**I. First Set**

Before going to Elkies' nonic, we start with something a bit simpler. There is a list of j-function formulas in this [MSE post][1]. For example, for prime levels $p = 5,7,13,$ we have,

$$j=\frac{(x^2+10x+5)^3}x$$
$$j=\frac{(x^2+5x+1)^3(x^2+13x+49)}x$$
$$j=\frac{(x^4+7x^3+20x^2+19x+1)^3(x^2+5x+13)}x$$

Assume $j$ as **any** number, hence a free parameter. Then these equations have groups $\text{PGL}(2,5), \text{PGL}(2,7), \text{PGL}(2,13),$ respectively, so generally is not solvable in radicals. However, $x$ is solvable if $j$ is the $j$-function **OR** if we do this, 

$$\frac{(x^2+10x+5)^3}x =\frac{(n^2+10n+5)^3}n$$

for **any** $n$. For example, let $n=2$, 

$$\frac{(x^2+10x+5)^3}x =\frac{29^3}2$$

Its irreducible $5$th-deg factor has Frobenius group $F(5)=4\times5=20$ hence is now solvable. Likewise for its higher siblings which have $F(7)=6\times7=42$ and $F(13)=12\times13=156$. 

So what is the reason they now have solvable groups?

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**II. Second Set**

Searching the literature for similar "formulas", the paper *[Galois Number Fields with Small Root Discriminant][2]* by J. Jones and D. Roberts was a gold mine. They had degrees for $p \leq 10$ and even for $p=17$ which I mentioned in an old [MO post][3]. But this equating of,

$$P(j,x) = P(j,n)$$

was not there. A small sample for deg 9,

$$j = \frac{(x^3 + 4x^2 + 10x + 6)^3}{(4x^2 + 13x + 32)}$$

Again, assume $j$ as a free parameter. This equation in Magma notation is $\text{9T32}$ and has order $9\times12\times14 = 1512$ (unsolvable). However,

$$\frac{(x^3 + 4x^2 + 10x + 6)^3}{(4x^2 + 13x + 32)} = \frac{(n^3 + 4n^2 + 10n + 6)^3}{(4n^2 + 13n + 32)}$$

(after removing the linear factor) is $\text{8T36}$ and has order $12\times14 = 168$ (solvable). Elkies' nonic is also $\text{9T32}$ but a bit trickier,

$$j^2+(9x^4 - 42x^3 - 675x^2 - 1485x - 441)j - (x^3 - 9x^2 - 69x - 123)^3 =0$$
$$j^2+(9n^4 - 42n^3 - 675n^2 - 1485n - 441)j - (n^3 - 9n^2 - 69n - 123)^3 =0$$

Eliminating $j$ between the two using resultants and getting rid of two linear factors yields a $16$-deg equation with order $12\times14\times16 = 2688$ (and solvable according to Magma).

**Note:** For consistency, I still used the variable $j$ for the second set, but they are not to be understood as level-9 formulas for the j-function.

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**III. Question**

So why is it that for these two sets (and some others) that eliminating the common variable $j$ between $P(j,x) = P(j,n) = 0$ suddenly yields a parametric equation with a solvable Galois group?


  [1]: https://math.stackexchange.com/a/4580319/4781
  [2]: http://cda.morris.umn.edu/~roberts/dpr/Research_files/lowgrd.pdf
  [3]: https://mathoverflow.net/questions/226449/whats-so-special-about-these-17th-deg-equations