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mathwiz

Mar 2019

1110012 Posts Quote:
 Originally Posted by Alberico Lepore I try to give you a demonstration
No. Demonstration here means "show you can factor a 100+ digit semiprime with your method".   2020-02-24, 21:01   #24
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

2×3×5×7×19 Posts Quote:
 Originally Posted by mathwiz No. Demonstration here means "show you can factor a 100+ digit semiprime with your method".
I'd settle for 40 digits. At even 20 digits, OP would show evidence that they're not a complete waste of time. I haven't seen OP tackle even a 10 digit number, but I have seen clear evidence that OP isn't able to actually DO anything with his equations.

At 20 digits, OP would notice how terrible his methods are, and that even if his methods could factor numbers, they're worse than trial division.   2020-02-26, 09:08   #25
Alberico Lepore

May 2017
ITALY

192 Posts Quote:
 Originally Posted by mathwiz No. Demonstration here means "show you can factor a 100+ digit semiprime with your method".
I implemented it and found that my hypothesis was incorrect.
Now I will test this

k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1)
,
3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8

I will assign k integers I, xk = I-1 && yk = I

I have to test

if
xk < x0 && k < k0
then
x > xk && y > yk

if
xk > x0 && k < k0
then
x < xk && y < yk

if
xk > x0 && k > k0
then
x < xk && y < yk

the problem is this:

as you can see from the link they are huge equations

https://www.wolframalpha.com/input/?...%2F8+%2C+x%2Cy

Last fiddled with by Alberico Lepore on 2020-02-26 at 09:38   2020-02-26, 20:39   #26
Alberico Lepore

May 2017
ITALY

5518 Posts Quote:
 Originally Posted by Alberico Lepore I implemented it and found that my hypothesis was incorrect. Now I will test this k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1) , 3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8 I will assign k integers I, xk = I-1 && yk = I I have to test if xk < x0 && k < k0 then x > xk && y > yk if xk > x0 && k < k0 then x < xk && y < yk if xk > x0 && k > k0 then x < xk && y < yk the problem is this: as you can see from the link they are huge equations https://www.wolframalpha.com/input/?...%2F8+%2C+x%2Cy
you have tried it
it would seem to work from the first hand tests   2020-02-27, 04:57   #27
CRGreathouse

Aug 2006

16DE16 Posts Quote:
 Originally Posted by Alberico Lepore you have tried it it would seem to work from the first hand tests
Great! We look forward to a demonstration.   2020-02-27, 07:14   #28
Alberico Lepore

May 2017
ITALY

192 Posts Quote:
 Originally Posted by CRGreathouse Great! We look forward to a demonstration.
The following may be wrong but it may also be right

In summary

Given an N = p * q in the form (p + q-4) mod 8 = 0

with p and q not necessarily prime but odd numbers

then

given the system

solve

k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1)
,
3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8

you will have that

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

Now I will test this

k=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1)
,
3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8

I will assign k integers I, xk = I-1 && yk = I

I have to test

if
xk < x0 && k < k0
then
x > xk && y > yk

if
xk > x0 && k < k0
then
x < xk && y < yk

if
xk > x0 && k > k0
then
x < xk && y < yk

it would seem to work from the first hand tests

the problem is this:

as you can see from the link they are huge equations

https://www.wolframalpha.com/input/?...%2F8+%2C+x%2Cy

I don't understand that much demonstrations, now I try

solve (k)=(x+1)*(2*(x-y+1)+1)+(x-y)*(x-y+1)

->

x=(1/3)*[sqrt(3*k+(y-3)*y)+2*y-3]

if
xk > x0 && k < k0
then
x < xk && y < yk

if
xk > x0 && k > k0
then
x < xk && y < yk

solve x=(1/3)*[sqrt(3*k+(y-3)*y)+2*y-3],3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8
,
k=x+2
,
y>=k

impossible

if
xk < x0 && k < k0
then
x > xk && y > yk

solve x=(1/3)*[sqrt(3*k+(y-3)*y)+2*y-3],3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=(3*N-1)/8
,
k=x
,
y<=k

-> x<=-1.5   2020-02-27, 07:24 #29 CRGreathouse   Aug 2006 2·2,927 Posts I was hoping for more along the lines of "here is a big number and how I factored it with my method". But here's a small number to get you started: 596085571408853. Could you show us how your method factors it? It takes less than a millisecond with Shanks' SQUFOF.   2020-02-27, 07:32   #30
Alberico Lepore

May 2017
ITALY

192 Posts Quote:
 Originally Posted by CRGreathouse I was hoping for more along the lines of "here is a big number and how I factored it with my method". But here's a small number to get you started: 596085571408853. Could you show us how your method factors it? It takes less than a millisecond with Shanks' SQUFOF.
Condition:

(p + q-4) mod 8 = 0

k>0 if p>= (p+q)/4

Last fiddled with by Alberico Lepore on 2020-02-27 at 07:35   2020-02-27, 09:53   #31
Alberico Lepore

May 2017
ITALY

192 Posts Quote:
 Originally Posted by Alberico Lepore Condition: (p + q-4) mod 8 = 0 k>0 if p>= (p+q)/4
I have tried with larger numbers and I have seen that it was another wrong hypothesis but I have found this

range 1/2 *[sqrt[2*[(3*N-1)/8-1]/3+1]-1]<= x <= 1/6*[sqrt[3*N]-3]

By assigning k integer numbers > 0

if
xk = k-1 < x0

then

x > 1/6*[sqrt[3*N]-3]

I don't know if it's another wrong hypothesis, I'll test it

************************** UPDATE

This hypothesis is also wrong

Last fiddled with by Alberico Lepore on 2020-02-27 at 10:25   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Robert Holmes Factoring 19 2010-11-08 18:46 kurtulmehtap Math 25 2010-09-12 14:13 dleclair NFSNET Discussion 1 2006-03-21 05:11 Wacky NFSNET Discussion 1 2006-03-20 23:43 Jeff Gilchrist NFSNET Discussion 7 2005-02-23 19:46

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