Consider the function $f:\mathbb{R^+} \to \mathbb{R^+}$ such that $f\left(\frac{f(x)}{x}+y\right)=1+f(y)$. Find all such functions.
My try:
Plug $x=y=1$ and assume $f(1)=k$, we get $$f(k+1)=k+1$$ Now assuming $f$ is injective, we have $$\begin{aligned} f\left(\frac{f(x)}{x}+y\right) & =f\left(\frac{f(y)}{y}+y\right)=1+f(y) \\ \Rightarrow \quad \frac{f(x)}{x} & =\frac{f(y)}{y} \\ \Rightarrow \frac{f(1)}{1}=\frac{f(k+1)}{k+1} \Rightarrow k(k+1)=k+1 \Rightarrow k=1 \end{aligned}$$ Thus $f(x)=x$. But what is left is to prove $f$ is Injective a Priori.
We have $f(nf(1)+y) = n + f(y) > n$, for every $y > 0, n \ge 1$. Hence $\lim_{y \to +\infty} f(y) = +\infty$. If there exists $x_1 \ne x_2$ such that $a_1 := f(x_1)/x_1 > a_2 := f(x_2)/x_2$, then, for every $y > 0$, $f(a_1 + y) = f(a_2 + y)$. This means $f$ is periodic. Indeed, for every $n$, $f(a_2 + n(a_1 - a_2)) = f(a_1 + (n-1)(a_1-a_2)) = f(a_2 + (n-1)(a_1-a_2))$. This is a contradiction. Hence there exists a constant $c$ such that $f(x) = cx$ for every $x > 0$. The rest of the argument is easy.