IMO 2019/P1

Discussion on International Mathematical Olympiad (IMO)
tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

IMO 2019/P1

Unread post by tanmoy » Thu Jul 18, 2019 10:57 pm

Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$,

$f(2a)+2f(b)=f(f(a+b)).$

Proposed by Liam Baker, South Africa
"Questions we can't answer are far better than answers we can't question"

thczarif
Posts: 17
Joined: Mon Sep 25, 2017 11:27 pm
Location: Dhaka,Bangladesh

Re: IMO 2019/P1

Unread post by thczarif » Fri Jul 19, 2019 4:19 pm

tanmoy wrote:
Thu Jul 18, 2019 10:57 pm
Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$,

$f(2a)+2f(b)=f(f(a+b)).$

Proposed by Liam Baker, South Africa
$f(x)=0$ or $2x+c$ are the only solution to this function where $c$ is an integer.

$Proof:$
I will first prove that it's a linear function.

plugging in $b=0$ we get, $f(2a) + 2f(0)=f(f(a))$
switching $a$ and $b$ we get $f(0) + 2f(a)=f(f(a))$
so, $f(2a)+2f(0)=f(0)+2f(a) \rightarrow f(2a)=2f(a)-f(0)$
plugging in the value of $f(2a)$ in the original equation we get,
$2f(a)+2f(b)-f(0)=f(f(a+b))$
now if, a+b=c+d then $2f(a)+2f(b)-f(0)=f(f(a+b))=f(f(c+d))=2f(c)+2f(d)-f(0)$
$\rightarrow f(a)+f(b)=f(c)+f(d)$
Now, (n+2)+n=(n+1)+(n+1),so, $f(n+2)+f(n)=f(n+1)+f(n+1) \rightarrow f(n+2)-f(n+1)=f(n+1)-f(n)$
Thus the function is linear.
let $f(x)=ax+b$ for some integer $a,b$. Now, plugging in the values in the original equation will give $a=2$ or, $a=b=0$

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