Joint with Qing-Hu Hou at Tianjin Univ., we seek for explicit criteria via coefficients for the solvability of an algebraic equation by radicals. In this direction, we formulate the following conjecture.


**Conjecture.** Let $p\ge5$ be a prime. Suppose that
$$f(x)=ax^p+bx^{p-1}+cx+d$$ is irreducible over $\mathbb Q$, where $a,b,c,d$ are pairwise coprime integers with $a\not=0$. Then the equation $f(x)=0$ over $\mathbb Q$ is not solvable by radicals.


We have checked this for $p=5,7$ and $|a|+|b|+|c|+|d|\le 108$. For example, the Galois group of the polynomial $$x^5+3x^4+5x+23$$ over $\mathbb Q$ is the alternating group $A\_5$ which is not solvable, and the Galois group of the polynomial $$x^7+x^6-49x-25$$ over $\mathbb Q$ is the alternating group $A\_7$.


**QUESTION.** Can one prove or disprove the above conjecture?


Your comments are welcome!



On the solvability of the equation $ax^p+bx^{p-1}+cx+d=0$ by radicals
This answer is experimentally driven. I tried to construct as many as possible solvable irreducible polynomials of the shape $f=x^5+bx^4+cx+d$ with integers $b,c,d$, then search for a message or a counterexample in the produced experimental list. The condition of having *pairwise coprime* integers turned out,
after seeing the examples,
to be very restricting for them, many examples have by the statistical "pseudo-law of small numbers" some common prime divisor of two coefficients.
(This condition leads formally immediately to $b,c,d\ne 0$, so it rules out something like $x^p-2$. The condition $a\ne 0$ was explicitly given for $f=ax^p+bx^{p-1}+cx+d$, and if i correctly get the message, it is superfluous.)


The counterexample found by chance was
$$
\color{blue}{
f = x^5 +3x^4+23x+1301\ ,
}
$$
the principal coefficient is one, the others are different prime numbers.
Computer check:



```
R.<x> = PolynomialRing(QQ)
f = x^5 + 3*x^4 + 23*x + 1301
G = f.galois_group()

print(f'f = {f}')
print(f'G = Gal(f) has order {G.order()}')
print(f'G has structure {G.structure_description()}')
print(f'Is G solvable? {G.is_solvable()}')

```

Results:



```
f = x^5 + 3*x^4 + 23*x + 1301
G = Gal(f) has order 10
G has structure D5
Is G solvable? True

```

The reader in hurry may please stop here.




---




---


Since i was trying to understand structurally as much as possible, and since the byproduct examples can be interesting also elsewhere, here are some details on the way the above $f$ was produced.
The search for solvable irreducible $f$ polynomials with $p=5$, $a=1$, was done along the usual path. While searching for a reference i could join, i found:


* [LJF2004] the master thesis of Lau Jing Feng [On Solvable Septics](https://core.ac.uk/download/pdf/48627385.pdf)
* [AS2022] a preprint of Abdeljalil Saghe, researchgate provides the link <https://www.researchgate.net/publication/358595406> for it.
* [SW1996] Spearman and Williams, [On Solvable Quintics...](https://projecteuclid.org/journals/rocky-mountain-journal-of-mathematics/volume-26/issue-2/On-Solvable-Quintics-X5aXb-and-X5aX2b/10.1216/rmjm/1181072083.full), Rocky Mountain Journal of Mathematics, Volume 26, Number 2, 753-772, 1996. (The link provides a [Download PDF] button.) See also the wiki note on this: [Quintics in Bring-Jerrard form](https://en.wikipedia.org/wiki/Quintic_function#Quintics_in_Bring%E2%80%93Jerrard_form).


It is a good idea to check first the **Proposition** from [SW1996] at page 754. This is cited with variations in the other sources. And [SW1996] further delegated the proof at an AMM article from 1994.


For the convenience of the reader i am reproducing here from [AS2022] a theorem, it also appears as a short mention in the same form in [LJF2004]. The latter points explicitly to [SW1996]. Although it is formulated for trinomial quintics (with a missing term in $x^4$), it may be useful to have an explicit example, and see the structure of the solution and the implications for the coefficients in the trinomial. Please skip if this feels not related to the question.




