I am finding this functional equation from a past high school mathematics competition rather tricky.
Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}^+$ such that:
$f\big(x+f(x)\big)=f(x)\quad \forall x\in\mathbb{R}$
and
$f(2012)=2012$.
(One must prove that the trivial constant solution is the only solution.)
By induction, $f(x+nf(x))=f(x)$. Hence as you observe, each $x+nf(x)$ is injective, since $$x+nf(x)=y+nf(y) \implies f(\cdots)=f(\cdots) \implies f(x) = f(y) \implies x + \cdot = y + \cdot$$ and therefore monotonically increasing, since $x+nf(x)\to\infty$ as $x\to\infty$.
Now for $x<y$, we have $x+nf(x)<y+nf(y)$ for all $n$ and so $f(y)-f(x)\ge0$.
But $f(x+nf(x))=f(x)$ and $f(x)>0$ implies that for all $x$ there exist points arbitrarily far away which take the same $f$ value, and hence as $f$ is non-decreasing, $f$ must be constant.