Quote:
Originally Posted by Dr Sardonicus
Factor (2x^{3} + y^{3})^{2}  y^{6} 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