagorean triples $(3,4,5)$ and $(5,12,13)$, we can check that $(3\cdot 5,4\cdot 5,4\cdot 12,3\cdot 12)=(15,20,48,36)$ is a solution, and the proof is complete.