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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!

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When you say "unique non-zero integers", do you mean "distinct non-zero integers"? –  Álvaro Lozano-Robledo Dec 15 '11 at 17:02
    
Thanks, I meant "distinct" (I've edited the question). @AndréNicolas: Discriminant[d((x+1)(x+2)(x+3))/dx] is 12 which is not a perfect square. –  Gnubie Dec 16 '11 at 18:52

1 Answer 1

up vote 4 down vote accepted

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 $14^2$.

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That's brilliant! Thank you very much. –  Gnubie Dec 16 '11 at 18:55

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