mersenneforum.org "New" same approach that isn't factorization
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2020-09-10, 09:21   #12
Alberico Lepore

May 2017
ITALY

5×97 Posts

before understanding this

Quote:
 Originally Posted by retina Well go on then, have at it.
I have to understand this

Quote:
 Originally Posted by Alberico Lepore What characteristic must N have for that equality to be true for example for N = 121 this 2 * 121 + 2 * 1 ^ 2 + y ^ 2- (22) ^ 2 = 0 it's not true that is, y is not integer
would you give me a little clue

2020-09-10, 09:48   #13
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2·32·73 Posts

Quote:
 Originally Posted by Alberico Lepore I have to understand this would you give me a little clue
That is your equation, you invented it. So you should be the one to understand it. I have no idea what you are trying to do, it all looks like nonsense to me.

It's up to you to show us how you factor things, not the other way around. Your claim, you prove it.

2020-09-10, 11:38   #14
Alberico Lepore

May 2017
ITALY

7458 Posts

Quote:
 Originally Posted by retina That is your equation, you invented it. So you should be the one to understand it. I have no idea what you are trying to do, it all looks like nonsense to me. It's up to you to show us how you factor things, not the other way around. Your claim, you prove it.
So maybe I found when it's true:
(a + b) mod 3 = 0
in two cases

M=[(a+b)/2-((a+b)/6-1)/2]*[2*[(a+b)/2-((a+b)/6-1)/2]-3]
and
M=[(a+b)/2-((a+b)/6+1)/2]*[2*[(a+b)/2-((a+b)/6+1)/2]+3]

so I tried to bring back a generic number (a + b) mod 3 = 0

in

M=[(a+b)/2-((a+b)/6-1)/2]*[2*[(a+b)/2-((a+b)/6-1)/2]-3]

but I didn't get any useful results

Example

N=161
,
2*(N+(n/2)^2-((a+b)/6-1)^2)+2*a^2+((b-a)/2)^2=((3*a+b)/2)^2
,
a*b=(N+(n/2)^2-((a+b)/6-1)^2)
,
2*(N+(n/2)^2-((a+b)/6-1)^2)+2*1^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0

but I will continue to study

Last fiddled with by Alberico Lepore on 2020-09-10 at 11:42

 2020-09-10, 12:20 #15 Alberico Lepore     May 2017 ITALY 5×97 Posts Bruteforce could be attempted for a multiple of 9 : 9*F N=161 , 2*(N*9*F)+2*a^2+((b-a)/2)^2=((3*a+b)/2)^2 , a*b=(N*9*F) , 2*(N*9*F)+2*1^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0 , F=15 -> a=105 GCD(105,161)=7 but I think this is very RANDOM
 2020-09-10, 16:48 #16 Alberico Lepore     May 2017 ITALY 1111001012 Posts If we solve F as a function of a and N solve 2*(N*9*F)+2*a^2+((b-a)/2)^2=((3*a+b)/2)^2 , a*b=(N*9*F) , 2*(N*9*F)+2*1^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0 ,F,b -> 9*N*F=2*a^2-3*a multiplying by 2 and imposing 2 * a = A we will have 18*N*F=A^2-3*A A0 < sqrt(18*N) is it possible to apply the Coppersmith method? https://en.wikipedia.org/wiki/Coppersmith_method
 2020-09-10, 21:42 #17 mathwiz   Mar 2019 2·5·17 Posts Please, stop posting.
2020-09-18, 17:51   #18
Alberico Lepore

May 2017
ITALY

5×97 Posts

Quote:
 Originally Posted by Alberico Lepore If we solve F as a function of a and N solve 2*(N*9*F)+2*a^2+((b-a)/2)^2=((3*a+b)/2)^2 , a*b=(N*9*F) , 2*(N*9*F)+2*1^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0 ,F,b -> 9*N*F=2*a^2-3*a multiplying by 2 and imposing 2 * a = A we will have 18*N*F=A^2-3*A A0 < sqrt(18*N) is it possible to apply the Coppersmith method? https://en.wikipedia.org/wiki/Coppersmith_method
I have found other equations where, perhaps, the Coppersmith method is applicable

I don't know with what efficiency

solve (N*F-1)/8=(X^2-1)/8-2*((b-a)/8)^2 ,a*b=(N*F) , 2*(N*F)+2*1^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0

8*X^2-6*X-9=F*N*9

8*X^2+6*X-9=F*N*9

multiplying everything by 2 and imposing A = 4 * X and B = 4 * X

are obtained

A^2-3*A-18=F*N*9*2

B^2+3*B-18=F*N*9*2

solve (65*F-1)/8=(X^2-1)/8-2*((b-a)/8)^2 ,a*b=(65*F) , 2*(65*F)+2*1^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0

8*X^2-6*X-9=F*65*9

8*X^2+6*X-9=F*65*9

multiplying everything by 2 and imposing A = 4 * X and B = 4 * X

are obtained

A^2-3*A-18=F*65*9*2

B^2+3*B-18=F*65*9*2

and these are the first two

and then

solve (N*F-1)/8=x*(x+1)/2-2*((b-a)/8)^2 ,a*b=(N*F) , 2*(N*F)+2*1^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0 ,F

32*x^2+20*x-7=F*N*9

32*x^2+44*x+5=F*N*9

multiplying everything by 2 and imposing A = 8 * X and B = 8 * X

are obtained

A^2+5*A-14=F*N*9*2

B^2+11*B+10=F*N*9*2

and these are the other 2

2020-09-18, 17:58   #19
CRGreathouse

Aug 2006

3×1,993 Posts

Quote:
 Originally Posted by Alberico Lepore is it possible to apply the Coppersmith method?
Sure, if you have enough information about the factor. If you don't know anything about the factors then it's much slower than other methods.

2020-09-18, 19:28   #20
Alberico Lepore

May 2017
ITALY

5×97 Posts

Quote:
 Originally Posted by CRGreathouse Sure, if you have enough information about the factor. If you don't know anything about the factors then it's much slower than other methods.
What kind of information about the factors?

2020-09-20, 09:11   #21
Alberico Lepore

May 2017
ITALY

5·97 Posts

Quote:
 Originally Posted by Alberico Lepore A^2-3*A-18=F*N*9*2 B^2+3*B-18=F*N*9*2 A^2+5*A-14=F*N*9*2 B^2+11*B+10=F*N*9*2 and these are the other 2
Infinite equations of this type can be generated.

2020-09-20, 11:52   #22
Dr Sardonicus

Feb 2017
Nowhere

4,591 Posts

Quote:
 Originally Posted by Alberico Lepore Infinite equations of this type can be generated.
I am brought to mind of the following:

Quote:
-- from Chapter 5 of Gulliver's Travels by Jonathan Swift

But an infinity of nonsense! You've got the Professor at the grand Academy of Lagado beat, hands down.

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