A383068 The set of all integers k >= -1 with the property that there exist integers X and Y such that X*Y*(X+Y) is nonzero and X^3 - k*X^2*Y - (k+3)*X*Y^2 - Y^3 is a divisor of k^2 + 3*k + 9.
-1, 0, 1, 2, 3, 5, 12, 54, 66, 1259, 2389
Offset: 1
Examples
For each integer k >= -1 given in the sequence, the list of all pairs (X,Y) such that X*Y*(X+Y) is nonzero and such that F_k(X,Y) is a positive integer divisor of k^2 + 3k + 9 is as follows: If k = -1, then (X,Y) = (-1,-1), (-1,2), (2,-1), (5,4), (4,-9), (-9,5), (2,1), (1,-3), or (-3,2). If k = 0, then (X,Y) = (2,1), (1,-3), (-3, 2), (-1,-1), (-1, 2), or (2,-1). If k = 1, then (X,Y) = (-5,-2), (-2, 7), or (7,-5). If k = 2, then (X,Y) = (-7,-2), (-2, 9), or (9,-7). If k = 3, then (X,Y) = (-1,-1), (-1, 2), (2,-1), (-4,-1), (-1, 5), or (5,-4). If k = 5, then (X,Y) = (-1,-2), (-2, 3), (3,-1), (-4,-1), (-1, 5), (5,-4), (19, 3), (3,-22), or (-22, 19). If k = 12, then (X,Y) = (-1,-1), (-1, 2), (2,-1), (-13,-1), (-1, 14), (14,-13), (-4,-1), (-1, 5), or (5,-4). If k = 54, then (X,Y) = (-1,-2), (-2, 3), (3,-1), (-4,-1), (-1, 5), or (5,-4). If k = 66, then (X,Y) = (-5,-2), (-2, 7), or (7,-5). If k = 1259, then (X,Y) = (-4,-5), (-5, 9), (9,-4), (-13,-1), (-1, 14), (14,-13), (-3,-19), (-19, 22), or (22,-3). If k = 2389, then (X,Y) = (-7,-2), (-2, 9), or (9,-7).
Links
- A. Hoshi, On correspondence between solutions of a family of cubic Thue equations and isomorphism classes of the simplest cubic fields, J. Number Theory 131 (2011), no. 11, 2135-2150. See Table 1 on page 2138.
- T. Kashio and R. Sekigawa, The characterization of cyclic cubic fields with power integral bases, Kodai Math. J. 44 (2021), no. 2, 290-306.
- R. Okazaki, The simplest cubic fields are non-isomorphic to each other, Doshisha University, Kyoto, Japan. [Cached copy at the Wayback Machine]
- M. Tinková, R. Visser, and P. Yatsyna, Sums of two units in number fields, arXiv:2502.01345 [math.NT], 2025. See page 9.
Programs
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Magma
is_A383068 := function(k) R
:= PolynomialRing(Integers()); T := Thue(x^3 - k*x^2 - (k+3)*x - 1); for d in Divisors(k^2 + 3*k + 9) do S := Solutions(T, d); for s in S do if (s[1]*s[2]*(s[1]+s[2]) ne 0) then return true; end if; end for; end for; return false; end function; [k : k in [-1..3000] | is_A383068(k)];
Comments