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

[Paul D Hanna]

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