If N is not a prime number
that is c * d = N with c and d not necessarily prime numbers
if N mod 4 = 1
If we solve F as a function of a and N
solve 2(9NF)+2a^2+((b-a)/2)^2=((3a+b)/2)^2 , ab=(9NF) , 2(9NF)+21^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0 ,F,b
->
9NF=2a^2-3a
multiplying by 2 and imposing 2 * a = A
we will have 18NF=A^2-3*A
A0 < sqrt(18*N)
or
If we solve F as a function of b and N
solve 2(9NF)+2a^2+((b-a)/2)^2=((3a+b)/2)^2 , ab=(N9F) , 2(9NF)+21^2+((a+b)/2+1)^2-((3*a+b)/2)^2=0 ,F,a
->
18NF=b^2+3*b
b0 < sqrt(18*N)
is it possible to apply the Coppersmith method?
Example
N=105
18105F=b^2+3*b
https://math.stackexchange.com/questions/3821343/is-it-possible-to-apply-the-coppersmith-method