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Old 2021-11-18, 15:42   #6
RomanM
 
Jun 2021

638 Posts
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Quote:
Originally Posted by Dr Sardonicus View Post
Factor (2x3 + y3)2 - y6 as the difference of two squares.
And what?
nor m^5+1, m^2*(m^3+1)= m^5+m^2 I make an error here
if we take some big m and compute mod((2*m^3+1)^2, m^5+m^2)=1, than compute
mod((2*m^3+1)^2, m^5+m^2+1)=A, A will be relative small compare to m^5+m^2+1,
for p~10^270, A~10^54 compare this to QS 10^134
so instead of sieve we can build such n+eps=p, where p number to factor, eps - small number,
n - number for those we can build the left part as above i.e. B^2==1 mod n

Last fiddled with by RomanM on 2021-11-18 at 16:11 Reason: ***
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