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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.

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

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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.

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  • $\begingroup$ Thanks! In your example, $\gcd(a,n)=\gcd(85,5)>1$. Can you find a counterexample with $\gcd(a,n)=1$ or even $a=1$? $\endgroup$
    – Zhi-Wei Sun
    Commented Apr 25, 2022 at 21:52
  • $\begingroup$ The polynomial $416x^{5} - 9x + 2$ is a solvable quintic with $\gcd(a,n) = 1$. I don't think this family allows a situation where $a = 1$ and $\gcd(b,nac) = 1$, but it may be possible to find such in other degrees. $\endgroup$
    – Jeremy Rouse
    Commented Apr 25, 2022 at 22:12
  • $\begingroup$ We did check the conjecture seriously for $n\le 8$, $1\le a\le 20$ and $|b|,|c|\le500$. $\endgroup$
    – Zhi-Wei Sun
    Commented Apr 25, 2022 at 22:22
  • $\begingroup$ I note that the family of solvable quintic trinomials you give can only yield counterexamples with $bc<0$. Probably, our conjecture with bc≥0 has no counterexample. $\endgroup$
    – Zhi-Wei Sun
    Commented Apr 26, 2022 at 0:32
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    $\begingroup$ You might however also note that in degrees other than 5,6 and 8, there are only finitely many solvable irreducible trinomials altogether (up to obvious equivalence), see mathoverflow.net/questions/146769/… , so that this is largely a conjecture on quintic, sextic and octic polynomials. In those degrees, there should be finitely many parameterizations similar to the above capturing all but finitely many solvable cases, and then examples should tend to run out once the restrictions on a,b,c are tightened more and more. $\endgroup$
    – Joachim König
    Commented Apr 26, 2022 at 5:26

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