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Algebra Difficulty 8.5 Shortlist Prove it Turkey

Let $Q^+$ denote the set of positive rational numbers, and $\mathbb{Z}$ denote the set of all integers. Find all functions $f: Q^+ \to \mathbb{Z}$ that satisfy the conditions $f(1/x) = f(x)$ and $(x+1)f(x-1) = x f(x)$ for all $x \in Q^+$ such that $x > 1$ .

Solution

Substituting $x = 2$ in the second equation gives $3f(1) = 2f(2)$ . In particular, $f(1)$ is even. It follows by induction that, for $n \in \mathbb{Z}^+$ , $f(n) = (n+1)f(1)/2$ .

Now we show by induction on $p+q$ that for all $p, q \in \mathbb{Z}^+$ and $(p, q) = 1$ , $f(p/q) = (p+q)f(1)/2$ . If $p > q$ , $p/q \cdot f(p/q) = (p+q)/q \cdot f((p-q)/q) = (p+q)/q \cdot (p-q+q)f(1)/2$ ; and if $p < q$ , then $f(p/q) = f(q/p)$ and we are in the first case.

To summarize, $f$ satisfies the conditions of the problem if and only if $m$ is a positive integer and $f(p/q) = (p+q)m$ for all $p, q \in \mathbb{Z}^+$ with $(p, q) = 1$ .

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