Calculating the bi-tangents of $x^4+y^3+1=0$

Assume that $P(x)=x^4+(ax+b)^3+1=x^4+a^3x^3+3a^2bx^2+3ab^2x+b^3+1$ is a square, i.e. $P(x)=(x^2+\alpha x+\beta)^2=x^4+2\alpha x^3+(\alpha^2+2\beta)x^2+2\alpha\beta x+\beta^2$ .

We get a system of equations: $\begin{aligned} & a^3=2\alpha\\ &3a^2b=\alpha^2+2\beta\\ &3ab^2=2\alpha\beta\overset{(1)}{=}a^3\beta\\ &b^3+1=\beta^2 \end{aligned}$ Consider equation $(3)$ :

1. If $a=0$ then, by $(1)$ we have $\alpha=0$ and then, by $(2)$ also $\beta=0$ , so by $(4)$ , $b^3+1=0$ - and we have three solutions.

2. Otherwise, $a\neq0$ , so we can divide by $a$ to get $\beta=3\frac{b^2}{a^2}$ (from $(3)$ ) and also $\alpha=\frac12a^3$ . Substituting those into equtions $(2),(4)$ we get: $\begin{aligned} &3a^2b=\frac{a^6}4+6\frac{b^2}{a^2} \hspace{5pt}\Rightarrow\hspace{5pt} a^8-12a^4b+24b^2=0\\ &b^3+1=9\frac{b^4}{a^4} \end{aligned}$ By solving the first equation for $a^4$ we get: $a^4=\frac{12b\pm\sqrt{144b^2-96b^2}}{2}=6b\pm\sqrt{12b^2}=(6\pm2\sqrt{2})b$ . Substituting this to the second equation we have: $9b^4=(b^3+1)a^4=(b^3+1)(6\pm2\sqrt{2})b \hspace{5pt}\Rightarrow\hspace{5pt} (3\mp2\sqrt{3})b^4=(6\pm2\sqrt{3})b$ So either $b=0$ or $b^3=\frac{6\pm2\sqrt{3}}{3\mp2\sqrt{3}}$ (the signs match each other).

The option $b=0$ is impossible since it implies $a=0$ but that implies $b^3=-1$ as we have seen before. So we have the later option - which gives $6$ solutions for $b$ . Each of those gives $4$ solutions for $a$ (since the signs have to match). Hence we hve $6\cdot4=24$ solutions here.

So there are $3+24=\boxed{27}$ pairs.

Any smooth plane quartic curve has $28$ bitangents (see e.g. http://en.wikipedia.org/wiki/Bitangents of a_quartic). So the projective curve $x^4 + zy^3 + z^4 = 0$ has $28$ bitangents; the given curve is just the affine piece of it. The "line at infinity" $z = 0$ is indeed a bitangent to this curve (it actually hits the curve at $(1,0,0)$ with multiplicity $4$ ), so there are $\fbox{27}$ affine bitangents.

Computing them explicitly is a bit of a mess!