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Here is my attempted approach to prove that the discriminant $\triangle = 4a^3+27b^2$ of an elliptic curve in the form of $y^2 = x^3 + ax +b$ is zero. I have a problem at the end which doesn't bring me to the expected conclusion. Below is the process, please let me know the error in the process. I try to approach this from the definition of non-singularity (part of the definition of an elliptic curve), which is equivalent to the statement: the equation $y^2 = x^3 + ax +b$ is differentiable everywhere on the graph. From that definition, I try to derive the derivative of $y$ in respect to $x$ through implicit differentiation: \begin{align} y^2 &= x^3+ax+b\\ \label{ref1} 2yy' &= 3x^2+a\\ y' &= \frac{3x^2+a}{2y} \end{align} If the graph is singular, then $y'$ does not exist, in other words: \begin{align} 2y &= 0\\ 3x^2+a &\neq 0 \end{align} I replace $y = 0$ in the equation of elliptic curve which yields \begin{equation} 0 \ = \ x^3 + ax + b \end{equation} I apply Cardano's method, who realize that the form could be represented as \begin{equation} (\alpha-\beta)^3 + 3\alpha\beta(\alpha-\beta) = \alpha^3 - \beta^3 \end{equation} in which \begin{align} \alpha\beta &= \frac{a}{3} \\ \alpha^3 - \beta^3 &= -b \end{align} By substituting $\alpha = \frac{a}{3\beta}$ in the second equation from above, I obtain \begin{equation} (\frac{a}{3\beta})^3 - \beta^3 = -b \end{equation} I further simplify this by considering $\beta^3$ as a whole, i.e. \begin{align} \frac{a^3}{27}-\beta^6 &= -b\beta^3\\ (\beta^3)^2 - b\beta^3 -\frac{a^3}{27} &= 0 \end{align} By quadratic formula, \begin{equation} \beta^3 = \frac{b\pm \sqrt{b^2+\frac{4a^3}{27}}}{2} = \frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}} \end{equation} From $\alpha^3 - \beta^3 = -b$, I get \begin{equation} \alpha^3 = \beta^3 - b = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}} \end{equation} From $3x^2+a \neq 0$: \begin{equation} 3\left(\left(-\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3} - \left(\frac{b}{2} \pm \sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)^\frac{1}{3}\right)^2 + a \neq0 \end{equation} The equation above is from $3(\alpha-\beta)^2+a \neq 0$. By De Moivre's Formula there are two other equations, $3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$ and $3\left(\alpha(\frac{-1}{2}+\frac{\sqrt{3}i}{2})^2-\beta(\frac{-1}{2}+\frac{\sqrt{3}i}{2})\right)^2+a \neq 0$. These result in several representations of b in terms of a and one integer solution, $a=0,b=0$ shouldn't exist at the same time, $b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$. These two meet my expectation, $\triangle = 0$. However, there are other four representations that do not meet my expectation $b \neq \pm \sqrt{\frac{a^3}{54}\pm\frac{5ia^3}{6\sqrt{3}}}$ and $b \neq \pm \sqrt{\frac{a^3}{54}\mp\frac{5ia^3}{6\sqrt{3}}}$. I also try to see what I could get from \begin{equation} 3\left(\left(-\frac{b}{2}\right)^\frac{1}{3} - \left(\frac{b}{2}\right)^\frac{1}{3}\right)^2 + a \neq0 \end{equation} which yields that $a=0,b=0$ shouldn't exist at the same time, or $b\neq\pm \frac{2ia^{3/2}}{3\sqrt{3}}$ these satisfy $\triangle = 0$ and also a weird pair of $b \neq \pm \frac{ia^{3/2}}{12\sqrt{3}}$, which doesn't even satisfy $\triangle = 0$. Could someone explain this?

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When $3x^2 + a \ne0$ and $2y = 0$, the curve is still nonsingular, its just that the slope of the tangent line is infinite. The singularity occurs when $$3x^2 + a = 0 \text{ and } 2y =0$$ simultaneously i.e. $a = - 3x^2$ so $0 = x^3 + (-3x^2)x + b$ giving $$ b = 2x^3$$ this is what leads to the equation $$27b^2 + 4a^3 = 0$$

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