---


**Theorem 13.8** in [LJF2004]: Let $a$ and $b$ be rational numbers
such that the quintic trinomial
$
x^5 + ax + b
$ is irreducible.
Then, the equation
$$x^5 + ax + b = 0$$
is solvable
by radicals if and only if there exist rational numbers
$\epsilon$ among $\pm 1$, $c\ge 0$, and $e\ne 0$ such that
$$
\tag{$13.9$}
\begin{aligned}
a &= \frac 1D\cdot 5e^4(3 − 4\epsilon c)\ ,\\
b &= −\frac 1D\cdot 4e^5(11\epsilon + 2c)\ ,\text{ where $D$ is }\\
D &:= c^2 + 1\ ,
\end{aligned}
$$
in which case the roots of $x^5 + ax + b = 0$ are
$$
\tag{$13.10$}
x\_j = e\Big(\ \omega^j u\_1 
 + \omega^{2j} u\_2 
 + \omega^{3j} u\_3
 + \omega^{4j} u\_4\ \Big)\ ,\ j = 0, 1, 2, 3, 4,
$$
for $j$ among $0,1,2,3,4$. Here
$\omega$ is a primitive fifth root of the unit,
$\omega^5=1$, $\omega\ne 1$, and
$$
\tag{$13.11$}
$$
$$
u\_1 = \left(\frac{v\_1^2v\_3}{D^2}\right)^{1/5}\ ,\ 
u\_2 = \left(\frac{v\_3^2v\_4}{D^2}\right)^{1/5}\ ,\ 
u\_3 = \left(\frac{v\_2^2v\_1}{D^2}\right)^{1/5}\ ,\ 
u\_4 = \left(\frac{v\_4^2v\_2}{D^2}\right)^{1/5}\ ,
$$
and
$$
\tag{$13.12$}
$$
$$
\begin{aligned}
v\_1 &= +\sqrt D + \sqrt{D-\epsilon\sqrt D}\ ,\\
v\_2 &= -\sqrt D - \sqrt{D+\epsilon\sqrt D}\ ,\\
v\_3 &= -\sqrt D + \sqrt{D+\epsilon\sqrt D}\ ,\\
v\_4 &= +\sqrt D - \sqrt{D-\epsilon\sqrt D}\ .
\end{aligned}
$$




---


It is maybe best to have a concrete example for this situation,
my choice of the computational weapon is [sage](https://www.sagemath.org).



```
eps, c, e = 1, 3/5, 34
D = c^2 + 1
a, b = 5*e^4*(3 - 4*eps*c)/D, -4*e^5*(11*eps + 2*c)/D

R.<x> = PolynomialRing(QQ)
f = x^5 + a*x + b
G = f.galois_group()

print(f'f = {f}')
print(f'Is f irreducible? {f.is_irreducible()}.')
print(f'Let G be the Galois group associated to f. Its order is {G.order()}.')
print(f'Is G solvable? {G.is_solvable()}.')
print(f'Structurally, G is {G.structure_description()}.')

```

Results:



```
f = x^5 + 2947800*x - 1630329920
Is f irreducible? True.
Let G be the Galois group associated to f. Its order is 20.
Is G solvable? True.
Structurally, G is C5 : C4.

```

We can also ask for the field in between.



```
K.<v> = NumberField(f)
H.<w> = K.galois_closure()
subfields_data = H.subfields(degree=4)
subfield_data  = subfields_data[0]
F, mor, nothing = subfield_data
F.inject_variables()

print('Let H be the Galois closure of f.')
print('Then H has a subfield F of degree 4:')
print(F)

```

This gives:



```
Defining w0
Let H be the Galois closure of f.
Then H has a subfield F of degree 4:
Number Field in w0 with defining polynomial 
    x^4 + 51114852000000*x^2 + 188270112726637200000000000

```

We can ask for the roots of the above polynomial, in a dialog with the sage interpreter:



```
sage: var('T');
sage: pol = F.defining_polynomial()
sage: for sol in solve(pol(T) == 0, T, solution_dict=True):
....:     print(sol[T].simplify_full())
....: 
-173400*sqrt(5)*sqrt(11*sqrt(170) - 170)
173400*sqrt(5)*sqrt(11*sqrt(170) - 170)
-173400*I*sqrt(5)*sqrt(11*sqrt(170) + 170)
173400*I*sqrt(5)*sqrt(11*sqrt(170) + 170)

```

Yes, we have an irreducible, solvable $f$, but the coefficients come with
many common primes. This can be traced back to the way they appear after fixing the parameters $\epsilon,c,e$.




