Galois revealed that an algebraic equation $f(x)=0$ with coefficients in a field $K$ of zero characteristic is solvable by radicals if and only if the Galois group of $f(x)$ over $K$ is solvable. However, many mathematicions actually expect that one could give a criterion for solvability by radicals simply by coefficients.


Joint with Qing-Hu Hou at Tianjin University, we formulate the following conjecture based on our computation.


**Conjecture.** Suppose that $f(x)=ax^n+bx+c$ is irreducible over $\mathbb Q$ (the field of rational numbers), where $n,a,b,c\in\mathbb Z$, $n>0$ and $a\not=0$. Provided that $\gcd(b,nac)=1$,
the Galois group of $f(x)$ over $\mathbb Q$ is isomorphic to the symmetric group $S\_n$,
and hence the equation $f(x)=0$ is not solvable by radicals if $n\ge5$.


Via an internet search, we note that the conjecture in the case $a=1$ and $\gcd(n-1,c)=1$ is known to be true, see, e.g., [Osaka's JNT paper](https://doi.org/10.1016/0022-314X(87)90029-1).


**QUESTION.** Is the conjecture true? Can one prove it completely?



A criterion for the equation $ax^n+bx+c=0$ not solvable by radicals via $a,b,c$ and $n$
No, the conjecture is false at least for $n = 5$. The irreducible quintic trinomial $f(x) = 85x^{5} - 4x + 1$ satisfies $\gcd(b,5ac) = \gcd(-4,5 \cdot 85 \cdot 1) = 1$. However, the Galois group of $f(x)$ is solvable.


This can be found as follows. The family of solvable quintic trinomials is
$$
 f(x) = (4u^{2} + 16)x^{5} + (5u^{2} - 5) x + (4u^{2} + 10u + 6)
$$
with $u \in \mathbb{Q}$. (Apparently, this dates back to Weber from 1898, although there are many other modern sources.) Taking $u = -19/21$ and then scaling the polynomial gives rise to the example above.


