2025 USAJMO Problems/Problem 1

Problem

Let $\mathbb Z$ be the set of integers, and let $f\colon \mathbb Z \to \mathbb Z$ be a function. Prove that there are infinitely many integers $c$ such that the function $g\colon \mathbb Z \to \mathbb Z$ defined by $g(x) = f(x) + cx$ is not bijective. Note: A function $g\colon \mathbb Z \to \mathbb Z$ is bijective if for every integer $b$ , there exists exactly one integer $a$ such that $g(a) = b$ .

Solution

Note that $c=f(a)-f(a+1)$ makes $g(a)=g(a+1),$ so for all $c=f(a)-f(a+1)$ we have that $g(x)$ is not bijective. If there are infinitely many possible values of $c,$ we are done. Otherwise, there are only finitely many possible values of $f(a)-f(a+1)$ so there are only finitely many possible values of $-(f(a)-f(a+1))=f(a+1)-f(a).$ This means there is a minimum $d$ of $f(a+1)-f(a).$ It is clear for $x>0$ that $f(x)=f(0)+\sum_{a=0}^{x-1}(f(a+1)-f(a))\ge dx+f(0)$ and for $x<0$ that $f(x)=f(0)-\sum_{a=x}^{-1}\le dx+f(0).$ Then, let $c>-d+3$ be an integer. We have for integers $x>0,$ $x\ge 1,$ so $g(x)=f(x)+cx\ge (d+c)x+f(0)>3x+f(0)>3+f(0),$ if $x=0$ we have $g(x)=f(0),$ and if $x<0$ we have $g(x)=f(x)+cx\le (d+c)x+f(0)<3x+f(0)<f(0).$ This means that there is no $a$ such that $g(a)=f(0)+1,$ so $g(x)$ is not bijective. $\blacksquare$

~BS2012

2025 USAJMO (Problems • Resources)
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2025 USAJMO Problems/Problem 1

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