An instructive example: polynomials $x(t), y(t)$ of degrees $8, 12$ such that $x^3 - y^2$ is a nonzero polynomial of degree at most $6$ (and at least $5$ by the usual ABC/Wronskian/Riemann-Hurwitz argument). These correspond to rational functions $f = x^3/y^2$ of degree $24$ that give rational maps $f : {\bf P}^1 \to {\bf P^1}$ branched over $4$ points, with monodromy generators of type $3^8$ above $0,$ type $2^{12}$ above $\infty,$ and type $1^6 18$ above $1,$ plus a simple transposition above the image of the extra zero of $f'.$ On the elliptic-surface side, we get elliptic K3 surfaces $Y^2 = X^3 - 3 x X + 2 y$ with a singular fiber of type at least $I_{18}$ at $t=\infty,$ so the moduli space has dimension $20-\rho = 20 - (2+17) = 1$ which (as usual for K3’s) is consistent with the parameter count.

Start by scaling and translation of $t$ to put $x$ in the form $$ x = (t^4 + a_2 t^2 + a_3 t + a_4)^2 + 2 (r_5 t^3 + r_6 t^2 + r_7 t + r_8 $$ (this corresponds to what we called “$x = \zeta^2 + \eta$” in the setting of Hall’s conjecture). Then expand $x^{3/2}$ in a Laurent series about $t=\infty,$ getting $$ x^{3/2} = t^{12} + 3 a_2 t^{10} + 3 a_3 t^9 + 3 (a_4 + a_2^2) t^8 + (3 r_5 + 6 a_2 a_3) t^7 + (3 r_6 + 6 a_2 a_4 + 3 a_3^2 + a_2^3) t^6 + \cdots = \sum_{n=0}^\infty c_n t^{12-n} $$ for some polynomials $c_n$ $(n=0,1,2,3,\ldots)$ of weight $n$ in the $a_i$ and $r_j.$ We set $y = \sum_{n=0}^{12} c_n t^{12-n},$ and require $c_n=0$ for $n=13,14,15,16,17.$ The weights suggest that only the first of these will be linear in any of our parameters $a_i,r_j$, and indeed $c_{14}$ already includes a term $3r_j^2/2;$ but thanks to the special structure of this cube-minus-square system the equations $c_{13} = c_{14} = c_{15} = 0$ are simultaneously linear in $(a_3,a_4,r_8),$ so we solve them and clear denominators to get two equations in the remaining parameters $a_2,r_5,r_6,r_7.$

At this point we might expect to have a complicated pair of equations whose intersection gives a moduli curve that has many singularities and/or a high genus. But in fact we get the union of two simpler curves, and this is already visible in $c_{17},$ which is now just $r_5 (2 a_2 r_5^2 - 3 r_6^2 - 3 r_5 r_7).$ Consider first $r_5 = 0.$ Since $r_5$ occurs in the denominators of the formulas for $a_3,a_4,r_8$ we must go back to the original equations $c_n=0,$ but fortunately this is easy. First $c_{13}$ simplifies to $3 r_6 r_7,$ and if $r_6=0$ then $c_{14}=3 r_7^2/2$ so $r_7 = 0$ anyway. Then $c_{14} = 3 r_6 (2 r_8 - a_2 r_6) / 2,$ and if we again take $r_6=0$ then $c_{16} = 3 r_8^2/2$ which means that the $r_j$ all vanish and we are in the degenerate “$\eta=0$” situation where $x$ is a perfect square. So we must instead have $r_8 = a_2 r_6/2,$ and then $c_{15} = -3 a_3 r_6^2/2$ so $a_3 = 0$ (we already know $r_6 \neq 0),$ which makes $c_{17} = 0$ as expected and leaves us with just $c_{16} = 3 (a_2^2 - 4 a_4) r_6^2/8$ so $a_4 = a_2^2/4.$ In particular $x$ is an even function of $t,$ specifically $$ x = (t^2 + a)^4 + 2 r_6 (t^2 + a) $$ where $a = a_2/2$. We recognize this as the identity $(T^4 - 2rT)^3 - (T^6 - 3rT^3 + 3r^2/2)^2 = r^3 T^3 - 9 r^4/4$ corresponding to our Belyi function for indices $(2^6, 3^4, 1^3 9),$ composed with an arbitrary quadratic polynomial (which takes the form $t^2+{\rm constant}$ thanks to our initial normalizations). On the elliptic-surface side, our K3 is a quadratic base-change from a rational elliptic surface over the $T$-line. […] That still leaves the factor $2 a_2 r_5^2 - 3 r_6^2 - 3 r_5 r_7$ of $c_{17}.$ Solving for $r_7$ and substituting this into $c_{16}$ yields $$ 31104 r_6^{10} + 15552 a_2^{\phantom.} r_5^2 r_6^8 + 1728 a_2^2 r_5^4 r_6^6 - 2592 r_5^6 r_6^5 - 288 a_2^3 r_5^6 r_6^4 - 864 a_2^{\phantom.} r_5^8 r_6^3 - 72 a_2^4 r_5^8 r_6^2 - 72 a_2^2 r_5^{10} r_6^{\phantom.} + 27 r_5^{12} - 4 a_2^5 r_5^{10} = 0. $$ We observe that $a_2$ often appears with $r_5^2,$ so substitute $a_2 = A / a_5^2$ to get the equivalent $$ -4 A^5 - 72 r_6^2 A^4 - 288 r_6^4 A^3 + 1728 r_6^6 A^2 - 72 r_5^6 r_6 A^2 + 15552 r_6^8 A - 864 r_5^6 r_6^3 A + 31104 r_6^{10} - 2592 r_5^6 r_6^5 + 27 r_5^{12} $$ with $A$ of weight $12$. Since the other variables have weights $5,6,$ each monomial must have the exponent of $r_5$ divisible by $6,$ so we rewrite in terms of $R = r_5^6,$ which has weight $30$. This yields a quadratic in $R$ whose linear and constant coefficients factor nicely: $$ 27 R^2 - 72 r_6 (A + 6 r_6^2)^2 R - 4 (A - 6 r_6^2) (A + 6 r_6^2)^4 = 0. $$ We can thus take $R = u_6 (A + 6 r_6^2)^2$, clear the factor $(A + 6 r_6^2)^2$ (after checking that $A = - 6 r_6^2$ does not yield a solution to our original system), and solve for $A$: $$ A = 6 r_6^2 - 18 r_6 u_6 + 27 u_6^2/4. $$ […]