"Pretty" Calculus I Problem
Kees J. Boer
I'm trying to create a function in this form.
f(x) = ax^3 + bx^2 + cx + d
These are the prerequisites:
a, b, c, d are all integers.
If ax^3 + bx^2 + cx + d = 0, then the factors are also all integers. In
other words, each of the x-intercepts of the functions consists out of
integers.
When the derivative 3ax^2 + 2bx + c = 0, the solution to x is also an
integer.
When I set the second derivative 6ax + 2b to zero, the solution of x is also
an integer.
In other words, this is a "pretty" Calculus 1 problem.
Is there a way outside of the trial and error method that I can do this?
Thanks!
Kees
Virgil
f(x) = (x-n)^3 will work.
Robert Israel
>I'm trying to create a function in this form.
>f(x) = ax^3 + bx^2 + cx + d
>These are the prerequisites:
>a, b, c, d are all integers.
>If ax^3 + bx^2 + cx + d = 0, then the factors are also all integers. In
>other words, each of the x-intercepts of the functions consists out of
>integers.
You want three integer roots? Choose your integer roots roots r1, r2, r3,
and a=1, b = -(r1+r2+r3), c=r1 r2 + r1 r3 + r2 r3, d = - r1 r2 r3 are integers.
>When the derivative 3ax^2 + 2bx + c = 0, the solution to x is also an
>integer.
The derivative can be written as
3 (x - (r1+r2+r3)/3)^2 - (r1^2 + r2^2 + r3^2 - r1 r2 - r2 r3 - r1 r3)/3
so you want r1+r2+r3 to be divisible by 3 and
r1^2 + r2^2 + r3^2 - r1 r2 - r2 r3 - r1 r3 to be a square divisible by 9.
>When I set the second derivative 6ax + 2b to zero, the solution of x is also
>an integer.
The second derivative can be written as
6 x - 2 (r1 + r2 + r3), so again you want r1+r2+r3 to be divisible by 3.
Let a = (r1 + r2 + r3)/3, b = r1 - r2 and c = a - r3. Then
r1 = a + b/2 + c/2, r2 = a - b/2 + c/2, r3 = a - c, and
r1^2 + r2^2 + r3^2 - r1 r2 - r2 r3 - r1 r3 = (9 c^2 + 3 b^2)/4 must be
a square divisible by 9, say 9 d^2.
We need b divisible by 3, say b = 3B. And then c^2 + 3 B^2 = 4 d^2
or 3 B^2 = 4 d^2 - c^2 = (2d - c)(2d + c). To solve this we take
2d - c = 3 u^2, 2d + c = v^2, B = u v, u and v both even or both odd
so d = (v^2 + 3 u^2)/4 and c = (v^2 - 3 u^2)/2 are integers. Then
r1 = a + (3 v^2 + 6 u v - u^2)/4, r2 = a + (3 v^2 - 6 u v - u^2)/4,
r3 = a + (u^2 - 3 v^2)/2. For example, with a=0, u=1, v=3 we get
r1 = 11, r2 = 2, r3 = -13, and the polynomial is x^3 - 147 x + 286.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Bill Dubuque
"nice" cubic polynomials p(x) in Z[x] such that the roots of p, p', p''
are all integers. This and related problems have been frequently discussed,
for example see the links below, and especially the references listed
after the review below.
-Bill Dubuque
Al Cuoco: Meta-problems in mathematics
http://www.edc.org/CME/showcase/metaproblems.pdf
http://www.edc.org/CME/showcase/nctmslides.pdf
------------------------------------------------------------------------------
2001c:11035 11D25 (11D41 11G05 11G30)
Buchholz, Ralph H.; MacDougall, James A.(5-NEWC)
When Newton met Diophantus: a study of rational-derived polynomials
and their extension to quadratic fields.
J. Number Theory 81 (2000), no. 2, 210-233.
http://dx.doi.org/10.1006/jnth.1999.2473
------------------------------------------------------------------------------
This is an interesting paper, which surveys the problem of determining the
set D(n) of all "k -derived" univariate polynomials of degree n (where
a polynomial f in k[x] is k-derived if f and each of its successive
derivatives has all roots in the ground field k ). Define two polynomials
f_1, f_2 in k[x] to be equivalent if f_1(x) = r f_2(sx+t) for r,s,t in k,
r,s != 0. Then up to equivalence, the following is known about Q-derived
polynomials:
D(1) = {x};
D(2) = {x^2, x(x-1)};
D(3) = {x^3} \/ { x (x-1)(x-a) | a = {w(w-2)}/{w^2-1}, w in Q};
D(4)>= {x^4} \/ {x^2(x-1)(x-a) | a = {9(2w+z-12)(w+z)}/{(z-w-18)(8w+z)},
(w,z) in E(Q), E: z^2 = w(w-6)(w+18)};
D(n)>= {x^n, x^{n-1}(x-1)} for n >= 5.
The authors prove that determining D(n) in general devolves into two
conjectures: (1) that no quartic with four distinct roots is Q-derived;
(2) that no quintic of type x^3(x-a)(x-b), a != b, a,b != 0, is Q-derived.
