I don't belong to the Math department. Currently, I am studying Elliptic curves for cryptographic applications. While going through the definition of elliptic curves, it states that
Elliptic curve is a curve of the form $y^2 = p(x)$, where $p(x)$ is a cubic polynomial with no repeated roots.
But, the twisted Edwards curve which belongs to the Elliptic curve family has an equation $a\ X^2 + Y^2 = 1 + d\ X^2Y^2$, which is not a cubic equation.
Why is this equation considered an elliptic curve equation even though it does not go with the definition of the elliptic curves?
Consider a twisted Edwards curve:
$$a\ X^2 + Y^2 = 1 + d\ X^2 Y^2$$
Make the substitutions $X=\frac x y$ and $Y = \frac {x-1}{x+1}$
$$ \begin{align} a\,\left(\frac xy\right)^2 + \left(\frac {x-1}{x+1}\right)^2 &= 1 + d\,\left(\frac xy\right)^2 \left(\frac {x-1}{x+1}\right)^2 \\ a\, x^2 (x+1)^2 + y^2(x-1)^2 &= y^2 (x+1)^2 + d\, x^2 (x-1)^2 \\ y^2(4x) &=x^2\left( ax^2 + 2ax + a - dx^2 + 2dx - d \right) \\ \therefore y^2 &=\frac {a-d} 4 x^3 + \frac {a+d} 2 x^2 + \frac {a-d} 4 x \\ &= \textrm{a cubic polynomial } p(x) \end{align} $$ This is the equivalence to an elliptic curve. (Source.)