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!
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.)
... At least, that's where I would begin solving this problem ...