Answer:
See proof below
Step-by-step explanation:
One way to solve this problem is to "add a zero" to complete the required squares in the expression of xy.
Let [tex]x=m^2+n^2[/tex] and [tex]y=l^2+k^2[/tex] with [tex]m,n,l,k\in \mathbb{Z}[/tex]. Multiplying the two equations with the distributive law and reordering the result with the commutative law, we get [tex]xy=(m^2+n^2)(l^2+k^2)=m^2l^2+m^2k^2+n^2l^2+n^2k^2=n^2l^2+m^2k^2+m^2l^2+n^2k^2[/tex]
Now, note that [tex]0=2lkmn-2lkmn=2nkml-2nlmk[/tex] by the commutativity of rational integers. Add this convenient zero the the previous equation to obtain [tex]xy=n^2l^2-2nlmk+m^2k^2+m^2l^2+2nkml+n^2k^2=(nl-mk)^2+(ml+nk)^2[/tex], thus xy is the sum of the squares of [tex]nl-mk,ml+nk\in \mathbb{Z}[/tex].