[seqfan] Re: Fun little identity (to me anyway)

[Mitch Harris]

On Fri, Jan 23, 2009 at 2:33 AM, Paul D Hanna <pauldhanna at juno.com> wrote:
> Seqfans,
>
> This is directly related to an identity of Jacobi Elliptic functions:
>
> (a^2 + k^2*b^2)*(c^2 + k^2*d^2) = (a*c + k^2*b*d)^2 + k^2*(a*d - b*c)^2

To further the connections, that specializes to the very simple
Fibonacci identity, 'the product of the sum of squares is a sum of
squares',

  (a^2 + b^2)(c^2 + d^2) = (a c + b d)^2 + (a d - b c)^2

(which not coincidentally is what happens in multiplication of complex numbers).

So, for the generalization to quaternions (Euclid's 4-square identity:

(a^2 + b^2 + c^2 + d^2) (e^2 + f^2 + g^2 + h^2) =

(a e - b f - c g - d h)^2 +
(a f + b e + c h - d g)^2 +
(a g - b h + c e + d f)^2 +
(a h + b g - c f + d e)^2

I wonder what the corresponding nice Wilson-style identity is.

d^2 = h^2 = k? probably not...

(a h + b g - c f + d e)^2 = k? more likely.


And then whatever from that, what is the corresponding sequence?

-- 
Mitch Harris