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 2021-11-18, 17:43 #8 RomanM   Jun 2021 5110 Posts I'm slowly came to understand this! Thanks! this originlal formula. mod((a*m^3+1/2*a*C)^2,m^5+C*m^2-1/4*a^2*C^2-4/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 2021-11-18 at 18:10