---


The case of a more general quintic is similar. For a solvable, irreducible quintic
$$
f = x^5 +bx^4 + cx+d\ ,\qquad b,c,d\in\Bbb Q\ ,\ bcd\ne 0\ ,
$$
the Galois group is a subgroup of the Frobenius group
$F\_{5\cdot 4}=F\_{20}$.
Let $x\_1,x\_2,x\_3,x\_4,x\_5$ be the roots of $f$. Consider
$$
\theta\_1 :=\sum x\_2^2(x\_1x\_3+x\_4x\_5)\ ,
$$
the sum being cyclic. Consider $\theta\_1,\theta\_2,\dots,\theta\_6$ obtained from $\theta\_1$ by acting with permutations representing the cosets of $S\_5/F\_{20}$. Then the polynomial
$$
f\_{20}=f\_{20}(T)=\prod\_{1\le j\le6}(T-\theta\_j)
$$
is symmetric in the $f$-roots, and can be rewritten as a polynomial in $T;b,c,d$. (See also §12 in [AS2022].) Then $f$ is solvable iff the associated $f\_{20}$ has a rational root. Here is the explicit form of $f\_{20}$ computed in sage:
$$
\begin{aligned}
f\_{20}(T)
&=
T^6 + A\_1T^5 + A\_2T^4+ A\_3T^3+A\_4T^2 +A\_5T+A\_6\ ,\text{ where}\\
A\_1 &= 8c\ ,
\\
A\_2 &=-8b^3d + 40c^2\ ,
\\
A\_3 &=-2b^4c^2 - 44b^3cd - 50b^2d^2 + 160c^3
\\
A\_4 &=16b^6d^2 - 8b^4c^3 - 144b^3c^2d - 200b^2cd^2 + 400c^4
\\
A\_5 &=8b^7c^2d + 48b^6cd^2 + 3b^4c^4 - 56b^5d^3 - 364b^3c^3d - 550b^2c^2d^2 \\
&\qquad + 512c^5 + 2500bcd^3 - 3125d^4
\\
A\_6 &=b^8c^4 + 12b^7c^3d + 86b^6c^2d^2 + 17b^4c^5 - 468b^5cd^3\\
&\qquad - 276b^3c^4d + 625b^4d^4 - 1450b^2c^3d^2 + 256c^6 + 7500bc^2d^3 - 9375cd^4
\ .
\end{aligned}
$$
Then some random search for cases when $f\_{20}(T)$ has at least one rational root (this is the one and only thing we need) delivered the following tuples
$(a,b,c,d)$ with $a=1$, together with the corresponding $T$-root:
$$
\begin{array}{|c|c|c|c||c|}
\hline
a & b & c & d & T\\\hline
1 & 1 & -13 & 13 & 52\\\hline
1 & 1 & 30 & -10 & -80\\\hline
1 & 1 & 85 & -340 & -255\\\hline
1 & 1 & 173 & 1038 & -519\\\hline
1 & 1 & 197 & -1182 & 1182\\\hline
%
1 & 2 & -360 & 880 & 1120 \\\hline
1 & 2 & -208 & 416 & 832\\\hline
1 & 2 & 26 & 26 & -26\\\hline
1 & 2 & 208 & 104 & -416\\\hline
1 & 2 & 212 & -636 & - 636 \\\hline
1 & 2 & 320 & -240 & -880\\\hline
1 & 2 & 480 & -320 & -1280 \\\hline
1 & 2 & 728 & 624 & -1248 \\\hline
%
1 & 3 & -42 & -534 & 144\\\hline
1 & 3 & -13 & 13 & 52\\\hline
1 & 3 & 6 & -6 & 0\\\hline
1 & 3 & 13 & -39 & -26\\\hline
1 & 3 & 15 & 15 & 0\\\hline
\color{blue}{1} & \color{blue}{3} & \color{blue}{23} & \color{blue}{1301} & 1612\\\hline
1 & 3 & 54 & 90 & 0\\\hline
1 & 3 & 135 & -135 & 0\\\hline
%
1 & 4 & -40 & 160 & 160 \\\hline
1 & 4 & 116 & 174 & - 348 \\\hline
1 & 4 & 148 & 368 & - 408 \\\hline
1 & 4 & 208 &-208 & - 624 \\\hline
1 & 4 & 312 & 728 & - 104 \\\hline
1 & 4 & 416 & 832 & - 416 \\\hline
1 & 4 & 1010 & 909 & -2121 \\\hline
%
1 & 5 & -349 & 1396 & 2443\\\hline
1 & 5 & -65 & 143 & 520\\\hline
1 & 5 & -50 & 238 & 240\\\hline
1 & 5 & -25 & 275 & 700\\\hline
1 & 5 & -10 & 6 & 80\\\hline
1 & 5 & 5 & 3 & 30\\\hline
1 & 5 & 5 & 15 & 110\\\hline
1 & 5 & 30 & 46 & 160\\\hline
1 & 5 & 55 & 67 & 60\\\hline
1 & 5 & 61 & -61 & -122\\\hline
1 & 5 & 65 & 91 & 130\\\hline
1 & 5 & 80 & 48 & -240\\\hline
1 & 5 & 85 & -1717 & 1870\\\hline
1 & 5 & 85 & 34 & -255\\\hline
1 & 5 & 85 & 68 & -255\\\hline
1 & 5 & 95 & -705 & 840\\\hline
1 & 5 & 100 & 20 & -300\\\hline
1 & 5 & 100 & 100 & -300\\\hline
1 & 5 & 100 & 164 & 100 \\\hline
1 & 5 & 125 & 150 & -375 \\\hline
1 & 5 & 135 & 81 & -360 \\\hline
1 & 5 & 135 & 567 & 720 \\\hline
1 & 5 & 200 & 432 & -100 \\\hline
1 & 5 & 205 & -82 & -615 \\\hline
1 & 5 & 205 & 328 & -615 \\\hline
1 & 5 & 157 & 314 & -314 \\\hline
1 & 5 & 160 & -32 & -480 \\\hline
1 & 5 & 160 & 224 & -480 \\\hline
%
1 & 6 & -72 & -1008 & 864\\\hline
1 & 6 & 96 & -192 & 0\\\hline
1 & 6 & 120 & -420 & 360\\\hline
1 & 6 & 135 & 270 & 0\\\hline
%
1 & 7 & 37 & 37 & 222 \\\hline
%
1 & 10 & -160 & 192 & 1280\\\hline
1 & 10 & -40 & 176 & 1120\\\hline
1 & 10 & -10 & 126 & 530\\\hline
1 & 10 & 80 & 96 & 480\\\hline
1 & 10 & 135 & 162 & 720\\\hline
%
1 & 12 & -72 & 96 & 864 \\\hline
%
1 & 15 & -153 & 153 & 1836\\\hline
1 & 15 & -105 & 93 & 1440\\\hline
1 & 15 & -25 & 25 & 700 \\\hline
1 & 15 & -15 & 141& 1170\\\hline
1 & 15 & 15 & 69 & 1080\\\hline
1 & 15 & 85 & 204 & 1870\\\hline
\end{array}
$$
(The above list shows only some solutions, it is not systematic, i picked some entries for my taste and purposes.)




