Estonian Mathematical Olympiad

Let $u=\frac{a}{b}$ and $x=\frac{c}{d}$ be the reduced fractions of $u$ and $x$. We are looking for rational numbers $v$ such that $v=\frac{mc-a}{md-b}$, where $m$ is large enough integer such that $mc-a$ and $md-b$ are both positive. According to the definition, $x$ is the mediant of $u$ and $v$ as soon as $\frac{mc-a}{md-b}$ is irreducible. Let's show that there are infinitely many natural numbers $m$ such that $\frac{mc-a}{md-b}$ is irreducible. This will complete the solution.

Let us first prove a lemma: prime numbers which can be used to reduce the fractions are also the factors of $ad-bc$. Indeed, if $p\mid mc-a$ and $p\mid md-b$, then $p\mid amd-b-b(mc-a)=m(ad-bc)$, therefore $p\mid m$ or $p\mid ad-bc$. If $p\mid m$, then $p\mid a$ and $p\mid b$ which contradicts the irreducibility of the fraction $\frac{a}{b}$. Therefore $p\mid ad-bc$.

As $u$ and $x$ are different, $ad-bc\ne 0$. Therefore the number $ad-bc$ has a finite number of prime factors. Let $p_1,\dots ,p_l$ be all the different prime factors which can reduce the fraction $\frac{mc-a}{md-b}$ for at least one $m$ and for each $i=1,\dots ,l$ let $m_i$ be natural number such that the fraction $\frac{m_{i}c-a}{m_{i}d-b}$ is reducible with prime $p_i$.

Let $n$ be an arbitrary factor for which the fraction $\frac{nc-a}{nd-b}$ is not irreducible. This fraction must be reducible with some prime number $p_i$ which can also reduce the fraction $\frac{m_{i}c-a}{m_{i}d-b}$. Then $p_i\mid (n-m_i)c$ and $p_i\mid (n-m_i)d$. Therefore $p_i\mid n-m_i$ as otherwise $p\mid c$ and $p\mid d$ which contradicts the irreducibility of the fraction $\frac{c}{d}$. Therefore $n\equiv m_i(\mathrm{mod}{p}_i)$.

Therefore by choosing $n$ such that $n\equiv m_i+1(\mathrm{mod}{p}_i)$ for each $i=1,\dots ,l$ the fraction $\frac{nc-a}{nd-b}$ must be irreducible. According to Chinese remainder theorem there are infinitely many natural numbers $n$ which satisfy such congruence system.