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My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

Here you will find my work:

https://www.linkedin.com/feed/update/urn:li:ugcPost:6777160628965535744/

or

https://github.com/Piunosei/gerva_factorization/blob/main/Transformation-of-the-factorization-problem-into-a-problem-of-solving-a-non-linear-system%20(5).pdf

Is written:

• in an almost unpresentable way

• the case x odd is missing

• (*) and is redundant

(*) as it is possible to establish the range of log_2 (x) and log_2 (y)

My problem is to solve the resulting nonlinear system.

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

Here you will find my work:

https://www.linkedin.com/feed/update/urn:li:ugcPost:6777160628965535744/

or

https://github.com/Piunosei/gerva_factorization/blob/main/Transformation-of-the-factorization-problem-into-a-problem-of-solving-a-non-linear-system%20(5).pdf

Is written:

• in an almost unpresentable way

• the case x odd is missing

• (*) and is redundant

(*) as it is possible to establish the range of log_2 (x) and log_2 (y)

My problem is to solve the resulting nonlinear system.

var('x y M a b c d A B C D t u v S T U V')
eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0  

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 -(u-1)*(u+1)/8 == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0

eq18 = -M+9*((2*3367 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D - 3*d + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + (3*d*(d - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq28 = 3*((2*S - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0



solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq28],x,y,M,a,b,c,d,A,B,C,D,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])
print(sol0[13])
print(sol0[14])
print(sol0[15])
print(sol0[16])
print(sol0[17])

(x, y, M, a, b, c, d, A, B, C, D, t, u, v, S, T, U, V)
x == 24
y == 13
M == 2701
a == 13
b == 7
c == -3
d == -1
A == 3367
B == 3538
C == 3583
D == 3598
t == -1
u == -3
v == 7
S == 3601
T == 3598
U == 3583
V == 3538

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

Here you will find my work:

https://www.linkedin.com/feed/update/urn:li:ugcPost:6777160628965535744/

or

https://github.com/Piunosei/gerva_factorization/blob/main/Transformation-of-the-factorization-problem-into-a-problem-of-solving-a-non-linear-system%20(5).pdf

Is written:

• in an almost unpresentable way

• the case x odd is missing

• (*) and is redundant

(*) as it is possible to establish the range of log_2 (x) and log_2 (y)

My problem is to solve the resulting nonlinear system.

UPDATE1:

In practice, this non-linear system must be resolved to be acceptable as a function of the number of variables

the number of variables is

2 * IntegerPart [log_2 (y)] + 2 * IntegerPart [log_2 (x)] + 4

where 8 * x + 4 = p + q and 4 * y-2 = q-p

var('x y M a b c d A B C D t u v S T U V')
eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0  

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 -(u-1)*(u+1)/8 == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0

eq18 = -M+9*((2*3367 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D - 3*d + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + (3*d*(d - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq28 = 3*((2*S - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0



solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq28],x,y,M,a,b,c,d,A,B,C,D,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])
print(sol0[13])
print(sol0[14])
print(sol0[15])
print(sol0[16])
print(sol0[17])

OUTPUT

(x, y, M, a, b, c, d, A, B, C, D, t, u, v, S, T, U, V)
x == 24
y == 13
M == 2701
a == 13
b == 7
c == -3
d == -1
A == 3367
B == 3538
C == 3583
D == 3598
t == -1
u == -3
v == 7
S == 3601
T == 3598
U == 3583
V == 3538

var('x y a b c d t u v S T U V')
eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0  

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 -(u-1)*(u+1)/8 == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0

eq19 = d+1 == 0
eq20 = t+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq19,eq20],x,y,a,b,c,d,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])

(x, y, M, a, b, c, d, t, u, v, S, T, U, V, X)
x == 24
y == 13
a == 13
b == 7
c == -3
d == -1
t == -1
u == -3
v == 7
S == 3601
T == 3598
U == 3583
V == 3538

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

Here you will find my work:

https://www.linkedin.com/feed/update/urn:li:ugcPost:6777160628965535744/

or

https://github.com/Piunosei/gerva_factorization/blob/main/Transformation-of-the-factorization-problem-into-a-problem-of-solving-a-non-linear-system%20(5).pdf

Is written:

• in an almost unpresentable way

• the case x odd is missing

• (*) and is redundant

(*) as it is possible to establish the range of log_2 (x) and log_2 (y)

My problem is to solve the resulting nonlinear system.

UPDATE1:

In practice, this non-linear system must be resolved to be acceptable as a function of the number of variables

the number of variables is

2 * IntegerPart [log_2 (y)] + 2 * IntegerPart [log_2 (x)] + 4

where 8 * x + 4 = p + q and 4 * y-2 = q-p

var('x y M a b c d A B C D t u v S T U V')
eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0  

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 -(u-1)*(u+1)/8 == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0

eq18 = -M+9*((2*3367 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D - 3*d + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + (3*d*(d - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq28 = 3*((2*S - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0



solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq28],x,y,M,a,b,c,d,A,B,C,D,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])
print(sol0[13])
print(sol0[14])
print(sol0[15])
print(sol0[16])
print(sol0[17])

OUTPUT

(x, y, M, a, b, c, d, A, B, C, D, t, u, v, S, T, U, V)
x == 24
y == 13
M == 2701
a == 13
b == 7
c == -3
d == -1
A == 3367
B == 3538
C == 3583
D == 3598
t == -1
u == -3
v == 7
S == 3601
T == 3598
U == 3583
V == 3538

UPDATE2;

I think I have simplified the system

var('x y a b c d t u v S T U V')
eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0  

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 -(u-1)*(u+1)/8 == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0

eq19 = d+1 == 0
eq20 = t+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq19,eq20],x,y,a,b,c,d,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])

