# A cubic equation whose roots are distinct natural number having no common divisors

If the roots of the cubic equation $$ax^3+bx^2+cx+d=0$$ are three distinct natural numbers having no common divisors except $$1$$ (i.e. they are relatively prime), what could be the values of $$a,b,c,d$$ if:

• they are real?

• they are complex?

I have no idea how to begin solving this problem. Any help would be appreciated. THANKS!

• Do you mean "no common divisors greater than $1$"? – Andrew Chin Nov 8 at 16:37
• @AndrewChin "no common divisors except $1$ (i.e. they are relatively prime) – Hussain-Alqatari Nov 8 at 16:38

Let the roots be $$r_1$$, $$r_2$$, $$r_3$$. Divide you cubic through by $$a$$ and set \begin{align*} f(x) &= (x-r_1)(x-r_2)(x-r_3) \\ &= x^3 + \frac{b}{a} x^2 + \frac{c}{a} x + \frac{d}{a} \text{.} \end{align*}
From this, we see $$b/a$$, $$c/a$$, and $$d/a$$ are all integers (because they are the sums and products of the integers $$r_1$$, $$r_2$$, $$r_3$$, which can be seen explicitly by expanding the first line and comparing coefficients by degrees. From this, we see $$d/a$$ is minus the product of three distinct relatively prime integers, $$c/a$$ is the sum of the three products of pairs of three distinct relatively prime integers, and $$b/a$$ is minus the sum of three distinct relatively prime integers.
Since $$r_1$$, $$r_2$$, and $$r_3$$ are all integers (hence real numbers), the discriminant of $$f$$ is positive. (Notice we have arranged that the leading coefficient of our $$f$$ is $$1$$, and the cited article has not done so. It is frequently much easier to go a little further and convert to the depressed cubic because its discriminant is so much simpler.)