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.) | ||
| − | ===Solution=== | + | ===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>. | ||
<math>F(2x)=F(x)+[F(x)-F(0)]=2F(x)</math>, | <math>F(2x)=F(x)+[F(x)-F(0)]=2F(x)</math>, | ||
<math>F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)</math>. | <math>F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)</math>. | ||
| − | Easily, by induction, <math>F(nx)=nF(x)</math> for all integers <math> | + | Easily, by induction, <math>F(nx)=nF(x)</math> for all integers <math>n</math>. |
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>. | ||
| + | |||
| + | ===Solution 2=== | ||
| + | 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:Functional Equation Problems]] | ||
Latest revision as of 17:55, 26 September 2021
Problem
Find all functions 
such that![\[f(x)+f(t)=f(y)+f(z)\]](http://latex.artofproblemsolving.com/7/e/c/7ecb395c1e56e6d9726fc45f1bc7a66d98333da2.png)
for all rational numbers 
that form an arithmetic progression. (
is the set of all rational numbers.)
Solution 1
According to the given, 
, where x and a are rational. Likewise 
. Hence 
, namely 
. Let 
, then consider 
, where 
.
![$F(2x)=F(x)+[F(x)-F(0)]=2F(x)$](http://latex.artofproblemsolving.com/e/6/9/e6959dbbcdcc9d4c4602765e78279713dd5f480d.png)
,
![$F(3x)=F(2x)+[F(2x)-F(x)]=3F(x)$](http://latex.artofproblemsolving.com/7/7/6/7769164d1358b6785f508bf2266d7e9ee7fd31b4.png)
.
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 
.
Solution 2
We have ![\[f(x-3d)+f(x+3d)=f(x-d)+f(x+d)\]](http://latex.artofproblemsolving.com/b/f/4/bf4c165e7ec9a47cd97c1ebdfb582a98a6e907b9.png)
and ![\[f(x)+f(x+3d)=f(x+d)+f(x+2d).\]](http://latex.artofproblemsolving.com/2/9/4/294a82ec8b00aa4cd5fb9847f1b9dc34dfa4de27.png)
Subtracting these two and rearranging gives ![\[f(x-3d)+f(x+2d)=f(x)+f(x-d),\]](http://latex.artofproblemsolving.com/e/0/b/e0b78eecb4a543ab9d022eb4af3411847a4380ea.png)
and since 
we get ![\[f(x-3d)+f(x+d)=2f(x-d)\]](http://latex.artofproblemsolving.com/8/8/b/88b9697ccd11742ea1dff41ec876a51755ba8cf3.png)
from which we get ![\[f(x-d)+f(x+d)=2f(x).\]](http://latex.artofproblemsolving.com/3/8/7/3877b46f7ec129f2b00e3c0526fa7a6b9060752a.png)
Then we have 
. Setting 
, we let 
to get 
. This is Cauchy's functional equation, so it has solutions at 
, so the answer is 
.
