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Apr 25 at 22:09 comment added user49822 The value at $p=-1$ should correspond to the (combinatorial) euler characteristic of the set of real points of the variety
Apr 25 at 22:02 answer added user49822 timeline score: 4
Apr 25 at 7:54 vote accept Alexander Chervov
Apr 24 at 16:26 answer added Peter Taylor timeline score: 5
Apr 24 at 9:06 comment added Alexander Chervov @PeterTaylor Hi, Peter, thanks for the insight ! Would be happy if you share the polynoms, if possible for 10-tuples also. May be as an answer - would be happy to upvote it . Sometimes it is not unique for n-tuples, but can be unique for n+1 tuples. Calculation for p=7 might be within reach, but not so easy for me. As a note: algebraic computations goes only for 2-tuples, all other n-tuples - is just combinatorial enumeration without solving algebraic equations
Apr 24 at 8:40 comment added Peter Taylor I think you're going to need values of $f(7)$ to narrow it down to a single guess. With just $f(-1),f(1),f(3),f(5)$ there are three candidate polynomials for the 9-tuples of the form $9x^{11} + O(x^{10})$ and maximum absolute coefficient $9$, which is suspiciously smaller than the maximum absolute coefficients for 5- to 8-tuples.
Apr 24 at 3:19 comment added Denis T After all, the preimage of $-1$ of a commutator mapping looks not very much unlike Weil-restricted $\mathbb G_m$; that is, group scheme defined by $x^2 + y^2 = 1$, which has exactly $q - (1)^{\frac{q-1}{2}}$ points. Various power residue congruences can lead you to "fake" point count polynomials if you're only looking at the prime field $F_p$ and not $F_q$ — because you do not really know degree of (reduced part of) your hypersurface, and hence cannot distinguish between coefficients coming from Langrange interpolation/Jacobsthal sums contributions, and coefficients coming from Hodge numbers.
Apr 24 at 3:06 comment added Denis T It's very unlikely that your numerical approximations are not contaminated by parts of point count which look like $(-1)^{\frac{q-1}{2}}$. If an algebraic variety has a truly polynomial point count, then it puts heavy restriction on the geometry of complex points, see arxiv.org/abs/math/0612668v3
Apr 23 at 20:02 history edited Alexander Chervov CC BY-SA 4.0
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Apr 23 at 19:55 history edited Alexander Chervov CC BY-SA 4.0
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Apr 23 at 19:51 comment added Sam Hopkins MO vs. AI sounds like a modern day "John Henry" ( en.wikipedia.org/wiki/John_Henry_(folklore) )
Apr 23 at 19:47 history asked Alexander Chervov CC BY-SA 4.0