Re: pain in neck calculus problem
From: Brian VanPelt (bvanpelt_at_neo.rr.com)
Date: 06/27/04
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Date: Sun, 27 Jun 2004 05:16:37 GMT
I have not solved this problem, but I sort of played with it a little,
and here are some of my ideas.
Since we need 2 inflection points, I though about looking at
y ' ' = (x - p)(x - q) = x^2 - (p + q) x + pq
Therefore, integrating once gives (with not constant term)
y ' = (x^3) / 3 - ((p + q) x^2) / 2 + pqx
= (x / 6) (2x^2 - 3(p + q)x + 6pq)
If my calculations are correct, the solutions to
2x^2 - 3(p + q)x + 6pq = 0
are
[ 3(p + q) +- sqrt( (3p - q)(p - 3q) ) ] / 4
Then, in order to get integers, we at least need for
(3p - q)(p - 3q)
to be a perfect square. So, we can set up equations of the form
3p - q = k
p - 3q = k
I found that k = 12 will make integer solutions to the system of
equations.
3p - q = 12
p - 3q = 12
However, that will not solve the original problem.
I think that going along these lines may work. You can let k be any
multiple of 12 (I have no idea right now), or perhaps some other
integer that makes
[ 3(p + q) +- sqrt( (3p - q)(p - 3q) ) ] / 4
into at least one integer solution as well as have p and q be integer
solutions.
Just my thoughts,
Brian
On Sun, 27 Jun 2004 03:14:22 GMT, "Troubled" <mkajumap@hotmail.com>
wrote:
>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.
>
>
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