I believe this is a basic proof but just hoping someone could give me some feedback on my attempt.
$\sum_2:=\{x^2+y^2 | x,y \in \mathbb{Z}\}$
Since $a,b \in \sum_2$ it follows:
$a=u^2+v^2$
$b=n^2+m^2$ $\qquad$ (For $m,n,u,v \in \mathbb{Z}$)
$ab=(u^2+v^2)(n^2+m^2)=u^2n^2+u^2m^2+v^2n^2+v^2m^2$
This can be rewritten as: $ab= (un)^2+(um)^2+(vn)^2+(vm)^2$
We can see $(un)^2+(um)^2 \in \sum_2$ and $(vn)^2+(vm)^2 \in \sum_2$
Is this enough to conclude that $ab \in \sum_2$? The question is not worth many marks so I feel a small proof is all that is needed but this feels incomplete to me.
I'd appreciate any insights!
Saying $ab = S + T$ where $S = (un)^2 + (um)^2 \in \Sigma_2$ and $T = (vn)^2 + (vm)^2 \in \Sigma_2$ would be enough to conclude that $ab \in \Sigma_2$ only if you had shown that $\Sigma_2$ is closed under addition. This is not the case.
Instead what you want to do is think of this in terms of complex numbers. If $u, v \in \mathbb Z$ then define $N(u + iv) = u^2 + v^2$. It is a basic fact of complex arithmetic that $N(z)N(z') = N(zz')$ where $z, z'$ are complex. How do we apply this here? Well,
$$(u^2 + v^2)(m^2 + n^2) = N(u + vi)N(m + ni) = N((u + vi)(m + ni)) $$
and we have
$$ (u + vi)(m + ni) = (um - vn) + (un + vm)i $$
so
$$ (u^2 + v^2)(m^2 + n^2) = (um - vn)^2 + (un + vm)^2. $$
You should expand both sides to verify this.