To solve T^2-W*(2*h+1)^2=1 and Y^2-3*X^2=-(3*(2*h+1)^2*W+2) What procedure does sagemath use? And what is the computational cost?

edit

factorization and generalized Pell equation

In some cases if

    N=p*q

    3*N=M=6*G+3=(2*a+1)^2-(2*b)^2

then

    3*x^2-6*x-4*y^2-4*y=M

it is true that (x+y)= p or -p or q or -q



Example

    N=55

    3*x^2-6*x-4*y^2-4*y=165

    solve 3*x^2-6*x-4*y^2-4*y=165 ,x

    x-1=sqrt[(4*y^2+4*y+168)/3]

    Y=2*y+1 ; X=x-1

    ->

    Y^2-3*X^2=-167

    X=8 ; Y=5

    ->

    x=9 ; y=2

    x+y=11


if we can transform a generic number W in polynomial time into 3*(2*h+1)*W=3*N=3*x^2-6*x-4*y^2-4*y` with `x+y=p or `-p or q or -w
we have solved the factorization problem


so


If we can transforms a generic number W such that (2*h+1)*W=N satisfies

    3*T^2-1 =3*(2*h+1)*W+2

the factorization is quite easy 

to do this I thought of looking for instead of (2*h+1)*W=N to lead to the Pell equation (2*h+1)^2*W=N

    3*T^2-1 =3*(2*h+1)^2*W+2

    T^2-W*(2*h+1)^2=1

So it's Pell again

Example

    W=91

    T^2-W*(2*h+1)^2=1

    T^2-91*(2*h+1)^2=1

    ->

    h=82

    3*x^2-6*x-4*y^2-4*y=3*(2*h+1)^2*W

    3*x^2-6*x-4*y^2-4*y=3*(2*82+1)^2*91

    Y^2-3*X^2=-(3*(2*h+1)^2*W+2)

    Y^2-3*X^2=-(3*(2*82+1)^2*91+2)

    ->

    X=-1574 ; Y=1

    ->

    x=-1573 ;y=0

    gcd(1573,91)=13

Question:

To solve

T^2-W*(2*h+1)^2=1

and

Y^2-3*X^2=-(3*(2*h+1)^2*W+2)

knowing W

What procedure does sagemath use?

And what is the computational cost?