A371553 Consider primitive pairs of integers (b, c) with b > 0 such that x^5 + b*x + c = 0 is irreducible and solvable by radicals: sequence gives values of b.

11, 15, 15, 20, 120, 120, 280, 280, 312, 330, 330, 750, 750, 4095, 4095, 5700, 5700, 7800, 7800, 10140, 10140, 10564, 10564, 11102, 11275, 11275, 21970, 21970, 27248, 30758, 31000, 31000, 31146, 31350, 31350, 32955, 32955, 35490, 35490, 38360, 38360, 41236

Offset: 1

Author

Ben Whitmore, Mar 27 2024

Keywords

• nonn

Comments

Define the equivalence relation ~ on pairs of nonzero rational numbers by (b, c) ~ (b', c') if there exists a nonzero rational number k such that b' = k^4*b and c' = k^5*c. Every such pair (b, c) is equivalent to a unique pair of integers (b', c') with c' > 0 and |b'| as small as possible, which we call a primitive pair. If (b, c) ~ (b', c') and x^5 + b*x + c is irreducible and solvable by radicals, then so is x^5 + b'*x + c' by making the substitution x -> x/k and multiplying by k^5. Hence, every polynomial of the form x^5 + b*x + c with b, c nonzero rationals is equivalent to one with integer coefficients and positive constant coefficient.

An irreducible polynomial of the form x^5 + b*x + c for rational b, c is solvable by radicals if and only if its Galois group is a subgroup of the Frobenius group of order 20, which happens if and only if the resolvent sextic (x - b)^4*(x^2 - 6*b*x + 25*b^2) - 3125*c^4*x has a rational root. If b and c are integers, then such a rational root x must be an integer, by the rational root theorem. Therefore, given an integer b, we can find all such integers c by solving the quadratic Diophantine equation x^2 - (6*b + 5*y)*x + 25*b^2 = 0 for x and y, which has finitely many solutions. The values of c are then a subset of the values +-(x-b)*y^(1/4)/5.

Examples

			15 is in the sequence twice because x^5 + 15*x + 12 and x^5 + 15*x + 44 are both irreducible and solvable by radicals, and (15, 12) and (15, 44) are both primitive pairs.
176 is not in the sequence because there is no integer c for which (176, c) is primitive and x^5 + 176*x + c is irreducible and solvable by radicals. x^5 + 176*x + 1408 is irreducible and solvable by radicals, but (176, 1408) is not primitive because it is equivalent to (11, 44).
x^5 + (10/13)*x - 3/13 is solvable by radicals, and (10/13, -3/13) ~ (21970, 85683) which is primitive, so 21970 is in the sequence.
		

Links

• Ben Whitmore, Table of n, a(n) for n = 1..95

Crossrefs

For values of c see A371554.

Programs

Mathematica

pairs = Join @@ Table[
  Select[{b, Abs[#1 - b] #2/5} & @@@
    Sort[SolveValues[x^2 - (6b + 5y^4)x + 25b^2 == 0 && y > 0, {x, y}, Integers]],
    Max[Last /@ FactorInteger[GCD @@ #]] < 4 &&
    AllTrue[#, IntegerQ] &&
    IrreduciblePolynomialQ[x^5 + #1x + #2 & @@ #] &
  ],
  {b, 1, 1000}
];
pairs[[All, 1]]

Formula

x^5 + a(n)*x + A371554(n) is irreducible and solvable by radicals.