# "Pretty" Calculus I Problem

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

1. ### Kees J. BoerGuest

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

2. ### William ElliotGuest

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.
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