On the solvable septic quadrinomial $x^7-7x^4-14x^3-7=0$?

Question

The concept of sparse polynomials has its place, and solvable but irreducible quadrinomial examples such as,

$$x^7-7x^4-14x^3-7=0$$ $$x^8+x^7+29x^2+29=0$$ $$x^9-27x^4-9x^3-9^2=0$$ $$x^{12}-36x^5-12x^3-12^2=0$$

may be intriguing, especially the octic which needs the $29$th root of unity. For trinomials, there are infinitely many quintics, sextics, and octics that are solvable and irreducible, such as,

$$x^5-5x^2-3=0$$ $$x^6+3x+3=0$$ $$x^8+9x+9=0$$

For septics, there do not seem to be any as I asked in this ancient MO question. But there are nice reducible ones such as,

$$x^7+7x+8=0$$

where the sextic factor is solvable. For now, we focus on irreducible septic quadrinomials. Over the past ten years, I've only found TWO that do not arise from binomials. (Edit.) Peter Mueller recently found another two of the same kind,

$$x^7-7x^4-14x^3-7=0\tag1$$ $$x^7-98x^4+686x^2+7^3=0\tag2$$ $$x^7+7x^4−35x+54 = 0\tag3$$ $$x^7+7x^3+21x+50 = 0\tag4$$

In Magma notation, the first one is a $7T2$ while the other three are $7T4$.

Question: Are there in fact infinitely many septic quadrinomials (without scaling) that are solvable and irreducible?

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Update: Within a space of 24 hours, thanks to Jeremy Rouse, Joachim Konig, and yours truly, we have added 5 more quadrinomials, all of which have order $42$ so are $7T4$. However, these have a root in $\mathbb{Q}[\sqrt[7]z]$ where $z$ is an integer so arise from binomials.

Let $\alpha=14,$ $$x^7 - 6\alpha\, x^4 + \alpha^2\,x^3 - 2\alpha^2 = 0\tag1$$ $$x^7 - 2\alpha^3\, x^2 + 2\alpha^4\,x - 6\alpha^4 = 0\tag2$$

Let $\beta=21,$ $$x^7 - 10\beta\, x^4 + 70\beta^2\,x - 215\beta^2= 0\tag3$$

Let $m=5,$ $$x^7 + 14m^3\, x^3 - 42m^3\,x^2 - 13^2m^4 = 0\tag4$$

Let $n=57,$ $$x^7 + 14n^2\,x^4 - 2(7n)^3\,x^2 - 13^2(\sqrt7\,n)^4= 0\tag5\\\\$$

with fields $\mathbb{Q}[\sqrt[7]2]\,$, $\mathbb{Q}[\sqrt[7]{3^2\times5^3}]\,$, $\mathbb{Q}[\sqrt[7]5]\,$, $\mathbb{Q}[\sqrt[7]{3^2\times19^3}]$, respectively. Are there any more?

P.S. Incidentally, there are infinitely many irreducible but solvable quadrinomials of $9$th and $12$th degrees,

$$x^9-3abc\,x^4+a^3c\,x^3+b^3c^2 = 0$$ $$x^{12}-3abc\,x^5+a^3c\,x^3+b^3c^2 = 0$$

for arbitrary ($a,b,c$) and roots in $\mathbb{Q}[\sqrt[3]c],$ examples given at the start of the post, but a general form for the septic (if any) seems more difficult.

Answer 1

It turns out there are infinitely many solvable septic quadrinomials after all (though with a caveat), and the answer was under my nose all along. For example, recall that, $$e^{\pi\sqrt{163}} =640320^3+743.999999999999925\dots$$

Then the septic,

$$x^7+7\left(\tfrac{1-\sqrt{-7}}2\right)x^4+7\left(\tfrac{1+\sqrt{-7}}2\right)^3x = -640320$$

surprisingly is solvable in radicals. In general, this is my version of Klein's 7th order formula for the j-function (which fittingly is also known as Klein's invariant) and is given by,

$$x^7+7\left(\tfrac{1-\sqrt{-7}}2\right)x^4+7\left(\tfrac{1+\sqrt{-7}}2\right)^3x = \sqrt[3]{j(\tau)}$$

which is solvable in radicals and has a nice closed-form using Ramanujan's theta functions described here. And I found the RHS can have a polynomial version as,

$$x\left(x^6+7\left(\tfrac{1-\sqrt{-7}}2\right)x^3+7\left(\tfrac{1+\sqrt{-7}}2\right)^3\right) = \left(\frac{49n^2+13n+1}{n}\right)^{1/3}\left(\frac{n^2+5n+1}{n^2}\right)$$

solvable for any $n$. This septic can be solved by a rather complicated cubic so I will not include it for now.

One can get rid of the cube root by letting $x = m^{1/3}y$ where $m=\left(\frac{49n^2+13n+1}{n}\right)$ to get the slightly simpler,

$$m^2y^7+7\left(\tfrac{1-\sqrt{-7}}2\right)m\,y^4+7\left(\tfrac{1+\sqrt{-7}}2\right)^3y =\frac{n^2+5n+1}{n^2}$$

So the caveat is this quadrinomial contains a square root. It remains to be seen if there really is an infinite family with just rational coefficients.