OUTPUT

(x, y, M, a, b, c, d, t, u, v, S, T, U, V, X)
x == 24
y == 13
a == 13
b == 7
c == -3
d == -1
t == -1
u == -3
v == 7
S == 3601
T == 3598
U == 3583
V == 3538

var('x y a b c d e f g h p q r s t u v O P Q R S T U V')

eq1 = 3*(((2*481501-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -481501+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*s+1)/24)+3*x*(x+1)/2)+1-R == 0  
eq10 = S+3*s*(s-1)/2-12*x*(x+1)/2-1 == 0  
eq11 = 3*(((2*R-3*r+1)/24)+3*x*(x+1)/2)+1-Q == 0  
eq12 = R+3*r*(r-1)/2-12*x*(x+1)/2-1 ==0  
eq13 = 3*(((2*Q-3*q+1)/24)+3*x*(x+1)/2)+1-P == 0  
eq14 = Q+3*q*(q-1)/2-12*x*(x+1)/2-1 == 0  
eq15 = 3*(((2*P-3*p+1)/24)+3*x*(x+1)/2)+1-O == 0
eq16 = P+3*p*(p-1)/2-12*x*(x+1)/2-1 == 0  
eq17 = 3*(((2*O-3*1+1)/24)+3*x*(x+1)/2)+1-O == 0

eq26 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq27 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq28 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq29 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0
eq30 = e*(e-1)/2 -(d-1)*(d+1)/8 == 0
eq31 = f*(f-1)/2 -(e-1)*(e+1)/8 == 0
eq32 = g*(g-1)/2 -(f-1)*(f+1)/8 == 0
eq33 = h*(h-1)/2 -(g-1)*(g+1)/8 == 0
eq34 = h+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq17,eq26,eq27,eq28,eq29,eq30,eq31,eq32,eq33,eq34],x,y,a,b,c,d,e,f,g,h,p,q,r,s,t,u,v,O,P,Q,R,S,T,U,V)
sol = solutions[0]
print(sol[0])
print(sol[1])

(x, y, a, b, c, d, e, f, g, h, p, q, r, s, t, u, v, O, P, Q, R, S, T, U, V)
x == 300
y == 201

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

Here you will find my work:

https://www.linkedin.com/feed/update/urn:li:ugcPost:6777160628965535744/

or

https://github.com/Piunosei/gerva_factorization/blob/main/Transformation-of-the-factorization-problem-into-a-problem-of-solving-a-non-linear-system%20(5).pdf

Is written:

• in an almost unpresentable way

• the case x odd is missing

• (*) and is redundant

(*) as it is possible to establish the range of log_2 (x) and log_2 (y)

My problem is to solve the resulting nonlinear system.

UPDATE1:

In practice, this non-linear system must be resolved to be acceptable as a function of the number of variables

the number of variables is

2 * IntegerPart [log_2 (y)] + 2 * IntegerPart [log_2 (x)] + 4

where 8 * x + 4 = p + q and 4 * y-2 = q-p

var('x y M a b c d A B C D t u v S T U V')
eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0  

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 -(u-1)*(u+1)/8 == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0

eq18 = -M+9*((2*3367 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D - 3*d + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + (3*d*(d - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq28 = 3*((2*S - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0



solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq28],x,y,M,a,b,c,d,A,B,C,D,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])
print(sol0[13])
print(sol0[14])
print(sol0[15])
print(sol0[16])
print(sol0[17])

OUTPUT

(x, y, M, a, b, c, d, A, B, C, D, t, u, v, S, T, U, V)
x == 24
y == 13
M == 2701
a == 13
b == 7
c == -3
d == -1
A == 3367
B == 3538
C == 3583
D == 3598
t == -1
u == -3
v == 7
S == 3601
T == 3598
U == 3583
V == 3538

UPDATE2;

I think I have simplified the system

var('x y a b c d t u v S T U V')
eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0  

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 -(u-1)*(u+1)/8 == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0

eq19 = d+1 == 0
eq20 = t+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq19,eq20],x,y,a,b,c,d,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])

OUTPUT

(x, y, M, a, b, c, d, t, u, v, S, T, U, V, X)
x == 24
y == 13
a == 13
b == 7
c == -3
d == -1
t == -1
u == -3
v == 7
S == 3601
T == 3598
U == 3583
V == 3538

UPDATE3:

I reduced the system again

With my computer (a bit old) it takes 8 minutes to solve with N = 1284003

my computer

Intel(R) Core(TM)2 Duo CPU E7300 @ 2.66GHz

RAM 4,00 GB

var('x y a b c d e f g h p q r s t u v O P Q R S T U V')

eq1 = 3*(((2*481501-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -481501+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*s+1)/24)+3*x*(x+1)/2)+1-R == 0  
eq10 = S+3*s*(s-1)/2-12*x*(x+1)/2-1 == 0  
eq11 = 3*(((2*R-3*r+1)/24)+3*x*(x+1)/2)+1-Q == 0  
eq12 = R+3*r*(r-1)/2-12*x*(x+1)/2-1 ==0  
eq13 = 3*(((2*Q-3*q+1)/24)+3*x*(x+1)/2)+1-P == 0  
eq14 = Q+3*q*(q-1)/2-12*x*(x+1)/2-1 == 0  
eq15 = 3*(((2*P-3*p+1)/24)+3*x*(x+1)/2)+1-O == 0
eq16 = P+3*p*(p-1)/2-12*x*(x+1)/2-1 == 0  
eq17 = 3*(((2*O-3*1+1)/24)+3*x*(x+1)/2)+1-O == 0

eq26 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq27 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq28 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq29 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0
eq30 = e*(e-1)/2 -(d-1)*(d+1)/8 == 0
eq31 = f*(f-1)/2 -(e-1)*(e+1)/8 == 0
eq32 = g*(g-1)/2 -(f-1)*(f+1)/8 == 0
eq33 = h*(h-1)/2 -(g-1)*(g+1)/8 == 0
eq34 = h+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq17,eq26,eq27,eq28,eq29,eq30,eq31,eq32,eq33,eq34],x,y,a,b,c,d,e,f,g,h,p,q,r,s,t,u,v,O,P,Q,R,S,T,U,V)
sol = solutions[0]
print(sol[0])
print(sol[1])

OUTPUT

(x, y, a, b, c, d, e, f, g, h, p, q, r, s, t, u, v, O, P, Q, R, S, T, U, V)
x == 300
y == 201

Could you kindly try it on a new computer?

var('x y M a b c d e f g h A B C D E F G H p q r s t u v O P Q R S T U V')

eq1 = 3*(((2*481501-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -481501+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*s+1)/24)+3*x*(x+1)/2)+1-R == 0  
eq10 = S+3*s*(s-1)/2-12*x*(x+1)/2-1 == 0  
eq11 = 3*(((2*R-3*r+1)/24)+3*x*(x+1)/2)+1-Q == 0  
eq12 = R+3*r*(r-1)/2-12*x*(x+1)/2-1 ==0  
eq13 = 3*(((2*Q-3*q+1)/24)+3*x*(x+1)/2)+1-P == 0  
eq14 = Q+3*q*(q-1)/2-12*x*(x+1)/2-1 == 0  
eq15 = 3*(((2*P-3*p+1)/24)+3*x*(x+1)/2)+1-O == 0
eq16 = P+3*p*(p-1)/2-12*x*(x+1)/2-1 == 0  
eq17 = 3*(((2*O-3*1+1)/24)+3*x*(x+1)/2)+1-O == 0

