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 djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Coleman, Mark [mailto:Mark.Coleman at LibertyMutual.com] To: mathgroup at smc.vnet.net 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