From $abc$ is a square we can have $a=xy,b=yz,c=zx$. Now for $a^2+b^2+c^2$ we have \begin{align*} k^2&=x^2y^2+y^2z^2+z^2x^2\\ k^2+x^2&=x^2y^2+y^2z^2+z^2x^2+x^2\\ &=(x^2+y^2)(x^2+z^2)\\ &=(x^2+yz)^2+(xz-xy)^2 \end{align*}by Lagrange's Identity. We see that $(x,y,z,k)=(x,y,x+y,x^2+y^2+xy)$ satisfies the equation. For $a+b+c$, we have $xy+(x+y)^2=m^2$. Let $m=x+y+a$ such that $xy=2ax+2ay+a^2$. By setting $x=2a+1$ we have $y=5a^2+2a$. Since \begin{align*} \gcd(2a+1,5a^2+2a)&=\gcd(2a+1,5a+2)\\ &=\gcd(2a+1,a)\\ &=1, \end{align*}this solution set works for all $a$.