eq35 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq36 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq37 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq38 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq39 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq40 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq41 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq42 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq43 = 3*((2*D - 3*d + 1)/24 + (3*x*(x + 1))/2) + 1 - E == 0
eq44 = D + (3*d*(d - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq45 = 3*((2*E - 3*e + 1)/24 + (3*x*(x + 1))/2) + 1 - F == 0
eq46 = E + (3*e*(e - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq47 = 3*((2*F - 3*f + 1)/24 + (3*x*(x + 1))/2) + 1 - G == 0
eq48 = F + (3*f*(f - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq49 = 3*((2*G - 3*g + 1)/24 + (3*x*(x + 1))/2) + 1 - H == 0
eq50 = G + (3*g*(g - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq51 = 3*((2*H - 3*h + 1)/24 + (3*x*(x + 1))/2) + 1 - O == 0
eq52 = H + (3*h*(h - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq53 = 3*((2*O - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - O == 0
eq54 = -M+9*((2*481501 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq17,eq35,eq36,eq37,eq38,eq39,eq40,eq41,eq42,eq43,eq44,eq45,eq46,eq47,eq48,eq49,eq50,eq51,eq52,eq53,eq54],x,y,M,a,b,c,d,e,f,g,h,A,B,C,D,E,F,G,H,p,q,r,s,t,u,v,O,P,Q,R,S,T,U,V)
sol = solutions
print(sol)

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

Here you will find my work:

https://www.linkedin.com/feed/update/urn:li:ugcPost:6777160628965535744/

or

https://github.com/Piunosei/gerva_factorization/blob/main/Transformation-of-the-factorization-problem-into-a-problem-of-solving-a-non-linear-system%20(5).pdf

Is written:

• in an almost unpresentable way

• the case x odd is missing

• (*) and is redundant

(*) as it is possible to establish the range of log_2 (x) and log_2 (y)

My problem is to solve the resulting nonlinear system.

UPDATE1:

In practice, this non-linear system must be resolved all small variables belong to be acceptable as a function of the number of variables

the number of Z

all big variables is

2 belong to N

If N = p * IntegerPart [log_2 (y)] + 2 * IntegerPart [log_2 (x)] + 4

where q and p + q = 8 * x + 4 = p + q and then q-p = 4 * y-2 = q-pthen

p = (4x+1-2(y-1))

q = (4x+3+2(y-1))

let's study the case

x is even and y is odd

Examples N=8979

(x+1+1)/2 is even then a = 1-(x+1+1)/2

(x+1+1)/2 is even odd a = (x+1+1)/2

(|a|+1)/2 is even then b = 1-(|a|+1)/2

(|a|+1)/2 is even odd b = (|a|+1)/2

(|b|+1)/2 is even then c = 1-(|b|+1)/2

(|b|+1)/2 is even odd c = (|b|+1)/2

(|c|+1)/2 is even then -1 = 1-(|c|+1)/2

(|c|+1)/2 is even odd -1 = (|c|+1)/2

then

16 <= x <= 30

S>D

D>C

C>B

B>A

A>M

and

(y+1)/2 is even then v = 1-(y+1)/2

(y+1)/2 is even odd v = (y+1)/2

(|v|+1)/2 is even then u = 1-(|v|+1)/2

(|v|+1)/2 is even odd u = (|v|+1)/2

(|u|+1)/2 is even then -1 = 1-(|u|+1)/2

(|u|+1)/2 is even odd -1 = (|u|+1)/2

then

9 <= y <= 15

S>T

T>U

U>V

V>(3*N-1)/8

then this system will have a single solution

var('x var('N p q x y M a b c d A B C D t u v S T U V')
eq0 = N-8979 == 0

eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V 3*(((2*(3*N-1)/8-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 -(3*N-1)/8+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S 3*(((2*T+3+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 T+3-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0   

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 1 -(u-1)*(u+1)/8 == 0
 eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 1 -(c-1)*(c+1)/8 == 0

eq18 = -M+9*((2*3367 -M+9*((2*(3*N-1)/8 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D - 3*d + 3 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + (3*d*(d - 3 - (12*x*(x + 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq28 = 3*((2*S - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0

 eq29 = p - (4*x+1-2*(y-1)) == 0
eq30 = q - (4*x+3+2*(y-1)) == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq28],x,y,M,a,b,c,d,A,B,C,D,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])
print(sol0[13])
print(sol0[14])
print(sol0[15])
print(sol0[16])
print(sol0[17])
solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq28,eq29,eq30],N,p,q,x,y,M,a,b,c,A,B,C,D,u,v,S,T,U,V)
sol = solutions 
print(sol)

OUTPUT

(x, (N, p, q, x, y, M, a, b, c, d, A, B, C, D, t, u, v, S, T, U, V)
[
[N == 8979, p == 73, q == 123, x == 24
24, y == 13
13, M == 2701
2701, a == 13
13, b == 7
7, c == -3
d == -1
-3, A == 3367
3367, B == 3538
3538, C == 3583
3583, D == 3598
t == -1
3598, u == -3
-3, v == 7
7, S == 3601
3601, T == 3598
3598, U == 3583
3583, V == 3538
3538]
]

UPDATE2;

I think I have simplified the system

var('x y a b c d t u v S T U V')
eq1 = 3*(((2*3367-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -3367+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0  

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = t*(t-1)/2 -(u-1)*(u+1)/8 == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0

eq19 = d+1 == 0
eq20 = t+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq19,eq20],x,y,a,b,c,d,t,u,v,S,T,U,V)
sol0 = solutions[0] 
print(sol0[0])
print(sol0[1])
print(sol0[2])
print(sol0[3])
print(sol0[4])
print(sol0[5])
print(sol0[6])
print(sol0[7])
print(sol0[8])
print(sol0[9])
print(sol0[10])
print(sol0[11])
print(sol0[12])

OUTPUT

(x, y, M, a, b, c, d, t, u, v, S, T, U, V, X)
x == 24
y == 13
a == 13
b == 7
c == -3
d == -1
t == -1
u == -3
v == 7
S == 3601
T == 3598
U == 3583
V == 3538

UPDATE3:

