2019 IMO Problems/Problem 1

Problem

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

Solution 1

The only solutions are $f(x)=0, 2x+c.$ For some integer $c.$

Obviously these work. We prove these are the only linear solutions. Plug $a=0$ and $b=0$ separately to get that $f(2x)=2f(x)-f(0).$ Plug $(0, a+b)$ to see $f(0)+f(a+b)=f(a)+f(b),$ and subtracting $2f(0)$ from both sides shows $f(a)-f(0)$ to be additive thus linear by Cauchy since this is on integers. Thus, $f(a)$ is linear, and so we are done since it can be easily shown that $0$ and $2x+c$ are the only linear solutions by plugging $mx+n$ into the equation.

Solution 2

We claim the only solutions are $f\equiv0$ and $f(x)=2x+c$ for some integer $c$ ., which obviously work. Plugging in $(0,n)$ and $(1,n-1)$ give $f(0)+2f(n)=f(f(n))=f(2)+2f(n-1)$ , so $f(n)-f(n-1)=\frac{f(2)-f(0)}2$ . Since this difference is constant and $f\colon\mathbb Z\rightarrow\mathbb Z$ , we must have $f$ is linear (finite differences or induction). It is easy to see the only linear solutions are those specified above. $\blacksquare$

~ Ezra Guerrero

Solution 3

We just plug values to find basic properties of the function. If $a=0$ , $f(0)+2f(b)=f(f(b)).$ We know that $f(0)$ is constant and we can see that $f(b)$ is not constant (because it varies depends on the value of b), so let $f(b)$ be a variable. This means we are making this a function of f(b). This yields that $f((f(b)))=2(f(b))+f(0)$ . This is a linear function (in terms of f(b))! However, we are not done yet. We have to show that this works. When $b=0$ and plugging in our function yields: $4a+2f(0)= f(f(a) \implies 4a+2f(0)= f(2a+f(0)) \implies 4a+2f(0)=4a+2f(0)$ and the last part is indeed true so we are good. Don’t forget the 0 function this works also. Let us verify that the 0 function works. This means $0+0=0$ , which is true and a so called “trivial” solution. Now let us prove this rigorously. Take $(a,b)$ when they equal $(-1,x)$ and $(0,x-1)$ (I chose it that way since they are 1 apart from each other; Really you could have chosen $(y,x)$ and $(y+1,x-1)$ and that would work). We will show that this is linear or below. Substitute in the equations for both yields $f(0)+2f(x-1)=f(f(x-1)) \qquad \text{and} \qquad f(-1)+ 2f(x)=f(f(x-1))$ Equating and then subtracting yields $2(f(x)-f(x-1))=f(0)-f(-1)$ . From this equation how do we know this is linear? Well treat the f(x) and f(x-1) like variables and dividing by two we can see that $(f(-1)-f(0))/2$ is constant since -1 and 0 are fixed values. And since the difference is constant and $x$ and $x-1$ are one apart this tells you that it is a linear function as the difference is constant, i.e 5x+4 is linear since the first positive x’s yields are 4, 9, 14… and we can see the difference is the same. Therefore we proved that the functions we said are good.

Side note: I am referring to the linear function $2x+b$ where $b \in \mathbb{R}$ and the constant function which is $0$ .

~EaZ_Shadow

2019 IMO (Problems) • Resources
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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2019 IMO Problems/Problem 1

Contents

Problem

Solution 1

Solution 2

Solution 3

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