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MathGroup Archive 2006

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Re: Challenge problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71225] Re: [mg71123] Challenge problem
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 10 Nov 2006 06:39:04 -0500 (EST)
  • References: <200611090837.DAA15514@smc.vnet.net>

On 9 Nov 2006, at 17:37, Coleman, Mark wrote:

> I recently received the annual newsletter from the Math-Stats  
> department
> 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
>


I think I can prove that the answer consists of fractions of the form  
(9 n + 8)/3, where n is any integers. Indeed, you can check that  
these work as follows:

eq = 3q^3 + 10q^2 + 3q;


Simplify[eq /. q -> (9*n + 8)/3]


81*n^3 + 306*n^2 + 361*n + 136

Unfortunately proof that no other expressions work makes no use of  
Mathematica at all. I doubt that such a proof can be found but who  
knows, I have seen more surprising things proved by means of computer  
algebra systems. Anyway, here goes an elementary proof, with only a  
slight use of Mathematica to justify posting it here.

  Assume that we have a solution of the form m/n where m and n are  
relatively prime.
Now, look at the expression:


Factor[Together[3*q^3 + 10*q^2 + 3*q /. q -> m/n]]


(m*(3*m + n)*(m + 3*n))/n^3

Suppose that p is a prime that divides n, hence it has to divide the  
numerator. It can's divide m by assumption. It can't also divide m +  
3*n for the same reason. So it has t divide 3*m +n, and since it does  
not divide m it has to be 3. So n must be a power of 3. So suppose n  
= 3^s. It is easy to see that we must have s=1. So our fraction must  
have n=3. The expression now takes the form


(1/9)*m*(m + 1)*(m + 9)

So m*(m + 1)*(m + 9) must be divisible by 9.

Now m and m+9 are not divisible by 3, so m+1 must be divisible by 9,  
which means m is of the form 9k-1, which is the same as 9l +8.

Andrzej Kozlowski



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