I reduced the system again

With my computer (a bit old) it takes 8 minutes Will you kindly help me to solve with N = 1284003

my computer

Intel(R) Core(TM)2 Duo CPU E7300 @ 2.66GHz

RAM 4,00 GB

var('x y a b c d e f g h p q r s t u v O P Q R S T U V')

eq1 = 3*(((2*481501-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -481501+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*s+1)/24)+3*x*(x+1)/2)+1-R == 0  
eq10 = S+3*s*(s-1)/2-12*x*(x+1)/2-1 == 0  
eq11 = 3*(((2*R-3*r+1)/24)+3*x*(x+1)/2)+1-Q == 0  
eq12 = R+3*r*(r-1)/2-12*x*(x+1)/2-1 ==0  
eq13 = 3*(((2*Q-3*q+1)/24)+3*x*(x+1)/2)+1-P == 0  
eq14 = Q+3*q*(q-1)/2-12*x*(x+1)/2-1 == 0  
eq15 = 3*(((2*P-3*p+1)/24)+3*x*(x+1)/2)+1-O == 0
eq16 = P+3*p*(p-1)/2-12*x*(x+1)/2-1 == 0  
eq17 = 3*(((2*O-3*1+1)/24)+3*x*(x+1)/2)+1-O == 0

eq26 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq27 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq28 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq29 = d*(d-1)/2 -(c-1)*(c+1)/8 == 0
eq30 = e*(e-1)/2 -(d-1)*(d+1)/8 == 0
eq31 = f*(f-1)/2 -(e-1)*(e+1)/8 == 0
eq32 = g*(g-1)/2 -(f-1)*(f+1)/8 == 0
eq33 = h*(h-1)/2 -(g-1)*(g+1)/8 == 0
eq34 = h+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq17,eq26,eq27,eq28,eq29,eq30,eq31,eq32,eq33,eq34],x,y,a,b,c,d,e,f,g,h,p,q,r,s,t,u,v,O,P,Q,R,S,T,U,V)
sol = solutions[0]
print(sol[0])
print(sol[1])

OUTPUT

(x, y, a, b, c, d, e, f, g, h, p, q, r, s, t, u, v, O, P, Q, R, S, T, U, V)
x == 300
y == 201

Could you kindly try it on a new computer?

UPDATE4:

I think that once we have established the range of x and the range of y we have to exploit the inequalities.

For example this non-linear system in the underlying system we have to exploit:

256 <= x <= 510

129 <= y <= 257

O>P

P>Q

Q>R

R>S

S>T

T>U

U>V

V>481501

O>H

H>G

G>F

F>E

E>D

D>C

C>B

B>A

A>M

var('x y M a b c d e f g h A B C D E F G H p q r s t u v O P Q R S T U V')

eq1 = 3*(((2*481501-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -481501+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T-3*t+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3*t*(t-1)/2-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*s+1)/24)+3*x*(x+1)/2)+1-R == 0  
eq10 = S+3*s*(s-1)/2-12*x*(x+1)/2-1 == 0  
eq11 = 3*(((2*R-3*r+1)/24)+3*x*(x+1)/2)+1-Q == 0  
eq12 = R+3*r*(r-1)/2-12*x*(x+1)/2-1 ==0  
eq13 = 3*(((2*Q-3*q+1)/24)+3*x*(x+1)/2)+1-P == 0  
eq14 = Q+3*q*(q-1)/2-12*x*(x+1)/2-1 == 0  
eq15 = 3*(((2*P-3*p+1)/24)+3*x*(x+1)/2)+1-O == 0
eq16 = P+3*p*(p-1)/2-12*x*(x+1)/2-1 == 0  
eq17 = 3*(((2*O-3*1+1)/24)+3*x*(x+1)/2)+1-O == 0

eq35 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq36 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq37 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq38 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq39 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq40 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq41 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq42 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq43 = 3*((2*D - 3*d + 1)/24 + (3*x*(x + 1))/2) + 1 - E == 0
eq44 = D + (3*d*(d - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq45 = 3*((2*E - 3*e + 1)/24 + (3*x*(x + 1))/2) + 1 - F == 0
eq46 = E + (3*e*(e - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq47 = 3*((2*F - 3*f + 1)/24 + (3*x*(x + 1))/2) + 1 - G == 0
eq48 = F + (3*f*(f - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq49 = 3*((2*G - 3*g + 1)/24 + (3*x*(x + 1))/2) + 1 - H == 0
eq50 = G + (3*g*(g - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq51 = 3*((2*H - 3*h + 1)/24 + (3*x*(x + 1))/2) + 1 - O == 0
eq52 = H + (3*h*(h - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq53 = 3*((2*O - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - O == 0
eq54 = -M+9*((2*481501 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0

solutions = solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq17,eq35,eq36,eq37,eq38,eq39,eq40,eq41,eq42,eq43,eq44,eq45,eq46,eq47,eq48,eq49,eq50,eq51,eq52,eq53,eq54],x,y,M,a,b,c,d,e,f,g,h,A,B,C,D,E,F,G,H,p,q,r,s,t,u,v,O,P,Q,R,S,T,U,V)
sol = solutions
print(sol)

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

Here you will find my work:

https://www.linkedin.com/feed/update/urn:li:ugcPost:6777160628965535744/

or

https://github.com/Piunosei/gerva_factorization/blob/main/Transformation-of-the-factorization-problem-into-a-problem-of-solving-a-non-linear-system%20(5).pdf

Is written:

• in an almost unpresentable way

• the case x odd is missing

• (*) and is redundant

(*) as it is possible to establish the range of log_2 (x) and log_2 (y)

My problem is to solve the resulting nonlinear system.

all small variables belong to Z

all big variables belong to N

If N = p * q and p + q = 8 * x + 4 then q-p = 4 * y-2 then

p = (4x+1-2(y-1))

q = (4x+3+2(y-1))

let's study the case

x is even and y is odd

Examples N=8979

(x+1+1)/2 is even then a = 1-(x+1+1)/2

(x+1+1)/2 is even odd a = (x+1+1)/2

(|a|+1)/2 is even then b = 1-(|a|+1)/2

(|a|+1)/2 is even odd b = (|a|+1)/2

(|b|+1)/2 is even then c = 1-(|b|+1)/2

(|b|+1)/2 is even odd c = (|b|+1)/2

(|c|+1)/2 is even then -1 = 1-(|c|+1)/2

(|c|+1)/2 is even odd -1 = (|c|+1)/2

then

16 <= x <= 30

S>D

D>C

C>B

B>A

A>M

and

(y+1)/2 is even then v = 1-(y+1)/2

(y+1)/2 is even odd v = (y+1)/2

(|v|+1)/2 is even then u = 1-(|v|+1)/2

(|v|+1)/2 is even odd u = (|v|+1)/2

(|u|+1)/2 is even then -1 = 1-(|u|+1)/2

(|u|+1)/2 is even odd -1 = (|u|+1)/2

then

9 <= y <= 15

S>T

T>U

U>V

V>(3*N-1)/8

then this system will have a single solution

var('N p q x y M a b c A B C D u v S T U V')
eq0 = N-8979 == 0

eq1 = 3*(((2*(3*N-1)/8-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -(3*N-1)/8+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T+3+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3-12*x*(x+1)/2-1 == 0  
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0   

eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 == 0
eq12 = 1 -(u-1)*(u+1)/8 == 0

eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = 1 -(c-1)*(c+1)/8 == 0

eq18 = -M+9*((2*(3*N-1)/8 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D + 3 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + 3 - (12*x*(x + 1))/2 - 1 == 0
eq28 = 3*((2*S - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0

eq29 = p - (4*x+1-2*(y-1)) == 0
eq30 = q - (4*x+3+2*(y-1)) == 0

solutions = solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq28,eq29,eq30],N,p,q,x,y,M,a,b,c,A,B,C,D,u,v,S,T,U,V)
sol = solutions 
print(sol)

