WTF factorization. Do you think it is a good method?

[Alberico Lepore]

case:

N=p*q
&
q-p=n
&
n mod 8 = 0
&
p+n/2=3*(2*x+1)

->

288*X-n^2-32*((N-1)/8-1)=0
288*X-(n-8)^2-32*((N-1)/8+2*(n/4-1)-1)=0

->

Example N=377

X=1152 h^2 + 64 h + 6
=
1152 k² + 32 k + 6

288*(1152 h^2 + 64 h + 6)-n^2-32*(47-1)=0
,
288*(1152 k² + 32 k + 6)-(n-8)^2-32*(47+2*(n/4-1)-1)=0
,
(1152 h^2 + 64 h + 6)=(1152 k² + 32 k + 6)







given n = 576 + 8 * w

assigning w values from 0 to 71

we will have O (72 * 1152 ^ 2 * [Lenstra elliptic-curve factorization])



Do you think it is a good method?

[Viliam Furik]

If you show an example of it working, maybe it is good...

[retina]

Alberico Lepore: Using N=377 is silly, stupid and pointless.

Solve the 18 digit challenge first.
https://mersenneforum.org/showthread.php?t=25929

[mathwiz]

Originally Posted by Alberico Lepore

Do you think it is a good method?

No. But, by all means, keep posting random equations and asking people what they think.

[Dr Sardonicus]

Originally Posted by Alberico Lepore

case:

N=p*q

If N is an RSA number you may assume that N = p*q.

&
q-p=n
&
n mod 8 = 0

This implies that N == 1 (mod 8).

If N == 3, 5, or 7 (mod 8), you're done.

&
p+n/2=3*(2*x+1)

This implies that p+q is divisible by 3, and therefore (assuming that N isn't 9) that

N == 2 (mod 3).

If N == 1 (mod 3), you're done.

So, you're saying that you're only interested in N == 17 (mod 24).
<snip>

Do you think it is a good method?

Method? I don't see any "method."

[Alberico Lepore]

Originally Posted by Alberico Lepore

case:

N=p*q
&
q-p=n
&
n mod 8 = 0
&
p+n/2=3*(2*x+1)

->

288*X-n^2-32*((N-1)/8-1)=0
288*X-(n-8)^2-32*((N-1)/8+2*(n/4-1)-1)=0

->

Example N=377

X=1152 h^2 + 64 h + 6
=
1152 k² + 32 k + 6

288*(1152 h^2 + 64 h + 6)-n^2-32*(47-1)=0
,
288*(1152 k² + 32 k + 6)-(n-8)^2-32*(47+2*(n/4-1)-1)=0
,
(1152 h^2 + 64 h + 6)=(1152 k² + 32 k + 6)







given n = 576 + 8 * w

assigning w values from 0 to 71

we will have O (72 * 1152 ^ 2 * [Lenstra elliptic-curve factorization])



Do you think it is a good method?

Originally Posted by Viliam Furik

If you show an example of it working, maybe it is good...

I lowered to O(72 *[Lenstra elliptic-curve factorization])

case: x odd

N=377

solve (1152 h^2 + 64 h + 6)=x*(x+1)/2
,
2/3*[9*(1152 h^2 + 64 h + 6)+1+sqrt[8*[9*(1152 h^2 + 64 h + 6)+1]+1]*3*(x-1)-1]+1=Z^2
,
h,x

[CRGreathouse]

Originally Posted by Alberico Lepore

I lowered to O(72 *[Lenstra elliptic-curve factorization])

You understand that this isn't logarithmic, though, right?

[Uncwilly]