Note that if $c=a+b$, $$ (ab)^2 + (bc)^2 + (ca)^2=(a^2+ab+b^2)^2 $$is a perfect square. The rest is routine.

Note that $a^2b^2+a^2(a+b)^2+b^2(a+b)^2=(a^2+ab+b^2)^2$. So if there exist three elements in the set with $a+b=c$, then $(ab)^2+(bc)^2+(ca)^2=(a^2+ab+b^2)^2$ is a square. So all we have to show is that a subset of $\{1,2,3,...n \}$ having at least $\frac{n}{2}+1$ is not sum-free. If $n$ is even, consider the $\frac{n}{2}$ pairs of the form $\{i,n+1-i \}$ and apply PHP. If instead $n$ is odd, consider the $\frac{n-1}{2}$ sets of the form $\{i,n-i \}$ for $i<n$. Note that there will be at least $\frac{n}{2}$ elements in the set which are less than $n$, so we again apply PHP and show that the set is not sum-free. Proved