OUTPUT

(N, p, q, x, y, M, a, b, c, A, B, C, D, u, v, S, T, U, V)
[
[N == 8979, p == 73, q == 123, x == 24, y == 13, M == 2701, a == 13, b == 7, c == -3, A == 3367, B == 3538, C == 3583, D == 3598, u == -3, v == 7, S == 3601, T == 3598, U == 3583, V == 3538]
]

Will you kindly help me to solve this non-linear system in the lowest possible computational cost?

import time

Start_Time = time.time()
var('N p q x y M a b c A B C D u v S T U V')
eq0 = N-8979 == 0

eq1 = 3*(((2*(3*N-1)/8-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -(3*N-1)/8+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T+3+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3-12*x*(x+1)/2-1 == 0  


eq18 = -M+9*((2*(3*N-1)/8 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D + 3 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + 3 - (12*x*(x + 1))/2 - 1 == 0


eq29 = p - (4*x+1-2*(y-1)) == 0
eq30 = q - (4*x+3+2*(y-1)) == 0

eq31 =(3*N-1)/8+3*(y*(y-1)/2-(v^2-v)/2) -V == 0
eq32 =(3*N-1)/8+3*(y*(y-1)/2-(u^2-u)/2) -U == 0
eq33 =(3*N-1)/8+3*(y*(y-1)/2-1) - T == 0
eq34 =(3*N-1)/8+3*(y*(y-1)/2) - S == 0

eq35 =M+3*(x*(x+1)/2-(a^2-a)/2) -A == 0
eq36 =M+3*(x*(x+1)/2-(b^2-b)/2) -B == 0
eq37 =M+3*(x*(x+1)/2-(c^2-c)/2) -C == 0
eq38 =M+3*(x*(x+1)/2-1) -D == 0

solutions = solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq29,eq30,eq31,eq32,eq33,eq34,eq35,eq36,eq37,eq38],N,p,q,x,y,M,a,b,c,A,B,C,D,u,v,S,T,U,V)
sol = solutions 
Execution_Time = time.time() - Start_Time
print (Execution_Time)
print(sol)

(N, p, q, x, y, M, a, b, c, A, B, C, D, u, v, S, T, U, V)
1.2627530097961426
[
[N == 8979, p == 73, q == 123, x == 24, y == 13, M == 2701, a == 13, b == 7, c == -3, A == 3367, B == 3538, C == 3583, D == 3598, u == -3, v == 7, S == 3601, T == 3598, U == 3583, V == 3538]
]

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

Here you All the variables I will find my work:

https://www.linkedin.com/feed/update/urn:li:ugcPost:6777160628965535744/

or

https://github.com/Piunosei/gerva_factorization/blob/main/Transformation-of-the-factorization-problem-into-a-problem-of-solving-a-non-linear-system%20(5).pdf

Is written:

• in an almost unpresentable way

• the case x odd is missing

• (*) and is redundant

(*) as it is possible to establish the range of log_2 (x) and log_2 (y)

My problem is to solve the resulting nonlinear system.

all small variables belong to Z

all big variables belong to N

If mention are integers

Let N = p * q in the form 4 * G + 3 and p + q = 8 * x + 4 then and q-p = 4 * y-2 y-2

then

solving the system that I will show you we find p = (4x+1-2(y-1))

q = (4x+3+2(y-1))

let's study the case

x and q

How is the system composed?

If y is odd the first two lines

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0

If y is even and y is odd

Examples N=8979

(x+1+1)/2 is even then a = 1-(x+1+1)/2

(x+1+1)/2 is even odd a = (x+1+1)/2

(|a|+1)/2 is even then b = 1-(|a|+1)/2

(|a|+1)/2 is even odd b = (|a|+1)/2

(|b|+1)/2 is even then c = 1-(|b|+1)/2

(|b|+1)/2 is even odd c = (|b|+1)/2

(|c|+1)/2 is even then -1 = 1-(|c|+1)/2

(|c|+1)/2 is even odd -1 = (|c|+1)/2

then

16 <= x <= 30

S>D

D>C

C>B

B>A

A>M

and

(y+1)/2 is even then v = 1-(y+1)/2

(y+1)/2 is even odd v = (y+1)/2

(|v|+1)/2 is even then u = 1-(|v|+1)/2

(|v|+1)/2 is even odd u = (|v|+1)/2

(|u|+1)/2 is even then -1 = 1-(|u|+1)/2

(|u|+1)/2 is even odd -1 = (|u|+1)/2

then

9 <= y <= 15

S>T

T>U

U>V

V>(3*N-1)/8

then this system the first two lines

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(1-((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0

the other lines will have a single solutionbe

3*((2* Aj - 3* aj + 1)/24 + M) + 1 - A(j+1) == 0
Aj + (3* aj *( aj - 1))/2 - 3*M-1-M == 0

where j goes from 1 to n-1 = IntegerPartOf (log_2 (x + 2 * p) -1)

the last two lines will be

3*((2* An + 3 + 1)/24 + M) + 1 - 3*M-1-M == 0
An + 3 - 3*M-1-M == 0

Example

Esempio N= 731

17*43 =731

(43-13+2)/4=7

7+2*17=41

var('N p q x y M a b c A B C D u v S T U V')
a1 a2 a3 a4 a5 A1 A2 A3 A4 A5')


eq0 = N-8979 N-731 == 0

eq1 = 3*(((2*(3*N-1)/8-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = -(3*N-1)/8+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
3*((2*(3*((N-3)/8+p^2)+1)-3*(((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
3*((2* A1 - 3* a1 + 1)/24 + M) + 1 - A2 == 0
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
A1 + (3* a1 *( a1 - 1))/2 - 3*M-1-M == 0
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
3*((2* A2 - 3* a2 + 1)/24 + M) + 1 - A3 == 0
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
A2 + (3* a2 * ( a2 - 1))/2 - 3*M-1-M == 0
eq7 = 3*(((2*T+3+1)/24)+3*x*(x+1)/2)+1-S == 0  
3*((2* A3 - 3* a3 + 1)/24 + M) + 1 - A4 == 0
eq8 = T+3-12*x*(x+1)/2-1 == 0  
A3 + (3*a3*( a3 - 1))/2 - 3*M-1-M == 0
eq9 = 3*(((2*S-3*1+1)/24)+3*x*(x+1)/2)+1-S == 0   

