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Old 2020-11-03, 11:51   #1
Alberico Lepore
 
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Default WTF factorization. Do you think it is a good method?

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?
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Old 2020-11-03, 12:07   #2
Viliam Furik
 
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If you show an example of it working, maybe it is good...
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Old 2020-11-03, 12:11   #3
retina
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Alberico Lepore: Using N=377 is silly, stupid and pointless.

Solve the 18 digit challenge first.
https://mersenneforum.org/showthread.php?t=25929
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Old 2020-11-03, 15:07   #4
mathwiz
 
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Quote:
Originally Posted by Alberico Lepore View Post
Do you think it is a good method?
No. But, by all means, keep posting random equations and asking people what they think.
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Old 2020-11-03, 15:21   #5
Dr Sardonicus
 
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Quote:
Originally Posted by Alberico Lepore View Post
case:

N=p*q
If N is an RSA number you may assume that N = p*q.
Quote:
&
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.
Quote:
&
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>
Quote:
Do you think it is a good method?
Method? I don't see any "method."
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Old 2020-11-03, 18:52   #6
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
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?
Quote:
Originally Posted by Viliam Furik View Post
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
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Old 2020-11-03, 19:07   #7
CRGreathouse
 
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Quote:
Originally Posted by Alberico Lepore View Post
I lowered to O(72 *[Lenstra elliptic-curve factorization])
You understand that this isn't logarithmic, though, right?
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Old 2020-11-03, 19:26   #8
Uncwilly
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