Problem 1885

We prove by induction that all integers greater than or equal to $2$ are polypythagorean. Extend the definition of polypythagorean to $n = 2$ in the logical way. For the induction basis, take $(3, 4)$ for $n = 2$ and $(44, 117, 240)$ from the example for $n = 3$.

For the induction step, let $n \ge 4$ and assume for our induction hypothesis that all $2 \le k < n$ are polypythagorean. In particular, $n - 2$ is polypythagorean.
Let $(a_1, a_2, \dots, a_{n-2})$ be such that when put around a circle, the sum of squares of each pair of neighbouring integers is a square. Choose a prime number $p$ that does not divide any of the $a_i$. Then let $x = p^2 - 1$ and $y = 2p$ so that $x^2 + y^2 = (p^2 - 1)^2 + (2p)^2 = (p^2 + 1)^2$. By multiplying our $n-2$ integers by $x$, we can now add two integers: $(xa_1, xa_2, \dots, xa_{n-2}, ya_{n-2}, ya_1)$.

We simply check that
$$\begin{align*} (xa_i)^2 + (xa_{i+1})^2 &= x^2(a_i^2 + a_{i+1}^2) \\ (xa_{n-2})^2 + (ya_{n-2})^2 &= (x^2 + y^2)a_{n-2}^2 \\ (ya_{n-2})^2 + (ya_1)^2 &= y^2(a_{n-2}^2 + a_1^2) \\ (ya_1)^2 + (xa_1)^2 &= (y^2 + x^2)a_1^2 \end{align*}$$
and these are indeed all squares by the induction hypothesis and by construction of $x$ and $y$. As the integers $a_1, a_2, \ldots, a_{n-2}$ are all different, the numbers $xa_1, xa_2, \ldots, xa_{n-2}$ are also all different from each other. The numbers $ya_{n-2}$ and $ya_1$ are also different from each other. As $y$ is divisible by $p$, but $x$ and $a_i$ are not, neither $ya_{n-2}$ nor $ya_1$ can be equal to any of the $xa_i$. We conclude that $xa_1, xa_2, \ldots, xa_{n-2}, ya_{n-2}, ya_1$ are all different, so $n$ is polypythagorean. This completes the induction.

$\square$