Absolute One-to-One Function

Let $f$ be a function which satisfies these properties. Let

$|f(x_1) - x_1| = |f(x_2) - x_2|= \ldots = |f(x_n) - x_n| = a$

For each $i$ , $1 \leq i \leq n$ , we have

$f(x_i) = x_i+\epsilon_i a$

where $\epsilon=\pm 1$ . Adding these equalities for each $i$ yields

$\displaystyle \sum_{i=1}^n f(x_i) = \displaystyle \sum_{i=1}^n x_i + a \displaystyle \sum_{i=1}^n \epsilon_i$

However, both $\displaystyle \sum_{i=1}^n f(x_i)$ and $\displaystyle \sum_{i=1}^n x_i$ have the same value by definition. Thus, we must have $a \displaystyle \sum_{i=1}^n \epsilon_i = 0$ . However, since $n$ is an odd positive integer, $\displaystyle \sum_{i=1}^n \epsilon_i \neq 0$ by parity. Thus, $a=0$ . Therefore, the only satisfying function is $f(x)=x$ . There is only $1$ satisfying function.

Obviously f(n)=n is a solution, so we have at least 1 solution.

Now assume that for at least 1 n, f(n) does not equal n. This implies that for all n, f(n) does not equal n.

Now consider the smallest x1 in the set. Let it map to xk. Further, the thing that maps to x1 must be xk. By induction, we can show that integers must be mapped off 2 by 2. However, since there are an odd number of elements, we can't pair off all the elements in twos. Thus, we have a contradiction, and f(n)=n is the only solution.