Let $f:R\to R$ be a continous function, without fixed points,and $n,m$ positive integers. Let $f^k$ denote $f$ composed with itself $k$ times.Show that the system: $f^m(x)=y$ and $f^n(y)=x$ has no solutions.
Let's assume there exist $x,y$ that satisfy that system. By composition we get $f^{m+n}(x)=x$ and $f^{m+n}(y)=y$. And so the problem I wrote before it's basically this one.
Suppose $f$ has no fixed point. Then the graph of $f$ cannot touch the line $\{(x,x)\}_x$.
Assume $f(t) >t$ for all $t$.
Then $x = f^k(x) > f^{k-1}(t) > \cdots > f(x)$, which is a contradiction.
Similarly if we assume $f(t) <t $ for all $t$.