Free RSA based Ransomware Decryption [attempt]

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=(9NF) , 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