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One may calculate the set S(k) using the brief
Mathematica code
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sset[k_]
:= Module [
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{ddd},
ddd = (-4)Divisors[k^2];
Return[Union[Mod[ddd, 4k -
1]]];
];
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To see that this works...
Lemma For n > 0, let T(n) be
the set of least residues, mod 4n-1, of the elements of {
-4 D | D is a positive divisor of n2 }. Then
S(n) = T(n).
(For an integer n, the least residue
of n, mod m, is the unique integer in {0, 1, ..., m-1}
which is congruent to n. Here, m is a positive
integer.)
Proof
Suppose that A is in S(n), and let W and X be the
positive divisors of n such that AW+X == 0 (mod 4n-1).
Let B be the positive integer, B = n/W, and take D = BX.
D is a positive integer divisor of n2 and 4BW
== 4n == 1 (mod 4n-1), so
0 == 4B(AW+X) == 4ABW + 4BX == A + 4D (mod
4n-1),
so A == -4D (mod 4n-1), so A is the least residue of
-4D (mod 4n-1), so A is in T(n).
Conversely, suppose that A is in T(n), and let D be
the positive divisor of n2 such that A == -4D
(mod 4n-1). There are positive integer divisors B and X
of n such that D = BX; let W = n/B. Now W and X are
positive divisors of n and 4BW == 4n == 1 (mod 4n-1),
so
0 == A + 4D == 4ABW + 4BX == 4B(AW+X) (mod
4n-1),
so since both 4 and B are relatively prime to 4n-1,
AW+X == 0 (mod 4n-1). Since A is a least residue mod
4n-1, A is in S(n).
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