Talk:Bhaskara's lemma

This is my first page addition on wikipedia. I copied the references from the "parent" page this came from. Is that okay or should references be cited only if they were used to research the article? 05:59, 29 September 2007 User:800km3rk


 * References should help later readers verify statements in the article. So they need to be relevant to this article. --Rumping (talk) 13:57, 14 August 2008 (UTC)

First line of proof
I got lost in the first line of the proof. Where did
 * $$N(mx+y)^2-(my+Nx)^2=-(m^2-N)(y^2-Nx^2)$$

come from? I have provided what I think is a simpler alternative proof, and if there are no objections I will delete the original later. --Rumping (talk) 13:57, 14 August 2008 (UTC)


 * There has been no objection in six weeks so I have removed the original. It follows here in case anybody want to see it. --Rumping (talk) 22:36, 29 September 2008 (UTC)

Earlier version of proof
We begin with an identity, verified by expansion (or substitution into the Brahmagupta-Fibonacci identity with $$a=m,c=y,b=i\sqrt{N},c=ix\sqrt{N}$$) :
 * $$N(mx+y)^2-(my+Nx)^2=-(m^2-N)(y^2-Nx^2)\Longleftrightarrow \frac{N(mx+y)^2-(my+Nx)^2}{y^2-Nx^2}=-(m^2-N).$$

Since $$y^2-Nx^2=k$$, we have that:
 * $$\frac{N(mx+y)^2-(my+Nx)^2}{k}=-(m^2-N)\Longleftrightarrow\frac{N(mx+y)^2-(my+Nx)^2}{k^2}=\frac{-(m^2-N)}{k}.$$

Suitable re-arrangement of this equation yields Bhaskara's Lemma:
 * $$N\left(\frac{mx + y}{k}\right)^2 + \frac{m^2 - N}{k} = \left(\frac{my + Nx}{k}\right)^2.$$