What value of $b$ makes the graph of $y^2 = x^3-x+b$ self-intersection? I was just exploring a little bit on Desmos, and was trying to figure out something somewhat interesting. I'm familiar that this is an elliptic curve, but ALL I know about them is that they are of the form $y^2=x^3+ax+b$. Nothing else, really....
So, here's what I'm thinking. Give $a$ an easy value, say $-1$, and let $b$ range over the reals. There is a specific value of $b$ for which the graph moves from one curve, to a curve and a closed circle-like figure. This specific value of $b$ makes the graph kind of.... "self-intersect" in a nice way. I've estimated it to be approximately $0.384900179459750$. I put this into WolframAlpha and it didn't recognize anything important. Is this constant important? Does it have a name? What's its value? Thanks.
 A: For the curve to be self intersecting, we take the form
$y^2 = (x-p)^2 (x-q) = x^3 - (2p+q)x^2 + (2pq+p^2)x-p^2q \ $. Please note that the curve forms only for $x \geq q$
As $x^2$ term is zero, $2p+q = 0 \implies q = - 2p$.
So, $2pq + p^2 = - 3p^2, -p^2q = 2p^3$
We also note that if $p \lt 0, q \gt 0$ but $x$ must be $\geq q$ so there is no solution (self-intersecting curve) for $p \lt 0$. Also when $p = 0, q = 0$ and so there is no solution for $p = 0$ either.
That leads to,
$y^2 = x^3 -3p^2 x +2p^3, p \gt 0$
A: Let $f(x)=x^3-x+b$ and consider the curve $y^2=f(x)$.
To say that the curve "self-intersects" at $(x_0,0)$  means $f$ has a double zero at $x_0$ : $f(x_0)=f'(x_0)=0$ and $f"(x_0)\neq 0$. (And because we work with the real numbers, $y^2=f(x)\ge 0$ near $x_0$)
If $x_0$ is such a double zero, then near we have the taylor expansion $f(x)=\frac{f''(x_0)}{2}(x-x_0)^2 + O((x-x_0)^2)=3x_0(x-x_0)^2 + O((x-x_0)^2)$. Therefore $x_0>  0$
Next, $f'(x)=3x^2-1$ so $x_0= \frac{\sqrt{3}}{3}$.
Finally $f(x_0)=0\Longleftrightarrow b- \frac{2\sqrt{3}}{9}=0$
So  $$b=\frac{2\sqrt{3}}{9}$$
