IMO Day 1 Problem 1

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

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

Easy Math Editor

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

The answer is $f\equiv 0$ and $f(x) = 2x+c \forall x\in \mathbb{Z}$ for some $c$ constant.

Solution for anyone needing it is:

By taking $a = 0$ , we get $f(0) + 2f(b) = f(f(b))$ for all $b\in \mathbb{Z}$ . So, the problem becomes: $f(2a) + 2f(b) = f(0) + 2f(a + b)$ Taking $a = 1$ for this one, we quickly obtain that $2(f(b+1) - f(b)) = f(2) - f(0)$ , so $f$ is linear. The rest is checking together with $f(0) + 2f(b) = f(f(b))$

Bravo Eleven - 1 year, 11 months ago

Why is $f$ linear?

Ruilin Wang - 1 year, 10 months ago

Look at the last statement for $f(b+1)-f(b) = \dfrac{f(2)-f(0)}{2} \forall b \in \mathbb{R}$

Bravo Eleven - 1 year, 10 months ago

@Bravo Eleven – Okay, thanks a lot!

Ruilin Wang - 1 year, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$