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Bulgarian Mathematical Olympiad

Algebra Difficulty 5.7 AIME, harder Prove it Bulgaria

Problem:
Let $\mathbb{Q}^{+}$ be the set of positive rational numbers. Find all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{R}$ such that $f(1)=1$ , $f(1 / x)=f(x)$ for any $x \in \mathbb{Q}^{+}$ and $x f(x)=(x+1) f(x-1)$ for any $x \in \mathbb{Q}^{+}, x>1$ .

Solution

Solution:
Let $x=\frac{p}{q}$ , where $p$ and $q$ are coprime positive integers. We shall prove by induction on $n=p+q \geq 2$ that $f(x)$ is uniquely determined. This is true for $n=2$ (since $f(1)=1$ ).

Suppose that it is true for all integers less than a given $n \geq 3$ and consider $x=\frac{p}{q}$ , where $p+q=n$ . It follows from the first condition that we may assume that $p>q$ .

Now the second condition shows that $f\left(\frac{p}{q}\right)$ is uniquely determined by $f\left(\frac{p-q}{q}\right)$ and since $p-q+q<n$ it is uniquely determined by the induction hypothesis.

Note that the function $f\left(\frac{p}{q}\right)=\frac{p+q}{2}$ for $p$ and $q$ relatively prime fulfills the conditions of the problem.

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