Is there an irreducible but solvable septic trinomial $x^7+ax^n+b = 0$? 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 $|a,b|<50$.
 A: The good news is that in this book:
Generic Polynomials: Constructive Aspects of the Inverse Galois Problem
 By Christian U. Jensen, Arne Ledet, Noriko Yui (pp. 52 and up).
There is a complete criterion, as follows:
$x^7 + a x + b$ is solvable (we assume it's irreducible) if and only if the polynomial (which I am intentionally leaving in Mathematica input form, so you can type it in yourself).
P35[a_, b_, x_] := 
 x^35 + 40 a x^29 + 302 b x^28 - 1614 a^2 x^23 + 2706 a b x^22 + 
  3828 b^2 x^21 - 5072 a^3 x^17 + 2778 a^2 b x^16 - 
  18084 a b^2 x^15 + 36250 b^3 x^14 - 5147 a^4 x^11 - 
  1354 a^3 b x^10 - 21192 a^2 b^2 x^9 - 26326 a b^3 x^8 - 
  7309 b^4 x^7 - 1728 a^5 x^5 - 1728 a^4 b x^4 + 720 a^3 b^2 x^3 + 
  928 a^2 b^3 x^2 - 64 a b^4 x - 128 b^5
has factorization pattern [degrees of irreducible factors] one of (14, 21), (7, 7, 7, 21), (7, 7, 7, 14), (7, 7, 7, 7, 7)
The bad news is that I have run this for all pairs (a, b) where both coordinates are between -1000 and 1000, and there is not a single success.
A: 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 $16$.
