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A371553
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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.
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4
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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x^5 + a(n)*x + A371554(n) is irreducible and solvable by radicals.
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EXAMPLE
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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.
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MATHEMATICA
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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]]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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