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• Hello all and happy New Year, Could someone tell me, please, if this stands up ? All odd numbers of the form (a^2 + b^2) / 2 with both a and b being odd, that
Message 1 of 4 , Jan 2, 2003
Hello all and happy New Year,
Could someone tell me, please, if this "stands up"?

All odd numbers of the form (a^2 + b^2) / 2 with both a and b being odd,
that is those numbers that lie equidistant between two odd perfect squares,
are the sum of two other perfect squares, one odd and the other even and of
the form ((a+b)/2)^2 and ((a-b)/2)^2

n = (a^2 + b^2) / 2 = x^2 + y^2

let x = (a+b)/2
let y = (a-b)/2

so that

n = (a^2 + b^2) / 2

can be rewritten as

n = ((x + y)^2 + (x - y)^2) / 2

2n = (x + y)^2 + (x - y)^2

2n = (x + y) * (x + y) + (x - y) * (x - y)

2n = x^2 + xy + xy + y^2 + x^2 -xy -xy + y^2

2n = 2x^2 + 2y^2

n = x^2 + y^2

Cheers,
David Litchfield
• ... (03:22) gp (a^2+b^2)/2 - (((a+b)/2)^2+((a-b)/2)^2) %1 = 0 i.e. if n=(a^2+b^2)/2, then n=(((a+b)/2)^2+((a-b)/2)^2) Phil ===== The answer to life s mystery
Message 2 of 4 , Jan 2, 2003
--- David Litchfield <Mnemonix@...> wrote:
> Hello all and happy New Year,
> Could someone tell me, please, if this "stands up"?
>
> All odd numbers of the form (a^2 + b^2) / 2 with both a and b being odd,
> that is those numbers that lie equidistant between two odd perfect squares,
> are the sum of two other perfect squares, one odd and the other even and of
> the form ((a+b)/2)^2 and ((a-b)/2)^2

(03:22) gp > (a^2+b^2)/2 - (((a+b)/2)^2+((a-b)/2)^2)
%1 = 0

i.e. if n=(a^2+b^2)/2, then n=(((a+b)/2)^2+((a-b)/2)^2)

Phil

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

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• ... 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 3 of 4 , Jan 2, 2003
--- 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

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Do you Yahoo!?
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• 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 4 of 4 , Jan 3, 2003
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/