I'm slowly came to understand this! Thanks!
this originlal formula.
mod((a*m^3+1/2*a*C)^2,m^5+C*m^21/4*a^2*C^24/a^2*m)=A
a=2, C=1 and we have a silly me with super obvios (2m^3+1)^2=blah blah))
whatever, main idea still the same
instead of sieve we can build alot of such n+eps=p, where p number to factor, eps  small number, 1 is the best)),
n  number of some special form, for those we can build the left part as above i.e. B^2==1 mod n, then compute our beloved little residual, factor them and do the same math as in QS
Last fiddled with by RomanM on 20211118 at 18:10
