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Old 2021-11-18, 17:43   #8
RomanM
 
Jun 2021

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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
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