"Pretty" Calculus I Problem

Discussion in 'General Math' started by Kees J. Boer, Oct 27, 2003.

  1. Kees J. Boer

    Kees J. Boer Guest

    Hi, I was wondering if there is a Mathematical 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
     
    Kees J. Boer, Oct 27, 2003
    #1
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  2. and the solutions to
    f(x) = ax^3 + bx^2 + cx + d = 0
    f'(x) = 3ax^2 + 2bx + c = 0
    f"(x) = 6ax + 2b = 0
    are all integers. Thus for some integers p,q,r
    f(x) = (x-p)(x-q)(x-r) = 0

    Well heck, let f(x) = x^3 and be done.
    I doubt there's others.
    To look for others here's what to do

    f'(x) = (x-q)(x-r) + (x-p)(x-r) + (x-p)(x-q)
    = x^2 - x(q+r) + qr + x^2 - x(p+r) + pr + x^2 - x(p+q) + pq
    = 3x^2 - 3(p+q+r)x + pq + qr + pr = 0
    This require pq + qr + pr to be divisible by 3.
    Now use the quadradic formula to solve for x
    That will determine more restrictions on p,q,r.
    Namely that the discriminate is a square.
    The easiest is to require it to be zero.
    namely 3(p+q+r)^2 = 4(pq + qr + pr)
    3(p^2 + q^2 + r^2) + 2(pq + qr + pr) = 0
    which may not have integral solution.
    In that case p+q+r needs also be divisible by 6.

    f"(x) = x-r + x-q + x-r + x-p + x-q + x-p
    = 6x - 3(p+q+r) = 0
    2x = p+q+r requires p+q+r to be even which is
    pre-empted by requiring p+q+r divisible by 6.
     
    William Elliot, Oct 27, 2003
    #2
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