Mathological03 wrote: Find all functions $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ such that \[ f(x) + f \left( \frac{1}{x} \right) = 1 \ \text{and} \ f(1 + 2x) = \frac{1}{2} f(x) \]for all $x \in \mathbb{Q}^+$. Setting $x=1$ in first property, we get $f(1)=\frac 12$ 1) At most one such function may exists Let positive integers $p,q$ and $s=p+q$ Let sequences of positive integers $a_n$ defined as : $a_0=p$ If $a_n=s-a_n$, then $a_{n+1}=a_n$ If $a_n\ne s-a_n$, then $a_{n+1}=|2a_n-s|$ Let rational number sequences $x_n=\frac{a_n}{s-a_n}$ and $y_n=f(x_n)$. Previous definition implies : $x_0=\frac pq$ If $x_n=1$, then $x_{n+1}=1$ and $y_{n+1}=y_n=\frac 12$ If $x_n>1$, then $x_{n+1}=\frac{x_n-1}2$ and $y_{n+1}=2y_n$ If $x_n<1$, then $x_{n+1})\frac{\frac 1{x_n}-1}2$ and $y_{n+1}=2-2y_n$ If at some step, we get $x_n=1$, then we get $y_n=\frac 12$ and the value of $y_0$ is uniquely known backward If $x_n\ne 1$ at eachstep, then note that $a_n\in[1,s-1]$ $\forall n$ can only take a finite number of values and so $a_n$ is periodic from a minimal given point $u$ and a minimal positive integer period $T$ So that $a_{n+T}=a_n$ $\forall n\ge u$ But it is easy to see that $y_{u+T}=c\pm 2^Ty_u$ (for some integer $c$) and so that $y_u$ is uniquely got thru equation $y_u=c\pm 2^Ty_u$ And so $y_0$ is uniquely known backward Q.E.D. 2) Solution $\boxed{f(x)=\frac 1{x+1}}$ fits and since at most one solution can exist, this is the unique one.