$f(x)=ax+b$ if $x$ is integer and $f(x)$ is any if $x$ is not integer.

Does anyone have a complete solution?

Let $P(m,n)$ be the assertion of the function over integers $m,n$. First, observe that if $f(x)$ is a solution, then $g(x) = f(x) -k$, where $k$ is an integer, is also a solution. So WLOG $f(0)=0$. We will now prove that $f(x) = xf(1)$ for $x$ a positive integer by induction. Base case: $x$=2 $P(0,1) \iff f(2)=2f(1)$ Induction step: Let $f(x)=cx$ holds of integers $0 \leq x \leq t$. We will prove it for $x=t+1$. $P(t-1,1) \iff (t-1)f(1) + f(t+1) = 2tf(1) \iff f(t+1) = (t+1)f(1)$. So it is proved for $x$ a positive integer. Analog for $x$ negative, we will get too $f(x) = xf(1)$ for $x$ negative by substituting the above values with 1 changed to -1. So the solution is $f(x) = xf(1)$ if $f(0)=0$. However, since all $g(x) = f(x) +k$ will also be a solution, then the solution will be $f(x) = xf(1) +k$, in other words, $f(x) = ax+b$ as above.

Am I wrong or $f(x)=a_{+}x+b, x\geq 0$ and $f(x)=a_{-}x+b, x<0$ is also a solution? Cause $n$ determines the sign of each argument of $f$ and so I think that we can change the coefficient of $x$ with the change of sign of $x$.