Browse Groups

(4)
• NextPrevious
• ... It means I m running The PARI Group s gp package at 3:22am. It s free, it s incredibly powerful, and it answers these kinds of questions without even
Message 1 of 4 , Jan 2, 2003
View Source
--- David Litchfield <mnemonix@...> wrote:
> Cheers for the reply, Phil. Could you explain one thing:
> > (03:22) gp >
> I'm not sure exactly what this means.

It means I'm running The PARI Group's "gp" package at 3:22am.

It's free, it's incredibly powerful, and it answers these kinds of questions
without even breaking a sweat.

> Further, I take it all I had need to have done was
>
> > i.e. if n=(a^2+b^2)/2, then n=(((a+b)/2)^2+((a-b)/2)^2)

I was simply rewording the identity
(a^2+b^2)/2-(((a+b)/2)^2+((a-b)/2)^2) = 0
into a form more similar to your original question.

Phil

=====
The answer to life's mystery is simple and direct:
Sex and death. -- Ian 'Lemmy' Kilminster

__________________________________________________
Do you Yahoo!?
http://mailplus.yahoo.com
• So we can write: (a+b)/2)^2+((a-b)/2)^2 =a^2/4 + ab/2 + b^2/4 + a^2/4 - ab/2 + b^2/4 =a^2/2 + b^2/2 Jon Perry perry@globalnet.co.uk
Message 2 of 4 , Jan 3, 2003
View Source
So we can write:

(a+b)/2)^2+((a-b)/2)^2

=a^2/4 + ab/2 + b^2/4 + a^2/4 - ab/2 + b^2/4

=a^2/2 + b^2/2

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/