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Re: Solve or Reduce on a monstrosity of an expresssion (and a prize!)

I'm taking a few parts on faith (I guess technically we all are given the
computer-assisted nature of this proof!) but it sounds like you and
Andrzej have similar ideas about establishing at most 2 roots.  I don't
know what you have in mind though for how to establish "sufficiently many"
values of n.  n=2,3,4,5,6, and 7 all work but for n=8, GroebnerBasis does
not include a degree 2 polynomial.  (Still, FindInstance confirms no roots
in (0,(n-1)/n) for n=8, and in fact does so all the way up to 684.)

--- \/   FROM Daniel Lichtblau AT 05.05.25 13:18 (Today)   \/ ---

> Daniel Reeves wrote:
> > yikes.  ok, let's assume that all works.  So it remains to show that both
> > of the following expressions of n are zero for all n >= 2.  Strangely,
> > FullSimplify can do this for all specific n up to 113 but 114 seems to
> > stump it.
> >
> > here are the expressions:
> > [...]
> I can point out some aspects to a probabilistic argument. Making it into
> a full fledged proof would require figuring out some bounds that offhand
> I don't know how to do, then verifying for appropriately many cases.
> First you need to show that there are 2 roots for each x, at least when
> n>=2. For this it suffices to show it holds for generic n. Since we
> cannot seem to do that computationally, instead do it for "random" n.
> Calling your expression fp, do, say,
> fp7 = Simplify[fp /. n->7]
> gb7 = GroebnerBasis[fp7,x]
> Now check for a quadratic polynomial (in x) in gb7. If you find one,
> that means there are at most 2 solutions when n=7. This can of course be
> repeated for other values of n. The point is that, for integers larger
> than 2, this computation appears to be feasible. Offhand I think this
> idea (of verifying at sufficiently many values of n) suffices, though
> there may be a lurking branch cut issue. Note that showing there are two
> solutions for n=7 is not quite sufficient in and of itself because 7
> might not be generic. In this setting, nongeneric points would lie on a
> variety (roots of a polynomial), hence be a set of measure zero.
> Verification that x->(n-1)/n and x->-(n+1)^2/(n*(n-1)) are the two
> solutions can be done by making the substitutions into fp and verifying,
> for sufficiently many n>=2, that these vanish. The reason is that
> univariate algebraic functions vanishing at sufficiently many integer
> points vanish everywhere that branches don't change (e.g. x-Sqrt[x^2]
> vanishes on positive integers and indeed on all positive reals). I think
>    the branch cut issue goes away when you have the proviso that n>=2,
> because for the substitutions above all radicands are positive in that
> region. But maybe I'm missing something and there is more to it than this.
> I should mention that if you did not know those were the appropriate
> substitutions, they might be deduced by finding particular {x,n} pairs
> and getting more values via homotopy continuation (basically, setting up
> and solving an ODE via NDSolve). Values from this could then be used to
> interpolate an implicit relation. This was the gist of some work I
> described in a note last week:
> Offhand I do not know how well it would work for your particular
> algebraic expression, but it strikes me as a plausible approach. (A
> better one for this case might be to take an implicit plot of the zero
> set of fp and guess the relation).
> Daniel Lichtblau
> Wolfram Research

--  - -  google://"Daniel Reeves"

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choreography, and the dancers hit each other."
  -- Jack Handy, Deep Thoughts

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