3*((2* A4 - 3* a4 + 1)/24 + M) + 1 - A5 == 0
eq10 = v*(v-1)/2 -(y-1)*(y+1)/8 A4 + (3* a4 *( a4 - 1))/2 - 3*M-1-M == 0
eq11 = u*(u-1)/2 -(v-1)*(v+1)/8 3*((2* A5 - 3* a5 + 1)/24 + M) + 1 - 3*M-1-M == 0
eq12 = 1 -(u-1)*(u+1)/8 == 0

A5 + 3* a5 *( a5 -1)/2 - 3*M-1-M == 0
eq13 = a*(a-1)/2 -(x+1-1)*(x+1+1)/8 == 0
eq14 = b*(b-1)/2 -(a-1)*(a+1)/8 == 0
eq15 = c*(c-1)/2 -(b-1)*(b+1)/8 == 0
eq16 = 1 -(c-1)*(c+1)/8 == 0

eq18 = -M+9*((2*(3*N-1)/8 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D + 3 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + 3 - (12*x*(x + 1))/2 - 1 == 0
eq28 = 3*((2*S - 3*1 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0

eq29 = p - (4*x+1-2*(y-1)) == 0
eq30 = q - (4*x+3+2*(y-1)) a5 + 1 == 0

solutions = solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14,eq15,eq16,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq28,eq29,eq30],N,p,q,x,y,M,a,b,c,A,B,C,D,u,v,S,T,U,V)
solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13],N,p,M,a1,a2,a3,a4,a5,A1,A2,A3,A4,A5)
sol = solutions 
print(sol)

OUTPUT

(N, p, q, x, y, M, a, b, c, A, B, C, D, u, v, S, T, U, V)
a1, a2, a3, a4, a5, A1, A2, A3, A4, A5)
[
[N == 8979, 731, p == 73, q == 123, x == 24, y == 13, 17, M == 2701, a == 13, b == 7, c == -3, A == 3367, B == 3538, C == 3583, D == 3598, u == -3, v == 7, S == 3601, T == 3598, U == 3583, V == 3538]
900, a1 == 21, a2 == 11, a3 == -5, a4 == 3, a5 == -1, A1 == 2971, A2 == 3436, A3 == 3556, A4 == 3592, A5 == 3598],
[N == 731, p == (43/7), M == (34866/49), a1 == 21, a2 == 11, a3 == -5, a4 == 3, a5 == -1, A1 == (108643/49), A2 == (131428/49), A3 == (137308/49), A4 == (139072/49), A5 == (139366/49)]
]

Will you kindly help me Other information:

if ((-p^2+2p+N)/(4p)+2p)/2 is even then a1 =1-((-p^2+2p+N)/(4p)+2p)/2

if ((-p^2+2p+N)/(4p)+2p)/2 è dispari allora a1 =((-p^2+2p+N)/(4p)+2p)/2

in general then:

if aj is negative and (-aj + 1)/2 is odd then a(j+1)=(-aj + 1)/2

if aj is negative and (-aj + 1)/2 is even then a(j+1)=1-(-aj + 1)/2

if aj is positive and (aj + 1)/2 is odd then a(j+1)=(aj + 1)/2

if aj is positive and (aj + 1)/2 is even then a(j+1)=1-(aj + 1)/2

other:

if N=4G+1 multiply N1=N3 ,N2=N*15

if N=4G+3 multiply N1=N1 ,N2=N*5

N1 or N2 are in the form 4 * G + 3 with p + q = 8 * x + 4 and q-p = 4 * y-2

I would like to have the cooperation of the mathematicians and programmers of sagemath to solve this non-linear and implement the system solution ad hoc.

Hopefully in the lowest possible computational cost?

EDIT:

after @vdelecroix replied, your reply, I found this other way

import time

Start_Time = time.time()
var('N p q x y M a b c A B C D u v S T U V')
eq0 = N-8979 == 0

eq1 = 3*(((2*(3*N-1)/8-3*y+1)/24)+3*x*(x+1)/2)+1-V == 0  
eq2 = -(3*N-1)/8+3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2 == 0  
eq3 = 3*(((2*V-3*v+1)/24)+3*x*(x+1)/2)+1-U == 0  
eq4 = V+3*v*(v-1)/2-12*x*(x+1)/2-1 == 0  
eq5 = 3*(((2*U-3*u+1)/24)+3*x*(x+1)/2)+1-T == 0  
eq6 = U+3*u*(u-1)/2-12*x*(x+1)/2-1 == 0  
eq7 = 3*(((2*T+3+1)/24)+3*x*(x+1)/2)+1-S == 0  
eq8 = T+3-12*x*(x+1)/2-1 == 0  


eq18 = -M+9*((2*(3*N-1)/8 - 3*y + 1)/24+(y-1)*(y+1)/8)+1 == 0
eq19 = 3*((2*M - 3*(x+1) + 1)/24 + (3*x*(x + 1))/2) + 1 - A == 0
eq20 = -M + (3*x*(x + 1))/2 - (3*x*(x + 1))/2 + ((3*x + 1)*(3*x + 2))/2 == 0
eq21 = 3*((2*A - 3*a + 1)/24 + (3*x*(x + 1))/2) + 1 - B == 0
eq22 = A + (3*a*(a - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq23 = 3*((2*B - 3*b + 1)/24 + (3*x*(x + 1))/2) + 1 - C == 0
eq24 = B + (3*b*(b - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq25 = 3*((2*C - 3*c + 1)/24 + (3*x*(x + 1))/2) + 1 - D == 0
eq26 = C + (3*c*(c - 1))/2 - (12*x*(x + 1))/2 - 1 == 0
eq26 = 3*((2*D + 3 + 1)/24 + (3*x*(x + 1))/2) + 1 - S == 0
eq27 = D + 3 - (12*x*(x + 1))/2 - 1 == 0


eq29 = p - (4*x+1-2*(y-1)) == 0
eq30 = q - (4*x+3+2*(y-1)) == 0

eq31 =(3*N-1)/8+3*(y*(y-1)/2-(v^2-v)/2) -V == 0
eq32 =(3*N-1)/8+3*(y*(y-1)/2-(u^2-u)/2) -U == 0
eq33 =(3*N-1)/8+3*(y*(y-1)/2-1) - T == 0
eq34 =(3*N-1)/8+3*(y*(y-1)/2) - S == 0

eq35 =M+3*(x*(x+1)/2-(a^2-a)/2) -A == 0
eq36 =M+3*(x*(x+1)/2-(b^2-b)/2) -B == 0
eq37 =M+3*(x*(x+1)/2-(c^2-c)/2) -C == 0
eq38 =M+3*(x*(x+1)/2-1) -D == 0

solutions = solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq18,eq19,eq20,eq21,eq22,eq23,eq24,eq25,eq26,eq27,eq29,eq30,eq31,eq32,eq33,eq34,eq35,eq36,eq37,eq38],N,p,q,x,y,M,a,b,c,A,B,C,D,u,v,S,T,U,V)
sol = solutions 
Execution_Time = time.time() - Start_Time
print (Execution_Time)
print(sol)

