The concept of [sparse polynomials](https://en.wikipedia.org/wiki/Sparse_polynomial) 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](https://mathoverflow.net/questions/146769/). 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?




---


**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.



On the solvable septic quadrinomial $x^7-7x^4-14x^3-7=0$?
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](https://mathoverflow.net/a/326028/12905). 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.


