Re: pain in neck calculus problem

From: Stephen M. Fortescue (sfortescue_at_adelphia.net)
Date: 06/27/04 

Date: 27 Jun 2004 16:42:44 -0700

"Troubled" <mkajumap@hotmail.com> wrote in message news:<iYqDc.116$dc5.74@news01.roc.ny>...
> I need to find a 4th degree polynomials with 3 extrema and 2 points
> of inflection, all 5 numbers being integers. I started off with
> letting y' = x (x-p) (x-q) , with p and q integers (so the critical
> values will be 0,p,and q). Then y" = (x-p)(x-q) + x(x-q) + x(x-p).
> I concluded that y"=0 when [(p+q) +/- sqrt( (p+q)^2-3pq)]/3.
> I know that (p+q)^2-3pq must be a perfect square, I^2. I concluded
> that p= [q +/-sqrt( (2I)^2 - 3q^2)]/2. This is where I got stuck.
> At this point I'd pick values for q and I and see if p is an integer
> and if it was I'd check to see if it satisfied the equation above.
> But I can't find a p,q combo that works, nor do I like to randomly
> pick numbers. Why is this problem so hard? Why can't I find a 4th
> degree polynomial with any 5 integers (or any 5 numbers)that satisfy
> the statement? Any hints or even solutions at this point would be
> appreciated. Believe it or not this is not homework. It is the
> problem of the month at a college which I do not attend that I came
> across.

This problem has appeared here before in disguise.
Robert Israel posted one solution method, and I posted another:

http://groups.google.com/groups?threadm=bnid8d%245u9%241%40nntp.itservices.ubc.ca

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