Answer:
See proof below
Explanation:
One way to solve this problem is to "add a zero" to complete the required squares in the expression of xy.
Let
and
with
. Multiplying the two equations with the distributive law and reordering the result with the commutative law, we get
![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](https://img.qammunity.org/2020/formulas/mathematics/high-school/n5ziox8ezgahdtniwyka5apxwqae2kncmp.png)
Now, note that
by the commutativity of rational integers. Add this convenient zero the the previous equation to obtain
, thus xy is the sum of the squares of
.