OUTPUT on my old computer

(N, p, q, x, y, M, a, b, c, A, B, C, D, u, v, S, T, U, V)
1.2627530097961426
[
[N == 8979, p == 73, q == 123, x == 24, y == 13, M == 2701, a == 13, b == 7, c == -3, A == 3367, B == 3538, C == 3583, D == 3598, u == -3, v == 7, S == 3601, T == 3598, U == 3583, V == 3538]
]

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

All the variables I will mention are integers

Let N = p * q in the form 4 * G + 3 and p + q = 8 * x + 4 and q-p = 4 * y-2

then solving the system that I will show you we find p and q

How is the system composed?

If y is odd the first two lines

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0

If y is even the first two lines

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(1-((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0

the other lines will be

3*((2* Aj - 3* aj + 1)/24 + M) + 1 - A(j+1) == 0
Aj + (3* aj *( aj - 1))/2 - 3*M-1-M == 0

where j goes from 1 to n-1 = IntegerPartOf (log_2 (x + 2 * p) -1)

the last two lines will be

3*((2* An + 3 + 1)/24 + M) + 1 - 3*M-1-M == 0
An + 3 - 3*M-1-M == 0

Example

Esempio N= 731

17*43 =731

(43-13+2)/4=7

7+2*17=41

var('N p M a1 a2 a3 a4 a5 A1 A2 A3 A4 A5')


eq0 = N-731 == 0

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0
eq3 = 3*((2* A1 - 3* a1 + 1)/24 + M) + 1 - A2 == 0
eq4 = A1 + (3* a1 *( a1 - 1))/2 - 3*M-1-M == 0
eq5 = 3*((2* A2 - 3* a2 + 1)/24 + M) + 1 - A3 == 0
eq6 = A2 + (3* a2 * ( a2 - 1))/2 - 3*M-1-M == 0
eq7 = 3*((2* A3 - 3* a3 + 1)/24 + M) + 1 - A4 == 0
eq8 = A3 + (3*a3*( a3 - 1))/2 - 3*M-1-M == 0
eq9 = 3*((2* A4 - 3* a4 + 1)/24 + M) + 1 - A5 == 0
eq10 = A4 + (3* a4 *( a4 - 1))/2 - 3*M-1-M == 0
eq11 = 3*((2* A5 - 3* a5 + 1)/24 + M) + 1 - 3*M-1-M == 0
eq12 = A5 + 3* a5 *( a5 -1)/2 - 3*M-1-M == 0
eq13 = a5 + 1 == 0

solutions = solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13],N,p,M,a1,a2,a3,a4,a5,A1,A2,A3,A4,A5)
sol = solutions 
print(sol)

OUTPUT

(N, p, M, a1, a2, a3, a4, a5, A1, A2, A3, A4, A5)
[
[N == 731, p == 17, M == 900, a1 == 21, a2 == 11, a3 == -5, a4 == 3, a5 == -1, A1 == 2971, A2 == 3436, A3 == 3556, A4 == 3592, A5 == 3598],
[N == 731, p == (43/7), M == (34866/49), a1 == 21, a2 == 11, a3 == -5, a4 == 3, a5 == -1, A1 == (108643/49), A2 == (131428/49), A3 == (137308/49), A4 == (139072/49), A5 == (139366/49)]
]

Other information:

if ((-p^2+2p+N)/(4p)+2p)/2 is even then a1 =1-((-p^2+2p+N)/(4p)+2p)/2

if ((-p^2+2p+N)/(4p)+2p)/2 è dispari allora a1 =((-p^2+2p+N)/(4p)+2p)/2

in general then:

if aj is negative and (-aj + 1)/2 is odd then a(j+1)=(-aj + 1)/2

if aj is negative and (-aj + 1)/2 is even then a(j+1)=1-(-aj + 1)/2

if aj is positive and (aj + 1)/2 is odd then a(j+1)=(aj + 1)/2

if aj is positive and (aj + 1)/2 is even then a(j+1)=1-(aj + 1)/2

other:

if N=4G+1 multiply N1=NN1=N * 3 ,N2=N3 ,N2=N*1515

if N=4G+3 multiply N1=NN1=N * 1 ,N2=N1 ,N2=N*55

N1 or N2 are in the form 4 * G + 3 with p + q = 8 * x + 4 and q-p = 4 * y-2

I would like to have the cooperation of the mathematicians and programmers of sagemath to solve and implement the system solution ad hoc.

Hopefully in your reply, I cordially greet you

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

All the variables I will mention are integers

Let N = p * q in the form 4 * G + 3 and p + q = 8 * x + 4 and q-p = 4 * y-2

then solving the system that I will show you we find p and q

How is the system composed?

If y is odd the first two lines

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0

If y is even the first two lines

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(1-((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0

the other lines will be

3*((2* Aj - 3* aj + 1)/24 + M) + 1 - A(j+1) == 0
Aj + (3* aj *( aj - 1))/2 - 3*M-1-M == 0

where j goes from 1 to n-1 = IntegerPartOf (log_2 (x (y + 2 * p) -1)

the last two lines will be

3*((2* An + 3 + 1)/24 + M) + 1 - 3*M-1-M == 0
An + 3 - 3*M-1-M == 0

Example

Esempio N= 731

17*43 =731

(43-13+2)/4=7

7+2*17=41

var('N p M a1 a2 a3 a4 a5 A1 A2 A3 A4 A5')


eq0 = N-731 == 0

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0
eq3 = 3*((2* A1 - 3* a1 + 1)/24 + M) + 1 - A2 == 0
eq4 = A1 + (3* a1 *( a1 - 1))/2 - 3*M-1-M == 0
eq5 = 3*((2* A2 - 3* a2 + 1)/24 + M) + 1 - A3 == 0
eq6 = A2 + (3* a2 * ( a2 - 1))/2 - 3*M-1-M == 0
eq7 = 3*((2* A3 - 3* a3 + 1)/24 + M) + 1 - A4 == 0
eq8 = A3 + (3*a3*( a3 - 1))/2 - 3*M-1-M == 0
eq9 = 3*((2* A4 - 3* a4 + 1)/24 + M) + 1 - A5 == 0
eq10 = A4 + (3* a4 *( a4 - 1))/2 - 3*M-1-M == 0
eq11 = 3*((2* A5 - 3* a5 + 1)/24 + M) + 1 - 3*M-1-M == 0
eq12 = A5 + 3* a5 *( a5 -1)/2 - 3*M-1-M == 0
eq13 = a5 + 1 == 0

solutions = solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13],N,p,M,a1,a2,a3,a4,a5,A1,A2,A3,A4,A5)
sol = solutions 
print(sol)

