Re: Challenge problem

• To: mathgroup at smc.vnet.net
• Subject: [mg71184] Re: [mg71123] Challenge problem
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Fri, 10 Nov 2006 06:37:38 -0500 (EST)

```Using brute force:

Union[Select[Flatten[Table[{q=n/m,3q^3+10q^2+3q},{n,-40,40},{m,100}],1],
(Element[#[[1]],Rationals] &&
Not[Element[#[[1]],Integers]] &&
Element[#[[2]],Integers])&]]

{{-(37/3), -4144}, {-(28/3), -1596}, {-(19/3), -380},
{-(10/3), -10}, {-(1/3), 0}, {8/3, 136}, {17/3, 884},
{26/3, 2730}, {35/3, 6160}}

Clear[n,q];

Let q = (9n-1)/3

3q^3+10q^2+3q/.q->(9n-1)/3//Simplify

n*(81*n^2 + 63*n - 8)

which is integer for all integer n.

Bob Hanlon

---- "Coleman wrote:
> of my undergraduate alma mater. In part of the newsletter they posed the
> following challenge problem:
>
> "For which rational numbers q is 3q^3 + 10q^2 + 3q an integer?"
>
> The problem comes from an annual mathematics competition the school
> sponsors.  I fumbled around a bit, using Mathematica v5.2 to attempt an answer,
> but without much luck. Of course I'm a statistician, not an algebraist
> :-). But my curiosity is now piqued and I was wondering if someone might
> outline an elegant answer using Mathematica.
>
> Thanks,
>
> -Mark
>

```

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