IMO

# Australia TST 2013 Exam2 Q3

Find all values of $n \in \mathbb {N} ^+$ such that there exists a function $f: \mathbb {R} \to \mathbb {R}$ that satisfies the following properties:

1), $f(x)=0$ if and only if $x=0$.

2), $\forall x \in \mathbb R,\, f(x) + f(2x) + ... + f(nx) =0$.

Answer: $\forall n \geq 2, n\in \mathbb {N}^+$.

Proof: $n=1$ does not fit into the criteria because it requires $f(x)=0,\forall x \in \mathbb R$. Fix $n \geq 2,n\in \mathbb N ^+$.

Bertrand’s Postulate (a.k.a. Tchebyshev’s theorem): $\forall m \in \mathbb {N}^+, \exists p, m, with $p$ prime.

Now we aim to find a $p$ such that $p\leq n <2p$.

For $n=2$, take $p=2$. For $n$ odd, $n\geq 3$, apply Tchebyshev’s theorem to $\left\lceil\frac{n}{2}\right\rceil$, there exists a prime $p, \frac {n+1}{2} < p \leq n+1$. But now $n+1$ is even and $n+1 \geq 4$ thus $n+1$ is not prime, and thus we have $p \leq n$, which shows that $p$ is the desired prime number as $2p > n+1 >n$. For $n$ even, apply Tchebyshev’s theorem to $\frac {n}{2}$ to yield a prime $p, \frac {n}{2} which gives $p\leq n <2p$. The $p$ has been chosen for all $n$.

Let $\mathbb Q ^*= \mathbb Q \backslash \{0\}$, $\mathbb R ^*= \mathbb R \backslash \{0\}$. Let $\mathbb Q_n = \{q\in \mathbb Q^* | q \textnormal { can be represented as }\frac {\alpha} {\beta},\, \alpha, \beta \in \mathbb Z, \textnormal{ where all prime factors of }\alpha \textnormal{ and }\beta \textnormal { do not exceed } n\}$.

(E.g. let $n=3$, then $Q_3 = \{q|q=\pm \frac {2^\eta}{3^\xi}, \eta,\xi \in \mathbb N ^+ \cup \{0\}\} \cup \{q|q=\pm \frac {3^\eta}{2^\xi}, \eta,\xi \in \mathbb N ^+ \cup \{0\}\}$.

Define an equivalence relation on $R^*$: for $x,y \in \mathbb R^*$, $x \sim y$ if and only if $\frac {x} {y} \in \mathbb Q_n$. It is easy to verify that this is indeed an equivalence relation. Partition $R^*$ into equivalence classes $K_{\gamma}$, by this relation. Let $\gamma$ be indexed by the set $\Gamma$.

Axiom of Choice: If $K_{\gamma}$ is a family of sets indexed by the set $\Gamma$, then the Cartesian product $\prod _ {\gamma \in \Gamma} K_{\gamma}$ is non-empty.

Apply Axiom of Choice to the equivalence classes of $R^*$, and we get a family of elements $\kappa _\gamma$, with each $\gamma \in \Gamma$ corresponding to exactly one $\kappa _ {\gamma}$.

Let $v_p(q)$ be the $p$-adic evaluation of $q\in Q^*$. (That is, reduce $q$ to simplest form $\frac {\alpha}{\beta}$, $v_p(q)= \textnormal{highest index of }p \textnormal{ dividing } \alpha$ if $p$ does not divide $\beta$, and $v_p(q)= - \textnormal{highest index of }p \textnormal { dividing } \beta$ if $p|\beta$.

Let $f(\kappa _ \gamma) =1,\forall \kappa _\gamma$. For $x\in \mathbb R ^*$, suppose $x\in K_g, g \in \Gamma$. Define $f(x)= (-n+1)^{v_p(\frac{x}{\kappa _g})}$. (E.g. if $n=3, p=3, x= \frac {2\pi}{3}, x \in K_g, \kappa _g =\pi, f(x)= (-2)^{(v_3(\frac 23))}=-\frac 12$.Similarly, $f(2x)=f(\frac {4\pi}{3})= -\frac 12$. This gives $f(x)+f(2x)+f(3x)=(-\frac 12) + (-\frac 12) + 1=0$.) We set $f(0)=0.$

Indeed, we show first show that the function satisfies the conditions.

First, we see that $f(x)$ does not vanish for non-zero $x$.

We claim that if $x,y \in K_g, f(x)/f(y)= (-n+1)^{v_p(\frac{x}{y})}$. This is obvious by substituting the definitions of $of f(x)$ and $f(y)$, and then LHS$=(-n+1)^{v_p(\frac{x}{\kappa_g})-v_p(\frac{y}{\kappa_g})}=(-n+1)^{v_p(\frac{x}{y})}=$RHS.

Now, we verify the conditions. Condition 1 is already checked. If $x=0$, $f(x)+f(2x)+...+f(nx)=0+0+...+0=0$. If $x \ne 0$, let $x \in K_g$. For $jx$ with $j\ne p$, we have $v_p(\frac {jx}{x})=v_p(j)=0$, and thus $f(jx)=f(x)$. Since $p\leq n < 2p$, there are $(n-1)$ such $j'$s. $f(px)=(-n+1)^{v_p(\frac{px}{x})}f(x)=(-n+1)f(x)$.

Thus $f(x)+f(2x)+...+f(nx)= (n-1)f(x)+(-n+1)f(x)=0$. We have found the desired function, for each of $n\in \mathbb N^+,n\ge 2$.

Q.E.D.

Comment: to be written later.