# Cubic polynomial discriminant

I am stuck on a cryptography problem that pertains to Elliptic Curves.

The problem is stated as follows:

Assume the cubic polynomial $X^3+AX+B = (X-a)(X-b)(X-c)$

If $4A^3 + 27B^2 = 0$, then show two or all of $a,b,c$ are the same.

So far, I expanded the right side, so I get the following equations:

$0 = a + b + c$

$A = ab + ac + bc$

$B = -abc$

I can't seem to use the hypothesis in the correct way. I tried to compute $A^3$ but the expansion of it looks horrible. If any one has any tips how to approach this problem, that would be great.

-
Do you recognise that $4A^3+27B^2$ is the discriminant of the cubic? –  Andrew Jul 26 '12 at 13:44
Also, you might want to change the title and possibly the tags, as the question, though it will be related to elliptic curves and cryptography at some point, is really just about polynomials as stated. –  Tobias Kildetoft Jul 26 '12 at 13:54
@Andrew: Yes I do now. But I guess I'm missing some key fact about that that allows me to conclude what I'm trying to show. –  MathNewbie Jul 26 '12 at 14:05
@Tobias: Hopefully this will be much better of a tag. –  MathNewbie Jul 26 '12 at 14:07
In general, the discriminant of a polynomial is the product of the square of the differences of the roots. Hence, it is zero if and only if there is a multiple root. –  M Turgeon Jul 26 '12 at 14:09

The case $A=0$ must be treated separately: In this case, $X^3+B$ has a double root if and only if $27B^2=0$, i.e. if and only if $B=0$. That is obvious, so let us assume $A\ne 0$.

Lets set $f(X):=X^3+AX+B$ and consider $f'(X)=3X^2 + A$. A polynomial has a double root if and only if it shares this root with its derivative. Hence, let us check when that is the case. $f'(x)=0$ means $x^2 = -\frac 13 A$ and $f(x)=0$ means $0=x^3+Ax+B=-\frac{1}{3}Ax+Ax+B$, so $\frac{2}{3}Ax = -B$, i.e. $x=\frac{-3B}{2A}$. Plugging this back into $f'$, we get that $$0=f'(x)=3\cdot\frac{9B^2}{4A^2} + A,$$ which yields $27B^2 = -4A^3$. Hence, $f$ has a double root if and only if $27B^2+4A^3=0$.

-
Thanks, rattle. –  MathNewbie Jul 26 '12 at 17:50

Let $X^3+Ax+B=(X-p)(X-q)^2$

Comparing the coefficients of the different powers of X,

$p+2q=0=>p=-2q$

$2pq+q^2=A=>A=-3q^2$

$-pq^2=B=>B=-q^2(-2q)=2q^3$

Eliminating q, $4A^3+27B^2=0$ as $A^3=-27q^6$ and $B^2=4q^6$

Conversely, the parametric values of (A,B) can be written as $(-3s^2,2s^3)$

Then, the equation becomes $X^3-3s^2X+2s^3=0$

Clearly, s is one of the solutions.

On the division by (X-s), we get $X^2+sX-2s^2=0 =>X=s,-2s$

-