OUTPUT

(N, p, M, a1, a2, a3, a4, a5, A1, A2, A3, A4, A5)
[
[N == 731, p == 17, M == 900, a1 == 21, a2 == 11, a3 == -5, a4 == 3, a5 == -1, A1 == 2971, A2 == 3436, A3 == 3556, A4 == 3592, A5 == 3598],
[N == 731, p == (43/7), M == (34866/49), a1 == 21, a2 == 11, a3 == -5, a4 == 3, a5 == -1, A1 == (108643/49), A2 == (131428/49), A3 == (137308/49), A4 == (139072/49), A5 == (139366/49)]
]

Other information:

My name is Alberico Lepore and I am an aspiring self-taught amateur and for about six years I have been looking for factorization in computationally acceptable times.

I have come to the point of verifying that resolving the type of systems listed below resolves the factorization.

All the variables I will mention are integers

Let N = p * q in the form 4 * G + 3 and p + q = 8 * x + 4 and q-p = 4 * y-2

then solving the system that I will show you we find p and q

How is the system composed?

If y is odd the first two lines

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0

If y is even the first two lines

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(1-((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0

the other lines will be

3*((2* Aj - 3* aj + 1)/24 + M) + 1 - A(j+1) == 0
Aj + (3* aj *( aj - 1))/2 - 3*M-1-M == 0

where j goes from 1 to n-1 = IntegerPartOf (log_2 (y + 2 * p) -1)

the last two lines will be

3*((2* An + 3 + 1)/24 + M) + 1 - 3*M-1-M == 0
An + 3 - 3*M-1-M == 0

Example

Esempio N= 731

17*43 =731

(43-13+2)/4=7

7+2*17=41

var('N p M a1 a2 a3 a4 a5 A1 A2 A3 A4 A5')


eq0 = N-731 == 0

eq1 = 3*((p^2-4*p+N)/(8*p)+p)*((p^2-4*p+N)/(8*p)+p+1)/2-M == 0
eq2 = 3*((2*(3*((N-3)/8+p^2)+1)-3*(((-p^2+2*p+N)/(4*p)+2*p))+1)/24 + M)+1 - A1 == 0
eq3 = 3*((2* A1 - 3* a1 + 1)/24 + M) + 1 - A2 == 0
eq4 = A1 + (3* a1 *( a1 - 1))/2 - 3*M-1-M == 0
eq5 = 3*((2* A2 - 3* a2 + 1)/24 + M) + 1 - A3 == 0
eq6 = A2 + (3* a2 * ( a2 - 1))/2 - 3*M-1-M == 0
eq7 = 3*((2* A3 - 3* a3 + 1)/24 + M) + 1 - A4 == 0
eq8 = A3 + (3*a3*( a3 - 1))/2 - 3*M-1-M == 0
eq9 = 3*((2* A4 - 3* a4 + 1)/24 + M) + 1 - A5 == 0
eq10 = A4 + (3* a4 *( a4 - 1))/2 - 3*M-1-M == 0
eq11 = 3*((2* A5 - 3* a5 + 1)/24 + M) + 1 - 3*M-1-M == 0
eq12 = A5 + 3* a5 *( a5 -1)/2 - 3*M-1-M == 0
eq13 = a5 + 1 == 0

solutions = solve([eq0,eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13],N,p,M,a1,a2,a3,a4,a5,A1,A2,A3,A4,A5)
sol = solutions 
print(sol)

OUTPUT

(N, p, M, a1, a2, a3, a4, a5, A1, A2, A3, A4, A5)
[
[N == 731, p == 17, M == 900, a1 == 21, a2 == 11, a3 == -5, a4 == 3, a5 == -1, A1 == 2971, A2 == 3436, A3 == 3556, A4 == 3592, A5 == 3598],
[N == 731, p == (43/7), M == (34866/49), a1 == 21, a2 == 11, a3 == -5, a4 == 3, a5 == -1, A1 == (108643/49), A2 == (131428/49), A3 == (137308/49), A4 == (139072/49), A5 == (139366/49)]
]

Other information:

> if ((-p^2+2*p+N)/(4*p)+2*p)  is even
> and ((-p^2+2*p+N)/(4*p)+2*p)/2 is even
> then a1 =1-((-p^2+2*p+N)/(4*p)+2*p)/2
> 
> if ((-p^2+2*p+N)/(4*p)+2*p)  is even
> and ((-p^2+2*p+N)/(4*p)+2*p)/2 id odd
> then a1 =((-p^2+2*p+N)/(4*p)+2*p)/2
> 
> if ((-p^2+2*p+N)/(4*p)+2*p)  is odd
> and ((-p^2+2*p+N)/(4*p)+2*p+1)/2 is
> even then a1
> =1-((-p^2+2*p+N)/(4*p)+2*p+1)/2
> 
> if ((-p^2+2*p+N)/(4*p)+2*p)  is odd
> and ((-p^2+2*p+N)/(4*p)+2*p+1)/2 id
> odd then a1
> =((-p^2+2*p+N)/(4*p)+2*p+1)/2
> 
> in general then:
> 
> if aj is negative and (-aj + 1)/2 is
> odd then a(j+1)=(-aj + 1)/2
> 
> if aj is negative and (-aj + 1)/2 is
> even then a(j+1)=1-(-aj + 1)/2
> 
> if aj is positive and (aj + 1)/2 is
> odd then a(j+1)=(aj + 1)/2
> 
> if aj is positive and (aj + 1)/2 is
> even then a(j+1)=1-(aj + 1)/2
> 
> other:
> 
> if N=4*G+1 multiply N1=N * 3 ,N2=N*15
> 
> if N=4*G+3 multiply N1=N * 1 ,N2=N*5
> 
> N1 or N2 are in the form 4 * G + 3
> with p + q = 8 * x + 4 and q-p = 4 *
> y-2

I would like to have the cooperation of the mathematicians and programmers of sagemath to solve and implement the system solution ad hoc.

Hopefully in your reply, I cordially greet you