Difference between revisions of "2015 USAJMO Problems/Problem 4"
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− | 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). | + | 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>. |
− | F(2x)=F(x)+[F(x)-F(0)]=2F(x), | + | <math>F(2x)=F(x)+[F(x)-F(0)]=2F(x)</math>, |
− | F(3x)=F(2x)+[F(2x)-F(x)]=3F(x). | + | <math>F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)</math>. |
− | Easily, by induction, F(nx)=nF(x) for all integers k. | + | Easily, by induction, <math>F(nx)=nF(x)</math> for all integers <math>k</math>. |
− | Therefore, for nonzero integer m, (1/m)F(mx)=F(x) , namely F(x/m)=(1/m)F(x) | + | Therefore, for nonzero integer m, <math>(1/m)F(mx)=F(x)</math> , namely <math>F(x/m)=(1/m)F(x)</math> |
− | 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. | + | 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>. |
Revision as of 20:55, 31 May 2015
Problem
Find all functions such that
for all rational numbers
that form an arithmetic progression. (
is the set of all rational numbers.)
Solution
According to the given, , where x and a are rational. Likewise
. Hence
, namely
. Let
, then consider
, where
.
,
.
Easily, by induction,
for all integers
.
Therefore, for nonzero integer m,
, namely
Hence
. Let
, we obtain
, where
is the slope of the linear functions, and
.