TITLE: $f(x+y)=f(x)+f(y)+99$, $f(100)=101$, find $f(7)$.
QUESTION [6 upvotes]: Question :

Let $f:\mathbb{Q} \longrightarrow \mathbb{Q}$, and $\forall x,y \in \mathbb{Q}$, $f(x+y)=f(x)+f(y)+99$, $f(100)=101$, find $f(7)$.

Attempts :
I am pretty sure this is one of those questions which require an amazingly genius solution. However after having tried for half an hour, I still couldn’t get a clue. My problem is that I don’t really know how $f(7)$ can show up.
What I know:

$f(100)=2f(50)+99 \Longrightarrow f(50)=1$
$f(50)=2f(25)+99 \Longrightarrow f(25)=-49$
$f(x)=f(x)+f(0)+99 \Longrightarrow f(0)=-99$

From here I want to break it down to smaller numbers, but I can’t find a way. Any suggestions or hints will be appreciated.

REPLY [3 votes]: Suppose $f(1)=A$.
Then:
$$f(2)=2f(1)+99=2A+99$$
$$f(3)=f(2)+f(1)=3A+198$$
$$f(4)=f(1)+f(3)+99=4A+3(99)$$
We are given the general rule that for all $n\in\mathbb{Z}$, $f(n)=nA+99(n-1)$.  We can plug in $n=100$ and get that $101=100A+99^2$, which gives that $A=-97$.  Then, substituting $n=7$, we have that $f(7)=7(-97)+6\cdot99=-85$