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

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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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[[Category:Olympiad Algebra Problems]]
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[[Category:Functional Equation Problems]]

Revision as of 07:28, 19 July 2016

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

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 $f(x)=kx+C$.

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