Is this type of nonlinear system known to factorize N ? What is the best way (computationally speaking) to solve it ?
If N=8G+3 -> N=(4x+2)^2-(2y-1)^2 -> (4x+2)-(2y-1)=p && (4x+2)+(2*y-1)=q
2^i < x < 2^(i+1)
2^j < y < 2^(j+1)
the nonlinear system will have i+j+1 equations and i+j unknowns
and will be composed of the equations
sqrt(16x^2+16x+4-N)-(2*y-1) == 0
(one equation)
(N+(2y-1)^2-4)/4+1-2(2x+1) +1-(4x1+2)^2 == 0
(one equation)
((4x(h)+2)^2-4)/4+1-2(2x(h)+1) +1-(4x(h+1)+2)^2 == 0
with h from 1 to i-2
(i-2 equations)
((4x(i-1)+2)^2-4)/4+1-2(2*x(i-1)+1) +1-36 == 0
(one equation)
(N+(2y-1)^2-16(x-y)(x+y+1)-4)/4+1-2(2y+1) +1-(4y1+2)^2 == 0 (se y < x)
or
(N+(2y-1)^2-16(x-y)(x+y+1)-4)/4+1-2(2y+1) +1-(4y1+2)^2 == 0 (se y < x)
(one equation)
((4y(k)+2)^2-4)/4+1-2(2y(k)+1) +1-(4y(k+1)+2)^2 == 0
with k from 1 to j-2
(j-2 equations)
((4y(j-1)+2)^2-4)/4+1-2(2*y(j-1)+1) +1-36 == 0
(one equation)
Example N=18483 (system written in sagemath)
import time
Start_Time = time.time()
var('N x y x1 x2 x3 x4 y1 y2 y3')
eq0 = (4*35+2)^2-(2*21-1)^2 -N == 0
eq1 = sqrt(16*x^2+16*x+4-N)-(2*y-1) == 0
eq2 = (N+(2*y-1)^2-4)/4+1-2*(2*x+1) +1-(4*x1+2)^2 == 0
eq3 = ((4*x1+2)^2-4)/4+1-2*(2*x1+1) +1-(4*x2+2)^2 == 0
eq4 = ((4*x2+2)^2-4)/4+1 -2*(2*x2+1) +1-(4*x3+2)^2 == 0
eq5 = ((4*x3+2)^2-4)/4+1 -2*(2*x3+1) +1-(4*x4+2)^2 == 0
eq6 = ((4*x4+2)^2-4)/4+1-2*(2*x4+1) +1-36 == 0
eq7 = (N+(2*y-1)^2-16*(x-y)*(x+y+1)-4)/4+1-2*(2*y+1) +1-(4*y1+2)^2 == 0
eq8 = ((4*y1+2)^2-4)/4+1-2*(2*y1+1) +1-(4*y2+2)^2 == 0
eq9 = ((4*y2+2)^2-4)/4+1 -2*(2*y2+1) +1-(4*y3+2)^2 == 0
eq10 = ((4*y3+2)^2-4)/4+1-2*(2*y3+1) +1-36 == 0
solutions = solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10],N,x,y,x1,x2,x3,x4,y1,y2,y3)
sol = solutions
Execution_Time = time.time() - Start_Time
print (Execution_Time)
print(sol)
Is this type of nonlinear system known to factorize N ?
What is the best way (computationally speaking) to solve it ?