RE: Challenge problem

• To: mathgroup at smc.vnet.net
• Subject: [mg71203] RE: [mg71123] Challenge problem
• From: "David Park" <djmp at earthlink.net>
• Date: Fri, 10 Nov 2006 06:38:07 -0500 (EST)

```Mark,

I'm sure you will get an elegant solution from Andrzej Kozlowski and others.
Here is my more feeble attempt.

I take that we are only interested in non-integer rational solutions of the
form n/d.

f[q_] = 3q^3 + 10q^2 + 3q;

Factor[f[q]];
f2[n_, d_] = % /. q -> n/d
(n*(3 + n/d)*(1 + (3*n)/d))/d

f2 is the polynomial in terms of n and d. For the second and third factors,
both have to be integers and one has to be a multiple of d. It turned out
that making the third factor a multiple of d leads only to interger
solutions.  So we write and solve the following equations with the second
factor a multiple of d:

eqns1 = {3 + n/d == k1 d, 1 + 3n/d == k2};
sols1 = First@Solve[eqns1, {n, d}]

{n -> ((-1 + k2)*(8 + k2))/(9*k1),
d -> (8 + k2)/(3*k1)}

Substituting the solutions into f2[n,d] we see that one of the factors must
be multiples of 9. The only choice that gave rational non-integer solutions
was putting k2 -> 9p

f2[n, d] /. sols1 // Simplify
% /. k2 -> 9p
(1/9)*(-1 + k2)*k2*(8 + k2)
p (-1 + 9 p) (8 + 9 p)

The rational numbers themselves, some solutions and the corresponding values
of the polynomial are given by:

n/d /. sols1 /. k2 -> 9p
Table[%, {p, -20, 20}]
f /@ %

(1/3)*(-1 + 9*p)

{-(181/3), -(172/3), -(163/3),
-(154/3), -(145/3), -(136/3),
-(127/3), -(118/3), -(109/3),
-(100/3), -(91/3), -(82/3),
-(73/3), -(64/3), -(55/3),
-(46/3), -(37/3), -(28/3),
-(19/3), -(10/3), -(1/3), 8/3,
17/3, 26/3, 35/3, 44/3, 53/3,
62/3, 71/3, 80/3, 89/3, 98/3,
107/3, 116/3, 125/3, 134/3, 143/3,
152/3, 161/3, 170/3, 179/3}

{-622640, -532684, -451836, -379610, -315520, -259080, -209804, -167206, \
-130800, -100100, -74620, -53874, -37376, -24640, -15180, -8510, -4144, \
-1596, -380, -10, 0, 136, 884, 2730, 6160, 11660, 19716, 30814, 45440,
64080, \
87220, 115346, 148944, 188500, 234500, 287430, 347776, 416024, 492660, \
578170, 673040}

David Park

From: Coleman, Mark [mailto:Mark.Coleman at LibertyMutual.com]
To: mathgroup at smc.vnet.net

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
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

```

• Prev by Date: Merge of Matrices 2
• Next by Date: Bairstow Method
• Previous by thread: Re: Challenge problem
• Next by thread: Re: Merge of Matrices