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+((ba)/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^23*a
multiplying by 2 and imposing 2 * a = A
we will have 18*N*F=A^23*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*F1)/8=(X^21)/82*((ba)/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^26*X9=F*N*9
8*X^2+6*X9=F*N*9
multiplying everything by 2 and imposing A = 4 * X and B = 4 * X
are obtained
A^23*A18=F*N*9*2
B^2+3*B18=F*N*9*2
solve (65*F1)/8=(X^21)/82*((ba)/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^26*X9=F*65*9
8*X^2+6*X9=F*65*9
multiplying everything by 2 and imposing A = 4 * X and B = 4 * X
are obtained
A^23*A18=F*65*9*2
B^2+3*B18=F*65*9*2
and these are the first two
and then
solve (N*F1)/8=x*(x+1)/22*((ba)/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*x7=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*A14=F*N*9*2
B^2+11*B+10=F*N*9*2
and these are the other 2