The first conjecture can be solved by determining all rational points on a
hyperelliptic surface of degree 10. The second conjecture can be solved by
determining all rational points on a curve of genus 2 (E. V. Flynn ["On
Q-derived polynomials", Preprint; per revr.] has now proved this second
conjecture). The authors also discuss briefly the situation of K-derived
polynomials for quadratic extensions K of Q; there is, for example, the
quartic y^2 = x^2(x-1)(x-(37-20\sqrt(3))/13) which is a Q(\sqrt{3})-
derived polynomial.
Reviewed by Andrew Bremner
References
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2. Enrico Bombieri, The Mordell conjecture revisted, Ann. Scuola
Norm. Pisa Cl. Sci. (4) 17 (1990), 615-640. [MR1093712 (92a:11072)]
3. T. Bruggeman and T. Gush, Nice cubic polynomials for curve sketching,
Math. Mag. 53 (1980), 233-234.
4. R. H. Buchholz and S. M. Kelly, Rational derived quartics,
Bull. Austral. Math. Soc. 51, No. 1 (1995), 121-132. [MR1313118 (95k:11033)]
5. Jim Buddenhagen, Charles Ford, and Mike May, Nice cubic polynomials,
pythagorean triples, and the law of cosines, Math. Mag. 65, No. 4
(Oct. 1992), 244-249. [MR1184558]
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23, No. 22 (1990), 36-38.
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Amer. Math. Monthly 96, No. 2 (Feb. 1989), 129-130.
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9, No. 3 (1987), 43.
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Arithmetic of Curves of Genus 2," London Math. Soc. Lecture Note
Ser. 230, Cambridge Univ. Press, 1996. [MR1406090 (97i:11071)]
10 M. Chapple, A cubic equation with rational roots such that it and
its derived equation also has rational roots, Bull. Math. Teachers
Secondary Schools 11 (1960), 5-7. (recently re-published in
Aust. Senior Math. J. 4, No. 1 (1990), 57-60).
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No. 4 (Dec. 1994), 367-394. [MR1334430 (96f:11137)]
12 L. E. Dickson, "History of the Theory of Numbers," Vol. 2, Chelsea, 1952.
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LXVI.1 (1994), 23-43. [MR1262651 (95g:11057)]
14 E. V. Flynn, Private correspondence, Dec. 1995.
15 Bill Galvin, "Nice" cubic polynomials with "Nice" derivatives,
Austral. Senior Math. J. 4, No. 1 (1990), 17-21.
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The Sequel, Austral. Senior Math. J. 8, No. 1 (1994), 23-27.
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96, No. 10 (Dec. 1989), 907-908.
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quartic polynomial curves, College Math. J. 23, No. 3 (May 1992), 186-195.
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G. A. Edgar
<kees...@integrity-computing.net> wrote:
In the problem section of Mathematics Magazine a while back there was
something much like this!
You can see the solution in your university library. If you or your
institution subscribe to JSTOR, you can see it on-line there.
Problem 1072 "A Tailored Polynomial", solution September, 1980, p. 247.
They had f of degree 4, didn't require factoring f, but only f' and f''.
The best one is: x^4+8x^3-270x^2.
For your problem, use the derivative of that: f(x) = 4x^3+24x^2-540x
or divide by 4, f(x) = x^3+6x^2-135.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Lynn Kurtz
.....and the polynomial is x^3 - 147 x + 286.
>
Nice. And to take it one step farther put x -> x-1 to get
f(x) = x^3 - 3x^2- 144x + 432
which has all the terms and all nonzero integer factors.
--Lynn
Stephen M. Fortescue
On 26 Oct 2003, Kees J. Boer wrote:
>
>Hi, I was wondering if there is a Mathimatical Way to do this.
>
>I'm trying to create a function in this form.
>
>f(x) = ax^3 + bx^2 + cx + d
>
>These are the prerequisites:
>
>a, b, c, d are all integers.
>
>If ax^3 + bx^2 + cx + d = 0, then the factors are also all integers.
>In other words, each of the x-intercepts of the functions consists
>out of integers.
>
>When the derivative 3ax^2 + 2bx + c = 0, the solution to x is also an
>integer.
>
>When I set the second derivative 6ax + 2b to zero, the solution of x
>is also an integer.
>
>In other words, this is a "pretty" Calculus 1 problem.
>
>Is there a way outside of the trial and error method that I can do this?
>
>Thanks!
>
>Kees
Suppose that the three roots of the cubic are s, t and 0.
Then the cubic is:
x^3 - (s+t) x^2 + s t x = 0
and its derivative is:
3 x^2 - 2 (s+t) x + s t = 0
The roots of the derivative are:
x = [ (s+t) +- sqrt( (s+t)^2 - 3 s t ) ] / 3
So the problem reduces to finding s and t such that
(s^2 - s t + t^2) is a square,
or finding a triangle with a 60 degree angle and integer sides.
Such solutions can be found by squaring Eisenstein integers.
(2 + 3 w)^2 = (4 + 12 w + 9 w^2) = (-8 - 3 w^2)
since w^3 = 1, and 1 + w + w^2 = 0.
Using 3 and 8 for s and t produces 7^2, and 4/3 or 6 for x.
Triple to get integers: s=9, t=24, x1=4, x2=18, inflection at 11.
x^3 - 33 x^2 + 216 x = 0
Another example:
(5 + w)^2 = (25 + 10 w + w^2) = (13 - 2 w - 11 w^2)
(x + 13)(x - 2)(x - 11) = x^3 - 147 x + 286
which is Robert Israel's example.