Is there any odd $n\in \mathbb{N}$ satisfying $n=e^2-f^2=g^2-h^2=k^2-l^2$ and $e^2f^2=g^2h^2+k^2l^2$ where $e, f, g, h, k, l~(>1)\in \mathbb{N}$?

Question

Is there any odd natural number $n$ satisfying $n=e^2-f^2=g^2-h^2=k^2-l^2$ and $e^2f^2=g^2h^2+k^2l^2$ where $e, f, g, h, k, l~(>1)$ are natural numbers? The problem is related to famous open problem ( around for several centuries) on Pythagorean triples.

Answer 1

$(a^2-b^2)=(c^2-d^2)=(e^2-f^2)$

Above has parametric form:

$(a,b,c,d,e,f)$=

$[mn+uv,mn-uv,mu+nv,mu-nv,mv+nu,mv-nu]$ -(1) We need:

$(ef)^2=(ab)^2-(cd)^2$ ---(2)

Substituting, (1) in (2) we get: $(m^2n^2+u^2v^2)^2=(m^4+n^4)(u^4+v^4)$

We have numerical solution,

& $(m,n)^4=(u,v)^4$

$(134,133)^4=(158,59)^4$ --(3)

Hence: $(m^2n^2+u^2v^2)=(m^4+n^4)$ ---(4)

After substituting, (3) in (4) we find: $(m^2n^2+u^2v^2)$ is not equal to $(m^4+n^4)$.

Hence in all probability the equation proposed by "OP" does not have a numerical solution.