Diophantus’s sum of squares identities

[Guest]

Diophantus’s sum of squares identities

I have these notes written from class, but i dont understand them... can someone help to explain them? (where there is a 2, its to the power 2.)

Diophantus’s sum of squares identities. He Stated in words the fact that if two numbers are sums of two second power, then their product is also a sum of two second powers, and in fact in two ways. The identites are (a2 + b2)(c2+d2) = (ac – bd) 2 + (ad +bc)2, and (a2 + b2 )(c2 + d2 ) = (ac + bd) 2 = (bc – ad) 2 , suitably adjusted if one of the differences here is negative. [We don’t know for certain how he discovered the identities. Probably he noticed patterns after studying many special examples. From a much later point of view these facts can be remembered in terms of the absolute values of complex numbers. The first identity states that |zw|2 = |z|2 |w|2 . The second one expands the square of the modulus of (z times the complex conjugate of w).]

 

[Gene]

^2 is the symbol for the second power.
If you expand the () terms you will see that they are equal (after you correct the typo in the second. (ac+bd)^2 + (bc–ad)^2

 

[soroban]

Re: Diophantus’s sum of squares identities

Hello, aagiants!

I've seen this relationship before, but I didn't know it was Diophantus'.

One of my books says:
. . The product of the sum of two squares by another sum of two squares
. . can be expressed as a sum of two squares in two ways.

And gives the same formula:
. . \(\displaystyle (a^2\,+\,b^2)(c^2\,+d^2)\;=\;(ac\,+\,bd)^2\,+\,(ad\,-\,bc)^2\)
. . . . . . . . . . . . . . . . . \(\displaystyle =\;(ac\,-\,bd)^2\,+\,(ad\,+\,bc)^2\)


For example: \(\displaystyle a=1,\,b=2,\,c=2,\,d=3\) gives us: \(\displaystyle 1^2+2^2\,=\,5\) and \(\displaystyle 2^2+3^2\,=\,13\)

. . Hence: \(\displaystyle \;5\cdot13\;=\;65\;=\;8^2+1^2\;=\;4^2+7^2\)

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Speaking of complex numbers, here's a cure for insomnia . . .

Take any complex number with integer coefficients, say, \(\displaystyle z\:=\:2\,+\,i\)

Take successive powers of \(\displaystyle z\) and calculate \(\displaystyle |z|^2\)

\(\displaystyle n=1:\;(2\,+\,i)^1\:=\:\;2\,\;+\,\;i\;\;\Rightarrow\;\;\;\;2^2\,+\,1^2\:\;\,=\:\;\,5\;=\;5^1\)

\(\displaystyle n=2:\;(2\,+\,1)^2\:=\:\;3\,+\,\;4i\;\;\Rightarrow\;\;\;\;3^2\,+\,4^2\:\;=\:\,25\;=\;5^2\)

\(\displaystyle n=3:\;(2\,+\,i)^3\:=\:\;2\,+\,11i\;\;\Rightarrow\;\;\;\;2^2\,+11^2\:\;=\:125\;=\;5^3\)

\(\displaystyle n=4:\;(2\,+\,i)^4\:=\:\)-\(\displaystyle 7\,+\,24i\;\;\Rightarrow\;\;(\)-\(\displaystyle 7)^2\,+\,24^2\:=\:625\;=\;5^4\)