Note that if $A=(a_1,\ a_2, ..., \, a_n)$ is polipythagorean then we can construct a secuence: $$A'=(a_1x,\ a_2x,....,\ a_nx, a_ny,\ a_1y)$$Such that $x^2+y^2$ is a perfect square, then if we can ensure that in A' no terms are repeated then A' is polypythagorean. Take $(x,y)$ such that $(x,y)=(m^2-1, 2m)$. Note that: $$\frac{x}{y}=\frac{m}{2}-\frac{1}{2m}>\frac{m}{2}-\frac{1}{2}$$Let $P$ be the maximum value of $\frac{a_i}{a_j}$ over all possible choices of $i,j$ then if $m>2p+1$ we have: $$\frac{x}{y}>p+\frac{1}{2}>p$$Then, for there does not exist $i,j$ such that $xa_i=ya_j$ with which we can finally conclude that $A'$ is polypythagorean. Then if $n$ is polypythagorean then $n+2$ is polypitagorhean. Construct a polypythagorean tuple for $n=4$ and an $$(3,4)\to (105,\ 140,\ 48,\ 36)$$ and the example gives that $n=3$ is polypythagorean so since $n=3$ and $n=4$ are polypythagorean and if $n$ is polypythagorean then $n+2$ is polypythagorean then we can conclude that for all $n \geq 3$ we have that $n$ is polypythagorean, with which we finish the problem.