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Old 2020-09-18, 17:51   #18
Alberico Lepore
 
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May 2017
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Quote:
Originally Posted by Alberico Lepore View Post
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
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