Bump. I'm just curious.

From $(m,n,a,b)$ satisfying $m^2-5n^2=a^2,m^2+5n^2=b^2$ we can generate a new quadrupel $(m',n',a',b')$ by \begin{align*} m'=25n^4+m^4\\ n'=2mnab\\ a'=b^4-2m^4\\ b'=b^4-50n^4 \end{align*}We get from $(41,12,31,49)$ the new quadrupel $(3344161,1494696,113279,4728001)$.

let $m^2-5n^2=x^2, m^2+5n^2=y^2$ let $k=(x+y)/2, l=(y-x)/2$ then $m^2=k^2+l^2,10n^2=4kl$ let $m=s^2+t^2,k=2st,l=s^2-t^2$ then $5(n/2)^2=st(s^2-t^2)$ let $s=u^2,t=5v^2$ then it is enough to show that $u^4-25v^4=w^2$ has a big root $41^4-25*12^2=1519^2$ QED

This is a case of the Congruent number problem. Hence the diophantine equation is equivalent to rational points on the elliptic curve $y^2=x^3-25x$, and a new rational point can be constructed by calculating the intersection with the tangent line at a point(doubling a point). It is a curious fact that the two methods(tangent line/the methods above) give the same answer. Can someone give any insight on why this is the case?