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Show that if x and y may both be written as the sum of the squares of two rational integers, then their product xy may also be written as the sum of the squares of two rational integers.

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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
x=m^2+n^2 and
y=l^2+k^2 with
m,n,l,k\in \mathbb{Z}. 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

Now, note that
0=2lkmn-2lkmn=2nkml-2nlmk by the commutativity of rational integers. Add this convenient zero the the previous equation to obtain
xy=n^2l^2-2nlmk+m^2k^2+m^2l^2+2nkml+n^2k^2=(nl-mk)^2+(ml+nk)^2, thus xy is the sum of the squares of
nl-mk,ml+nk\in \mathbb{Z}.

User Bufh
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