APMO 2019 P1

Discussion on Asian Pacific Mathematical Olympiad (APMO)
samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

APMO 2019 P1

Unread post by samiul_samin » Thu Aug 15, 2019 8:41 pm

Let $\mathbb{Z}^+$ be the set of positive integers. Determine all functions $f : \mathbb{Z}^+\to\mathbb{Z}^+$ such that $a^2+f(a)f(b)$ is divisible by $f(a)+b$ for all positive integers $a,b$.

Monovariant
Posts: 1
Joined: Sat Dec 26, 2020 4:16 pm

Re: APMO 2019 P1

Unread post by Monovariant » Sat Dec 26, 2020 4:26 pm

The only solution is $f(a)=a$ for all $a \in \mathbb{Z}^+$. It is easy to see that this solution works. We now prove that this is the only solution.

Let $P(a,b)$ denote the assertion in the problem. First, we have
\begin{align*}f(a)+b &|a^2+f(a)f(b)\\ \Longrightarrow f(a)+b &|a^2+f(a)f(b)-f(b)\left(f(a)+b\right)\\ &=a^2-bf(b).\\ \end{align*}
Denote the last assertion by $Q(a,b)$. $Q(1,1)$ gives $$f(1)+1|1-f(1),$$
i.e. for some $k\in\mathbb{Z}$ we have
\begin{align*}k\left(f(1)+1\right) &=1-f(1)\\ \Longrightarrow (k+1)f(1) &=1-k.\end{align*}
The only value of $k$ for which this equality holds is $0$. This gives $f(1)=1$.
Now, $P(1,a)$ gives
\begin{align*} 1+a &|1+f(a)\\ \Longrightarrow 1+a &|1+f(a)-(1+a)\\ &=f(a)-a, \end{align*}
i.e. for all $a\in\mathbb{Z}^+$ there exists an integer $m$ such that
\begin{align}f(a)-a &=m(1+a)\\ \Longrightarrow f(a) &=(m+1)a+m \end{align}
Now, $P(a,1)$ gives
\begin{align*}f(a)+1 &|a^2+f(a)\\ \Longrightarrow (m+1)a+m+1 &|a^2+(m+1)a+m\\ \Longrightarrow (a+1)(m+1) &|(a+m)(a+1)\\ \Longrightarrow m+1 &|a+m\\ \Longrightarrow m+1 &|a+m-(m+1) \\ &=a-1.\end{align*}
Plugging $a=2$ above we get $m+1|1$, i.e. $m\in\{0,-2\}$. Plugging the values of $m$ in $(2)$ we see that only $m=0$ gives us a function in $\mathbb{Z}^+$, and that is $f(a)=a$. $\square$

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