The following *irreducible* trinomials are solvable:


$$x^5-5x^2-3 = 0$$


$$x^6+3x+3 = 0$$


$$x^8-5x-5=0$$


Their Galois groups are isomorphic to ${\rm D}\_5$, ${\rm S}\_3 \wr {\rm C}\_2$ and
$({\rm S}\_4 \times {\rm S}\_4) \rtimes {\rm C}\_2$, respectively.


**Question:** Is there an irreducible septic trinomial $x^7+ax^n+b=0$ with
solvable Galois group, where $n \in \{1, \dots, 6\}$, $a, b \in \mathbb{Z} \setminus \{0\}$? 


**P.S**. For $n=1$, I did a search only for those with 1 real root and, if I did it correctly, there are none with integer $|a,b|<50$.



Is there an irreducible but solvable septic trinomial $x^7+ax^n+b = 0$?
If such polynomials exist, there will only be finitely many of them, up to composing on both sides with scalar polynomials $\alpha x$ with $\alpha\in\mathbf{Q}$. More generally, Guralnick and Shareshian proved that if $d=7$ or $d>8$ then there are only finitely many equivalence classes of irreducible degree-$d$ trinomials in $\mathbf{Q}[x]$ whose Galois group is not $A\_d$ or $S\_d$, where two trinomials $f(x)$ and $g(x)$ are called equivalent if $f(x)=\alpha\cdot g(\beta x)$ for some $\alpha,\beta\in\overline{\mathbf{Q}}$. They proved an analogous result over number fields, and also for reducible trinomials. See Theorem 1.4.3 of "Symmetric and Alternating Groups as Monodromy Groups of Riemann Surfaces I: Generic Covers and Covers with Many Branch Points", AMS Memoirs (2007), vol. 189, no. 886.


This explains why you were able to find the examples you found. The only pairs $(d,n)$ for which there are infinitely many equivalence classes of irreducible degree-$d$ trinomials in $K[x]$ with terms of degrees $0,n,d$ and Galois group not $A\_d$ or $S\_d$ (for some number field $K$) are those pairs with $0<n<d$ and $d\le 8$ and $d\ne 7$ and either $d=4$ or $\gcd(n,d)=1$. Also Guralnick-Shareshian listed the groups which occur for infinitely many equivalence classes over a number field.


In order to answer the question you asked, one can use the following approach by Elkies. Let $H=\text{AGL}(1,7)$ be the group of all linear maps on $\mathbf{F}\_7$. For each $n\in\{1,2,\dots,6\}$, Guralnick-Shareshian say that Elkies told them that there is a bijection between the set of equivalence classes of solvable degree-$7$ trinomials in $\mathbf{Q}[x]$ with terms of degrees $0,n,7$ and the set of rational points on a certain Riemann surface $X$. The information Guralnick-Shareshian use about this Riemann surface is that it admits a degree-$120$ branched cover $X\to\mathbf{P}^1$ with monodromy group $S\_7$ and three branch points, two of which have branch cycles $(1,n+1)$ and $(1,2,\dots,7)$. I haven't thought about how to prove this bijection. Anyway, assuming it's correct, then your question amounts to asking whether any of six specific Riemann surfaces have any rational points.


**Added later:** as Noam notes in his comment, there are really three Riemann surfaces, because the ones for $n$ and $7-n$ are the same. By my computations, all of these Riemann surfaces $X$ are defined over $\mathbf{Q}$, and the genus of $X$ is $17$, $10$, or $16$ according as $n=1$, $2$, or $3$. So the original question amounts to asking whether there are rational points on any of three specific curves over $\mathbf{Q}$, having genera $17$, $10$ and $16$.


