not to difficult

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Nov 18th 2006, 04:25 PM
putnam120

not to difficult

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.
Nov 18th 2006, 05:57 PM
Soroban

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.

Quote:

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 $M \:=\:a^2+b^2,\;N \:=\:c^2+d^2$
. . then: . $M\!\cdot\!N \;= \;(a^2+b^2)(c^2+d^2) \;= \;\begin{Bmatrix}(ac+bd)^2 + (ad-bc)^2 \\ (ac-bd)^2 + (ad+bc)^2\end{Bmatrix}$

Example: . $a = 1,\:b = 2,\:c = 2,\:d = 3$

We have: . $\begin{array}{cc}M \:= \:1^2+2^2\:=\\ N \:=\:2^2+3^2\:=\end{array}<br /> \begin{array}{cc}5\\13\end{array}$

Then: . $65 \:=\:5\cdot13\:=\:\begin{Bmatrix}8^2+1^2 \\ 4^2+7^2\end{Bmatrix}$
Nov 18th 2006, 06:01 PM
ThePerfectHacker

Soroban it is an error to say "IN TWO WAYS" because what will happen if $N,M$ are representable as sum of two squares in more then 1 way :eek:.

The purpose of this is two show that is it possible not how many times.
Nov 18th 2006, 08:51 PM
putnam120

To Soroban: nice well here is a proof that does not use complex numbers.

$(a^2+b^2)(x^2+y^2)=a^2x^2+a^2y^2+b^2x^2+b^2y^2$

then we just add 0 to this and have

$(a^2x^2+2abxy+b^2y^2)+(b^2x^2-2abxy+a^2y^2)=(ax+by)^2+(bx-ay)^2$

you can do the same thing to get the other form.