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RomanM · 2021-12-22, 13:59

Hi.
Both QS and NFS stuck with famouse A^2==B^2 mod p.
If we look on the A^3==B^3 mod p that do the split also (30% vs 50% for quadratic)
and in general to A^n==B^n*m & B^m=y mod p?? (1)
as a result, for n*m=2*3*5*7*11... the amount of (1) grow outstanding fast.

It is necessary to assess and verify the possibility of applying this approach.
QS is not good i.e. (A+t)^n-p>>sqrt(p) for n>2 so this lead us to huge FB, long sieve and as result - very very tiny number of smoth numbers.

NFS in spite of this sieve the Linear (!!!) things ((a-b*m)&&(a-b*θ)), and sieve for different n*m is nor do not impossible, but likely not super hard.
LA will be harder, complicated and interesting, thought

Root of polynomial, root of m degree, m>=2.
We have a problem here...
How do You think, Is this idea viable or not?

2021-12-22, 13:59	#1
RomanM Jun 2021 3×17 Posts	An extension of NFS Hi. Both QS and NFS stuck with famouse A^2==B^2 mod p. If we look on the A^3==B^3 mod p that do the split also (30% vs 50% for quadratic) and in general to A^n==B^nm & B^m=y mod p?? (1) as a result, for nm=235711... the amount of (1) grow outstanding fast. It is necessary to assess and verify the possibility of applying this approach. QS is not good i.e. (A+t)^n-p>>sqrt(p) for n>2 so this lead us to huge FB, long sieve and as result - very very tiny number of smoth numbers. NFS in spite of this sieve the Linear (!!!) things ((a-bm)&&(a-bθ)), and sieve for different n*m is nor do not impossible, but likely not super hard. LA will be harder, complicated and interesting, thought Root of polynomial, root of m degree, m>=2. We have a problem here... How do You think, Is this idea viable or not?