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#12 |
May 2017
ITALY
52·19 Posts |
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before understanding this
I have to understand this would you give me a little clue |
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#13 | |
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6,113 Posts |
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It's up to you to show us how you factor things, not the other way around. Your claim, you prove it. |
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#14 | |
May 2017
ITALY
52·19 Posts |
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(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 |
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#15 |
May 2017
ITALY
52×19 Posts |
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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 |
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#16 |
May 2017
ITALY
52·19 Posts |
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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 |
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#17 |
Mar 2019
2248 Posts |
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Please, stop posting.
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#18 | |
May 2017
ITALY
47510 Posts |
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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 |
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#19 |
Aug 2006
32×5×7×19 Posts |
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#20 |
May 2017
ITALY
52×19 Posts |
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#21 |
May 2017
ITALY
1110110112 Posts |
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#22 | |
Feb 2017
Nowhere
3×1,481 Posts |
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I am brought to mind of the following:
Quote:
But an infinity of nonsense! You've got the Professor at the grand Academy of Lagado beat, hands down. |
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