Skolem Proof

For which \(n \in \mathbb{N}\) is it possible to construct Skolem sequences?
A Skolem sequence of order \(n\) is a permutation of the sequence: \(S = [1, 1, 2, 2, ... , n, n]\).
Each number \(i = 1...n\) appears twice in the sequence, the position of the first instance of \(i\) we will call \(x_i\) and the second \(y_i\).
A Skolem sequence also has the property that each pair of numbers \(x_i\) and \(y_i\) are separated in position equal to their value, \(i\). That is, \(y_i = x_i + i\).
Proof: Let S be a Skolem sequence, with the position of each number given as: \(P = [{x_1,y_1,x_2,y_2,...,x_n,y_n}] = [1, 2, ... , 2n]\).
Consider the sum of positions \(x_i\) and \(y_i\) in the sequence Skolem sequence \(S\): \[\sum P = \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} y_i = 1 + 2 + ... + 2n\]

Since \(y_i = x_i + i\). we can rewrite the left side of the expression: \[\sum_{i=1}^{n} x_i + \sum_{i=1}^{n} y_i = \sum_{i=1}^{n} x_i + \sum_{i=1}^{n} (x_i + i) = 2\sum_{i=1}^{n} x_i + \sum_{i=1}^{n} i\]

Now our expression can be written as: \[2\sum_{i=1}^{n} x_i + \sum_{i=1}^{n} i = 1 + 2 + ... + 2n\]

We can use the fact that the sum of numbers 1, 2, .. n = \(\frac{n(n+1)}{2}\). \[2\sum_{i=1}^{n} x_i + \frac{n(n+1)}{2} = \frac{2n(2n+1)}{2}\]

Next, let's rearrange terms to isolate the sum of the \(x_i\)s. \[\sum_{i=1}^{n} x_i = \frac{2n(2n+1) - n(n+1)}{4}\]

Multiplying out the right side of the expression we get a simplified equation: \[\sum_{i=1}^{n} x_i = \frac{4n^2 + 2n - n^2 - n}{4} = \frac{3n^2 + n}{4}\]
The \(\sum_{i=1}^{n} x_i\) must be an integer, therefore so must \(\frac{3n^2 + n}{4}\). We can split the equation into separate cases by considering \(3n^2 + n \bmod{4}\).
If \(n \bmod{4} = 0\), then \(n\) is a multiple of 4 and so is \(3n^2 + n\) since we can factor out a 4 from each term.
If \(n \bmod{4} = 1\), then \(3n^2 + n = 3(1+4i)^2 + 1 + 4i = 12i^2 + 28i + 4\) is also a multiple of 4.
If \(n \bmod{4} = 2\), then \(3n^2 + n = 3(2+4i)^2 + 2 + 4i = 12i^2 + 52i + 14\) is not divisible by 4.
If \(n \bmod{4} = 3\), then \(3n^2 + n = 3(3+4i)^2 + 3 + 4i = 12i^2 + 76i + 30\) is not divisible by 4.
Finally, we can conclude that Skolem sequences only exist for \(n \bmod{4} = 0, 1\).