Substitute $x: = y+1$ to the preposition to get $|f(y+1) - f(y)| \le |y+1-y| = 1$. Because $f$ is injective, $|f(y+1) - f(y)| = 1$. Now set $f(0) = a$. Because $|f(1) - f(0)| = 1$, either $f(1) = a+1$ or $f(1) = a-1$. Suppose $f(1) = a+1$. We induct that $f(n) = a+n$. For $n=0,1$, this is correct. Suppose for $n=k,k+1$, this claim is correct, then: $|f(k+2) - f(k+1)| = 1$ $f(k+2) = f(k+1) + 1 \vee f(k+1) - 1$ $f(k+2) = a+k+2 \vee a+k$ But $f(k) = a+k$, so since $f$ is injective, $f(k+2) = a+k+2$, and the claim is proven for $n \ge 0$. For $n=1,0$, this is correct. Suppose for $n=k+1,k$, this claim is correct, then: $|f(k-1) - f(k)| = 1$ $f(k-1) = f(k) + 1 \vee f(k) - 1$ $f(k-1) = a+k+1 \vee a+k-1$ But $f(k+1) = a+k+1$, so $f(k-1) = a+k-1$, and the claim is proven for $n \le 0$. Hence $f(n) = a+n$ for all integer $n$. The same argument can be applied to where $f(0) = a, f(1) = a-1$ to get $f(n) = a-n$. Hence we have our solutions: $f(x) = c+x$ for all $x$, or $f(x) = c-x$ for all $x$, for any $c$.

Let$f(0)=\lambda $ and $g:\mathbb{Z}\to \mathbb{Z},g\left( x \right)=f\left( x \right)-f\left( 0 \right)\Rightarrow g$ is injective $\left| g\left( x \right)-g\left( y \right) \right|\le \left| x-y \right|$ ,for any $x,y\in \mathbb{Z}$. Substitute $y=0$ to the preposition to get $|g(x)|\le |x|,\left( \forall \right)x\in \mathbb{Z}$. We induct that $g(n)=n,g(-n)=-n$. For $n=0,1$, this is correct. Suppose for $n\le k,k\ge 1$, this claim is correct Because $\left\{ g\left( -k \right),...,g\left( -1 \right),g\left( 0 \right),g\left( 1 \right),...,g\left( k \right) \right\}=\left\{ -k,...,-1,0,1,...,k \right\}$and $\left| g\left( k+1 \right) \right|\le \left| k+1 \right|,\left| g\left( -k-1 \right) \right|\le \left| k+1 \right|\Rightarrow g\left( k+1 \right),g\left( -k-1 \right)\in \left\{ -k-1,k+1 \right\}$ $\Rightarrow g\left( k+1 \right)=k+1,g\left( -k-1 \right)=-k-1$,because if $g\left( k+1 \right)=-k-1,g\left( -k-1 \right)=k+1\Rightarrow \left| g\left( k+1 \right)-g\left( k \right) \right|=2k+1>1,$false $\Rightarrow f(x)=f(0)+x$ for all $x$, or $f(x)=f(0)-x$ for all $x$. $\Rightarrow f(x)=a+x$ for all $x$, or $f(x)=a-x$ for all $x$, for any $a$.

Let $y = x-1$ to get $|f(x) - f(x-1)| \le n$. Because $f$ is injective, $f(x) = f(x-1) + n$ or $f(x) = f(x-1) - n$. Continue as in my proof, changing most occurrences of $1$ to $n$.

Generalization Let $n\in {{\mathbb{N}}^{*}}.$Find all injective functions$f:\mathbb{Z}\to \mathbb{Z}$ that satisfy: i) $n\left| \left( f\left( x \right)-f\left( y \right) \right) \right.$,for any $x,y\in \mathbb{Z}$. ii)$\left| f\left( x \right)-f\left( y \right) \right|\le n\left| x-y \right|$ ,for any $x,y\in \mathbb{Z}$.

Wait how did that post move down Anyway, my solution is my previous post. Should be post 4 if no posts are changing place. What is $n$ in condition 2 anyway? I assume that condition 2 is part of condition 1 (to make "exist $n$ such that $n$ divides $f(x)-f(y)$ for all $x,y$ and $|f(x)-f(y)| \le n |x-y|$ for all $x,y$").