Generalization of the Banach fixed point theorem 

If $T$ is a mapping from a complete metric space $(M, d)$ into itself such that $T^m$ is a contraction mapping for some $m \in \mathbb{N}$, then show that $T$ has a unique fixed point.

Answer

Suppose $T^m(x)=x$, and assume $x$ is the only fixed point of $T^m$. Then
$$T^m(T(x))=T(x),$$
 and hence $T(x)$ is also a fixed point of $T^m$. Since the fixed point of $T^m$ is unique, we must have
\[T(x)=x,\]
so $x$ is also a fixed point of $T$. It remains to show that $T$ does not have any other fixed point. Assume $T(y)=y$, for some $y \in M$. Then 
\[T^{m}(y)=T^{m-1}(y)= \dots =T(x)=y,\]
so $y$ is a fixed point of $T^m$, and therefore $y=x$, since $T^m$ has only one fixed point.

$T$ also has the unique fixed point $x$. 

The answer is accepted.