Prove that if two numbers can be written as the sum of two squares then the product of these numbers can also be written as the sum of two squares.
Hello, putnam120!
I happen to be familiar with this theorem,
. . but I don't know of a proof.
However, it also has an extra and amazing feature.
Prove that if two numbers can be written as the sum of two squares,
then the product of these numbers can also be written
as the sum of two squares in two ways.
Let $\displaystyle M \:=\:a^2+b^2,\;N \:=\:c^2+d^2$
. . then: .$\displaystyle M\!\cdot\!N \;= \;(a^2+b^2)(c^2+d^2) \;= \;\begin{Bmatrix}(ac+bd)^2 + (ad-bc)^2 \\ (ac-bd)^2 + (ad+bc)^2\end{Bmatrix} $
Example: .$\displaystyle a = 1,\:b = 2,\:c = 2,\:d = 3$
We have: .$\displaystyle \begin{array}{cc}M \:= \:1^2+2^2\:=\\ N \:=\:2^2+3^2\:=\end{array}
\begin{array}{cc}5\\13\end{array}$
Then: .$\displaystyle 65 \:=\:5\cdot13\:=\:\begin{Bmatrix}8^2+1^2 \\ 4^2+7^2\end{Bmatrix}$
To Soroban: nice well here is a proof that does not use complex numbers.
$\displaystyle (a^2+b^2)(x^2+y^2)=a^2x^2+a^2y^2+b^2x^2+b^2y^2$
then we just add 0 to this and have
$\displaystyle (a^2x^2+2abxy+b^2y^2)+(b^2x^2-2abxy+a^2y^2)=(ax+by)^2+(bx-ay)^2$
you can do the same thing to get the other form.