Discriminant of derivative of cubic equation being a perfect square Is it possible for the discriminant of the first derivative of a cubic polynomial $(x+a)(x+b)(x+c)$, where $a, b$ and $c$ are distinct non-zero integers (i.e. Discriminant $[d((x+a)(x+b)(x+c))/dx]$ in Wolfram|Alpha/Mathematica) to be a perfect square?
If so, what are the smallest absolute values of $a, b$ and $c$? Thanks!
 A: Let $p(x) = x^3 + (a + b + c)x^2 + ((b + c)a + bc)x + abc$. Then, 
$$p'(x) = 3x^2 + 2(a+b+c)x + ((b+c)a+bc).$$
The discriminant of $p'(x)$ is
$$\Delta = 4(a+b+c)^2 - 4\cdot 3\cdot ((b+c)a+bc)=4(a^2 - (b + c)a + (b^2 - bc + c^2)).$$
So you need to find values of $a,b,c$ that make $\Delta$ a square... Notice that of course $\Delta$ is symmetric in $a,b,c$:
$$\Delta = 4(a^2+b^2+c^2-ab-bc-ac).$$
The solution $a=1$, $b=4$, $c=-4$ seems to be the "smallest", with $a,b,c$ all distinct, and non-zero (because of the symmetry of $\Delta$, you can permute $a,b,c$ and get the same solution). This gives
$$p(x)=(x+1)(x+4)(x-4)$$
and
$$p'(x)=3x^2 + 2x - 16,$$
whose discriminant is $4-4\cdot 3\cdot (-16) = 196$.
Notice that the very first expression we found for the discriminant:
$$\Delta = 4((a+b+c)^2 - 3\cdot ((b+c)a+bc)),$$
gives an infinity family of rational solutions to your problem when $(b+c)a+bc=0$, that is, when $b\neq -c$ and 
$$a=-\frac{bc}{b+c}.$$
This is an integer if $(b+c)$ divides $bc$, but $bck=b+c$ implies that $b$ divides $c$, so $c=bf$ and $bc/(b+c)=b^2f/(b(1+f))=bf/(1+f)$. The numbers $f$ and $1+f$ are relatively prime, so we need $1+f$ to divide $b$, say $b=(1+f)g$. Then 
$$b=(1+f)g, \quad c=bf=(1+f)fg$$
and
$$\frac{bc}{b+c} = \frac{(1+f)^2g^2f}{(1+f)g+(1+f)fg}=\frac{(1+f)fg}{(1+f)}=fg.$$
Thus, we get an infinite family of integer solutions:
$$a=-fg, \quad b=(1+f)g,\quad c=(1+f)fg,$$
for any $f,g\in\mathbb{Z}$. For instance, $f=2$, $g=-1$, yields
$$a=2,\quad b=-3,\quad c=-6$$
so 
$$p(x)=(x+2)(x-3)(x-6)$$
with
$$p'(x)= 3x^2 - 14x$$
and the discriminant is $14^2$.
