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Difference between revisions of "2015 USAJMO Problems/Problem 4"

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Find all functions <math>f:\mathbb{Q}\rightarrow\mathbb{Q}</math> such that<cmath>f(x)+f(t)=f(y)+f(z)</cmath>for all rational numbers <math>x<y<z<t</math> that form an arithmetic progression. (<math>\mathbb{Q}</math> is the set of all rational numbers.)
 
Find all functions <math>f:\mathbb{Q}\rightarrow\mathbb{Q}</math> such that<cmath>f(x)+f(t)=f(y)+f(z)</cmath>for all rational numbers <math>x<y<z<t</math> that form an arithmetic progression. (<math>\mathbb{Q}</math> is the set of all rational numbers.)
  
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===Solution===
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===Solution 1===
 
According to the given, <math>f(x-a)+f(x+0.5a)=f(x-0.5a)+f(x)</math>, where x and a are rational. Likewise <math>f(x-0.5a)+f(x+a)=f(x+0.5a)+f(x)</math>. Hence <math>f(x+a)-f(x)= f(x)-f(x-a)</math>, namely <math>2f(x)=f(x-a)+f(x+a)</math>. Let <math>f(0)=C</math>, then consider <math>F(x)=f(x)-C</math>, where <math>F(0)=0,</math> <math>2F(x)=F(x-a)+F(x+a)</math>.  
 
According to the given, <math>f(x-a)+f(x+0.5a)=f(x-0.5a)+f(x)</math>, where x and a are rational. Likewise <math>f(x-0.5a)+f(x+a)=f(x+0.5a)+f(x)</math>. Hence <math>f(x+a)-f(x)= f(x)-f(x-a)</math>, namely <math>2f(x)=f(x-a)+f(x+a)</math>. Let <math>f(0)=C</math>, then consider <math>F(x)=f(x)-C</math>, where <math>F(0)=0,</math> <math>2F(x)=F(x-a)+F(x+a)</math>.  
  
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Therefore, for nonzero integer m, <math>(1/m)F(mx)=F(x)</math> , namely <math>F(x/m)=(1/m)F(x)</math>
 
Therefore, for nonzero integer m, <math>(1/m)F(mx)=F(x)</math> , namely <math>F(x/m)=(1/m)F(x)</math>
 
Hence <math>F(n/m)=(n/m)F(1)</math>. Let <math>F(1)=k</math>, we obtain <math>F(x)=kx</math>, where <math>k</math> is the slope of the linear functions, and <math>f(x)=kx+C</math>.
 
Hence <math>F(n/m)=(n/m)F(1)</math>. Let <math>F(1)=k</math>, we obtain <math>F(x)=kx</math>, where <math>k</math> is the slope of the linear functions, and <math>f(x)=kx+C</math>.
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===Solution 2===
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We have <cmath>f(x-3d)+f(x+3d)=f(x-d)+f(x+d)</cmath> and <cmath>f(x)+f(x+3d)=f(x+d)+f(x+2d).</cmath> Subtracting these two and rearranging gives <cmath>f(x-3d)+f(x+2d)=f(x)+f(x-d),</cmath> and since <math>f(x+2d)=f(x+d)+f(x)-f(x-d)</math> we get <cmath>f(x-3d)+f(x+d)=2f(x-d)</cmath> from which we get <cmath>f(x-d)+f(x+d)=2f(x).</cmath> Then we have <math>f(x)+f(y)=f(0)+f(x+y)=2f\left(\frac{x+y}{2}\right)</math>. Setting <math>f(0)=c</math>, we let <math>f(x)=g(x)+c</math> to get <math>g(x)+g(y)=g(x+y)</math>. This is Cauchy's functional equation, so it has solutions at <math>g(x)=kx</math>, so the answer is <math>\boxed{f(x)=kx+c}</math>.
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Functional Equation Problems]]
 
[[Category:Functional Equation Problems]]

Latest revision as of 16:55, 26 September 2021

Problem

Find all functions $f:\mathbb{Q}\rightarrow\mathbb{Q}$ such that\[f(x)+f(t)=f(y)+f(z)\]for all rational numbers $x<y<z<t$ that form an arithmetic progression. ($\mathbb{Q}$ is the set of all rational numbers.)

Solution 1

According to the given, $f(x-a)+f(x+0.5a)=f(x-0.5a)+f(x)$, where x and a are rational. Likewise $f(x-0.5a)+f(x+a)=f(x+0.5a)+f(x)$. Hence $f(x+a)-f(x)= f(x)-f(x-a)$, namely $2f(x)=f(x-a)+f(x+a)$. Let $f(0)=C$, then consider $F(x)=f(x)-C$, where $F(0)=0,$ $2F(x)=F(x-a)+F(x+a)$.

$F(2x)=F(x)+[F(x)-F(0)]=2F(x)$, $F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)$. Easily, by induction, $F(nx)=nF(x)$ for all integers $n$. Therefore, for nonzero integer m, $(1/m)F(mx)=F(x)$ , namely $F(x/m)=(1/m)F(x)$ Hence $F(n/m)=(n/m)F(1)$. Let $F(1)=k$, we obtain $F(x)=kx$, where $k$ is the slope of the linear functions, and $f(x)=kx+C$.

Solution 2

We have \[f(x-3d)+f(x+3d)=f(x-d)+f(x+d)\] and \[f(x)+f(x+3d)=f(x+d)+f(x+2d).\] Subtracting these two and rearranging gives \[f(x-3d)+f(x+2d)=f(x)+f(x-d),\] and since $f(x+2d)=f(x+d)+f(x)-f(x-d)$ we get \[f(x-3d)+f(x+d)=2f(x-d)\] from which we get \[f(x-d)+f(x+d)=2f(x).\] Then we have $f(x)+f(y)=f(0)+f(x+y)=2f\left(\frac{x+y}{2}\right)$. Setting $f(0)=c$, we let $f(x)=g(x)+c$ to get $g(x)+g(y)=g(x+y)$. This is Cauchy's functional equation, so it has solutions at $g(x)=kx$, so the answer is $\boxed{f(x)=kx+c}$.

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