---




---


I was also looking for examples in degree $p=7$.
This is hard and becomes harder and harder when $p$ gets bigger.
Still, i'll try to say some words on this case.


In the *solvable* case, for $p=7$, the related Galois group associated to $f$ is and must be a subgroup of the Frobenius group $F\_{42}$, this is [LJF2004] Corollary **2.1.7**. The following post was targeting the solvable trinomials $f=x^7+ax+b$: [Is there an irreducible but solvable septic trinomial...?](https://mathoverflow.net/questions/146769/is-there-an-irreducible-but-solvable-septic-trinomial-x7axnb-0/146780) Given the answers in *loc. cit.*, to obtain
more, we have to use the one more term in $bx^6$ allowed in
$a^7+bx^6+cx+d$, i will eventually search next days for a solution.
The problem is that the group of Galois symmetries should "drop" from $5040=7!=|S\_7|$ to such a small divisor like $42$ (or even below it).


In order to obtain examples, instead of the polynomial $f\_{20}=\prod(T-\theta\_i)$ of degree $|S\_5|/|F\_{20}|=5!/20=3!=6$ for the quintic case we have a polynomial $f\_{42}=\prod(T-\theta\_i)$ of degree $|S\_7|/|F\_{42}|=7!/42=5!=120$ for the septic case, one can compute it, then search for cases where it has a rational root.


Examples with a slightly bigger divisor like $168$ and decent coefficients can be given. A simple trinomial example with this bigger, but **not solvable** Galois order $168=4\cdot 42$ is $g = x^7 - 154x + 99$, computer check for it:



```
sage: g = x^7 -154*x + 99
sage: G = g.galois_group()
sage: G.order()
168
sage: G.structure_description()
'PSL(3,2)'
sage: G.is_solvable()
False

```

