does anyone have an idea how to proof this theorem?
Let $B$ be a balanced bipartite graph of order $2n$, $n ≥ 2$, as defined below and $s, t$ be two integers in $[0, n − 1]$. Then $D_s ∪ D_t$ forms a Hamilton cycle in $B$ if $\gcd(|t − s|, n) = 1$.
And needed theory: Balanced bipartite graph $B = (X ∪ Y, E)$ with vertex set $V = X ∪ Y$ , where $|X| = |Y|$, $X = {x_1, x_2, · · · , x_n}$ and $Y = {y_1, y_2, · · · , y_n}$. For $0 ≤ i ≤ n−1$, define $Di = \{x_jy_k : i ≡ j − k \pmod n\}$.
It's really a proposition of elementary number theory more than graph theory. I might mess up the signs of the indices a little, but this is the idea behind your argument.
Imagine the cycle of the graph that includes $y_1$. $y_1$ is adjacent via $D_t$ to $x_{t+1}$. It's other neighbor (via $D_s$) is $y_{t-s+1}$. The vertex in $Y$ that comes next is $y_{2(t-s)+1}$, and so on. If (and only if) $\gcd(t-s,n)=1$, then this sequence of vertices in $Y$ will hit all of the vertices of $Y$ before returning to $y_1$, leaving you with a Hamilton cycle. The cycle also hits all the vertices in $X$ by symmetry, since $X$ and $Y$ are balanced.