Let $S$ be a set of all numbers $k$ such that $(n, k) = 1, 1 \leq k \leq n.$
Of course, smallest element in $S$ is (by definition) $1$ and largest is $n – 1$ (since $\text{gcd}(n – 1, n) = 1$).
In few examples (specifically in cases $n = \{15, 20, 23\}$), I noticed that $S$ must be in form
$S = \{k_1 = 1, k_2, …, n – k_2 , k_r = n – 1)\}.$ In other words, $S$ always have even number of elements, $\varphi{(n)} = r \equiv 0 \pmod 2,$ (except trivial cases $n = \{1, 2\}$) and importantly sum of 1st and $r$th, 2nd and $(r – 1)$-th is always $n.$
How to prove it? Please use elementary methods, thanks.
Best Answer
Just follow your arguments. As S can be divided by two sets S1={ x : x in S, x< n/2}, and S2= { n-x: x in S, x