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

Paul D Hanna pauldhanna at juno.com
Fri Jan 23 08:33:18 CET 2009

```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

where

a = dn(u)   ,
b = sn(u)*cn(v)  ,
c = dn(v)  ,
d = sn(v)*cn(u)  ,

and k is the elliptic modulus:

cn(u)^2 + sn(u)^2 = 1,
dn(u)^2 + k^2*sn(u)^2 = 1.

This identity involves the Addition Theorem since:

a*c + k^2*b*d  =  dn(u-v)*(1 - k^2*sn(u)^2*sn(v)^2),

a*d - b*c  =  -sn(u-v)*(1 - k^2*sn(u)^2*sn(v)^2)

and

a^2 + k^2*b^2  =  c^2 + k^2*d^2  =  (1 - k^2*sn(u)^2*sn(v)^2) ,

dn(u-v)^2 + k^2*sn(u-v)^2 = 1.

So David's fun little identity is simple but not trivial.
Paul

-- avik roy <avik_3.1416 at yahoo.co.in> wrote:
This identity can be more generalized:

[(n+p)^2 + k][n^2 + k] = [n(n+p) + k]^2 + kp^2;

This identity can contribute to expressing the product of sums of two squares as a sum of two squares [if k is a square].

As an illustration, check:

(3^2 + 2^2)*(10^2 + 2^2)=34^2 + 14^2

```