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Difference between revisions of "2019 USAJMO Problems/Problem 2"
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We see that if <math>x=|b|n</math> then <math>f(x)\equiv f(0) \pmod{|a|}</math>, if <math>x=|b|n+1</math> then <math>f(x)\equiv f(1)\pmod{|a|}</math>, if <math>x=|b|n+2</math> then <math>f(x)\equiv f(2)\pmod{|a|}</math>... if <math>x=|b|(n+1)-1</math> then <math>f(x)\equiv f(|b|-1)\pmod{|a|}</math>. This means that the <math>b</math>-element collection <math>\{f(0), f(1), f(2), ... ,f(|b|-1)\}</math> contains all <math>|a|</math> residues mod <math>|a|</math> since <math>f</math> is surjective, so <math>|b|\ge |a|</math>. Doing the same to <math>g</math> yields that <math>|a|\ge |b|</math>, so this means that only <math>|a|=|b|</math> can work.
 
We see that if <math>x=|b|n</math> then <math>f(x)\equiv f(0) \pmod{|a|}</math>, if <math>x=|b|n+1</math> then <math>f(x)\equiv f(1)\pmod{|a|}</math>, if <math>x=|b|n+2</math> then <math>f(x)\equiv f(2)\pmod{|a|}</math>... if <math>x=|b|(n+1)-1</math> then <math>f(x)\equiv f(|b|-1)\pmod{|a|}</math>. This means that the <math>b</math>-element collection <math>\{f(0), f(1), f(2), ... ,f(|b|-1)\}</math> contains all <math>|a|</math> residues mod <math>|a|</math> since <math>f</math> is surjective, so <math>|b|\ge |a|</math>. Doing the same to <math>g</math> yields that <math>|a|\ge |b|</math>, so this means that only <math>|a|=|b|</math> can work.
  
−
For <math>a=b</math> let <math>f(x)=g(x)=x+\frac{a}{2}</math> and for <math>a=-b</math> let <math>f(x)=-x+\frac{a}{2}</math> and <math>g(x)=-x-\frac{a}{2}</math>, so <math>|a|=|b|</math> does work and are the only solutions, as desired.
+
For <math>a=b</math> let <math>f(x)=x</math> and <math>g(x)=x+a</math>, and for <math>a=-b</math> let <math>f(x)=-x+a</math> and <math>g(x)=-x-a, so </math>|a|=|b|$ does work and are the only solutions, as desired.
  
 
-Stormersyle
 
-Stormersyle

Revision as of 20:52, 21 April 2019

Problem

Let $\mathbb Z$ be the set of all integers. Find all pairs of integers $(a,b)$ for which there exist functions $f:\mathbb Z\rightarrow\mathbb Z$ and $g:\mathbb Z\rightarrow\mathbb Z$ satisfying \[f(g(x))=x+a\quad\text{and}\quad g(f(x))=x+b\] for all integers $x$.

Solution

We claim that the answer is $|a|=|b|$.

Proof: $f$ and $g$ are surjective because $x+a$ and $x+b$ can take on any integral value, and by evaluating the parentheses in different order, we find $f(g(f(x)))=f(x+b)=f(x)+a$ and $g(f(g(x)))=g(x+a)=g(x)+b$. We see that if $a=0$ then $g(x)=g(x)+b$ to $b=0$ as well, so similarly if $b=0$ then $a=0$, so now assume $a, b\ne 0$.

We see that if $x=|b|n$ then $f(x)\equiv f(0) \pmod{|a|}$, if $x=|b|n+1$ then $f(x)\equiv f(1)\pmod{|a|}$, if $x=|b|n+2$ then $f(x)\equiv f(2)\pmod{|a|}$... if $x=|b|(n+1)-1$ then $f(x)\equiv f(|b|-1)\pmod{|a|}$. This means that the $b$-element collection $\{f(0), f(1), f(2), ... ,f(|b|-1)\}$ contains all $|a|$ residues mod $|a|$ since $f$ is surjective, so $|b|\ge |a|$. Doing the same to $g$ yields that $|a|\ge |b|$, so this means that only $|a|=|b|$ can work.

For $a=b$ let $f(x)=x$ and $g(x)=x+a$, and for $a=-b$ let $f(x)=-x+a$ and $g(x)=-x-a, so$|a|=|b|$ does work and are the only solutions, as desired.

-Stormersyle

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

See also

2019 USAJMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
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