Difference between revisions of "2020 OIM Problems/Problem 5"
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== Solution == | == Solution == | ||
− | { | + | Clearly <math>\boxed{f(x)=x+1}</math> works; we prove it is the only solution. |
+ | |||
+ | Define <math>F(a,b)</math> to be plugging <math>x=a</math> and <math>y=b</math> into our functional equation above. First, we consider <math>F(0,y)</math>: | ||
+ | <cmath>f(0f(0-y))+yf(0)=0+y+f(0)</cmath> | ||
+ | <cmath>\Rightarrow f(0)+yf(0)=y+f(0)</cmath> | ||
+ | <cmath>\Rightarrow y(f(0)-1)=0</cmath> | ||
+ | Letting <math>y\ne0</math>, we must have <math>f(0)=1</math>. | ||
+ | |||
+ | Next, we try <math>F(1,1)</math>: | ||
+ | <cmath>f(f(0))+f(1)=2+f(1)</cmath> | ||
+ | <cmath>\Rightarrow f(1)=2</cmath> | ||
+ | where we use <math>f(0)=1</math> from above. | ||
+ | |||
+ | Now we try <math>F(1,y)</math>: | ||
+ | <cmath>f(f(1-y))+2y=1+y+2</cmath> | ||
+ | <cmath>\Rightarrow f(f(1-y))=(1-y)+2</cmath> | ||
+ | Since <math>1-y</math> can take on all real values, let <math>z=1-y</math>; then for all real <math>z</math>, | ||
+ | <cmath>f(f(z))=z+2</cmath> | ||
+ | |||
+ | If we substitute <math>f(z)</math> into <math>z</math>, we get: | ||
+ | <cmath>f(f(f(z)))=f(z)+2</cmath> | ||
+ | But if we simply apply our function to both sides of the functional equation: | ||
+ | <cmath>f(f(f(z)))=f(z+2)</cmath> | ||
+ | implying | ||
+ | <cmath>f(z+2)=f(z)+2</cmath> | ||
+ | In particular, using <math>f(0)=1</math> and <math>f(1)=2</math> from above, we can use induction to show that for all integers <math>n</math>, we have | ||
+ | <cmath>f(n)=n+1</cmath> | ||
+ | |||
+ | We return to our original functional equation. Consider <math>F(-1,y)</math>: | ||
+ | <cmath>f(-f(-1-y))+yf(-1)=-1+y+f(1)</cmath> | ||
+ | But we just found that <math>f(-1)=0</math>, so | ||
+ | <cmath>\Rightarrow f(-f(-1-y))=y+1</cmath> | ||
+ | Let <math>z=-y-1</math>; then, since the domain of <math>y</math> is all reals, the domain for <math>z</math> is also all reals; therefore, for all real <math>z</math>, | ||
+ | <cmath>f(-f(z))=-z</cmath> | ||
+ | If we apply our function to both sides, we get | ||
+ | <cmath>f(f(-f(z)))=f(-z)</cmath> | ||
+ | But from before, <math>f(f(z))=z+2</math>, so | ||
+ | <cmath>\Rightarrow -f(z)+2=f(-z)</cmath> | ||
+ | <cmath>\Rightarrow f(z)+f(-z)=2</cmath> | ||
+ | |||
+ | Now we define a new function <math>g(x)=f(x)-1</math>. Then, from the condition we just found, <math>g(x)+g(-x)=0</math>, implying that <math>g(x)</math> is odd. Substituting into the original functional equation: | ||
+ | <cmath>g(xg(x-y)+x)+1+yg(x)+y=x+y+g(x^2)+1</cmath> | ||
+ | <cmath>\Rightarrow g(xg(x-y)+x)+yg(x)=x+g(x^2)</cmath> | ||
+ | Clearly <math>g(0)=0</math>, so consider <math>G(x,y)</math>: | ||
+ | <cmath>g(x)+xg(x)=x+g(x^2)</cmath> | ||
+ | But if we consider <math>G(-x,-y)</math>: | ||
+ | <cmath>g(-x)-xg(-x)=-x+g(x^2)</cmath> | ||
+ | Subtracting the second equation from the first: | ||
+ | <cmath>g(x)-g(-x)+xg(x)+xg(-x)=2x</cmath> | ||
+ | But now we utilize the odd condition of <math>g(-x)=-g(x)</math>: | ||
+ | <cmath>\Rightarrow g(x)+g(x)+xg(x)-xg(x)=2x</cmath> | ||
+ | <cmath>\Rightarrow 2g(x)=2x</cmath> | ||
+ | <cmath>\Rightarrow g(x)=x</cmath> | ||
+ | Therefore, <math>f(x)=g(x)+1=x+1</math>, so we are done. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ||
== See also == | == See also == | ||
[[OIM Problems and Solutions]] | [[OIM Problems and Solutions]] |
Latest revision as of 23:39, 3 May 2025
Problem
Find all functions such that
for any real numbers .
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Clearly works; we prove it is the only solution.
Define to be plugging
and
into our functional equation above. First, we consider
:
Letting
, we must have
.
Next, we try :
where we use
from above.
Now we try :
Since
can take on all real values, let
; then for all real
,
If we substitute into
, we get:
But if we simply apply our function to both sides of the functional equation:
implying
In particular, using
and
from above, we can use induction to show that for all integers
, we have
We return to our original functional equation. Consider :
But we just found that
, so
Let
; then, since the domain of
is all reals, the domain for
is also all reals; therefore, for all real
,
If we apply our function to both sides, we get
But from before,
, so
Now we define a new function . Then, from the condition we just found,
, implying that
is odd. Substituting into the original functional equation:
Clearly
, so consider
:
But if we consider
:
Subtracting the second equation from the first:
But now we utilize the odd condition of
:
Therefore,
, so we are done.