Difference between revisions of "2020 OIM Problems/Problem 5"

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== Solution ==
 
== Solution ==
{{solution}}
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Clearly <math>\boxed{f(x)=x+1}</math> works; we prove it is the only solution.
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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>:
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<cmath>f(0f(0-y))+yf(0)=0+y+f(0)</cmath>
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<cmath>\Rightarrow f(0)+yf(0)=y+f(0)</cmath>
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<cmath>\Rightarrow y(f(0)-1)=0</cmath>
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Letting <math>y\ne0</math>, we must have <math>f(0)=1</math>.
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Next, we try <math>F(1,1)</math>:
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<cmath>f(f(0))+f(1)=2+f(1)</cmath>
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<cmath>\Rightarrow f(1)=2</cmath>
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where we use <math>f(0)=1</math> from above.
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Now we try <math>F(1,y)</math>:
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<cmath>f(f(1-y))+2y=1+y+2</cmath>
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<cmath>\Rightarrow f(f(1-y))=(1-y)+2</cmath>
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Since <math>1-y</math> can take on all real values, let <math>z=1-y</math>; then for all real <math>z</math>,
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<cmath>f(f(z))=z+2</cmath>
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If we substitute <math>f(z)</math> into <math>z</math>, we get:
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<cmath>f(f(f(z)))=f(z)+2</cmath>
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But if we simply apply our function to both sides of the functional equation:
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<cmath>f(f(f(z)))=f(z+2)</cmath>
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implying
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<cmath>f(z+2)=f(z)+2</cmath>
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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
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<cmath>f(n)=n+1</cmath>
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We return to our original functional equation. Consider <math>F(-1,y)</math>:
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<cmath>f(-f(-1-y))+yf(-1)=-1+y+f(1)</cmath>
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But we just found that <math>f(-1)=0</math>, so
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<cmath>\Rightarrow f(-f(-1-y))=y+1</cmath>
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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>,
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<cmath>f(-f(z))=-z</cmath>
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If we apply our function to both sides, we get
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<cmath>f(f(-f(z)))=f(-z)</cmath>
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But from before, <math>f(f(z))=z+2</math>, so
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<cmath>\Rightarrow -f(z)+2=f(-z)</cmath>
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<cmath>\Rightarrow f(z)+f(-z)=2</cmath>
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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:
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<cmath>g(xg(x-y)+x)+1+yg(x)+y=x+y+g(x^2)+1</cmath>
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<cmath>\Rightarrow g(xg(x-y)+x)+yg(x)=x+g(x^2)</cmath>
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Clearly <math>g(0)=0</math>, so consider <math>G(x,y)</math>:
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<cmath>g(x)+xg(x)=x+g(x^2)</cmath>
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But if we consider <math>G(-x,-y)</math>:
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<cmath>g(-x)-xg(-x)=-x+g(x^2)</cmath>
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Subtracting the second equation from the first:
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<cmath>g(x)-g(-x)+xg(x)+xg(-x)=2x</cmath>
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But now we utilize the odd condition of <math>g(-x)=-g(x)</math>:
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<cmath>\Rightarrow g(x)+g(x)+xg(x)-xg(x)=2x</cmath>
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<cmath>\Rightarrow 2g(x)=2x</cmath>
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<cmath>\Rightarrow g(x)=x</cmath>
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Therefore, <math>f(x)=g(x)+1=x+1</math>, so we are done.
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~ [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 $f : \mathbb{R} \to \mathbb{R}$ such that

\[f(xf(x-y))+yf(x)=x+y+f(x^2)\]

for any real numbers $x, y$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Clearly $\boxed{f(x)=x+1}$ works; we prove it is the only solution.

Define $F(a,b)$ to be plugging $x=a$ and $y=b$ into our functional equation above. First, we consider $F(0,y)$: \[f(0f(0-y))+yf(0)=0+y+f(0)\] \[\Rightarrow f(0)+yf(0)=y+f(0)\] \[\Rightarrow y(f(0)-1)=0\] Letting $y\ne0$, we must have $f(0)=1$.

Next, we try $F(1,1)$: \[f(f(0))+f(1)=2+f(1)\] \[\Rightarrow f(1)=2\] where we use $f(0)=1$ from above.

Now we try $F(1,y)$: \[f(f(1-y))+2y=1+y+2\] \[\Rightarrow f(f(1-y))=(1-y)+2\] Since $1-y$ can take on all real values, let $z=1-y$; then for all real $z$, \[f(f(z))=z+2\]

If we substitute $f(z)$ into $z$, we get: \[f(f(f(z)))=f(z)+2\] But if we simply apply our function to both sides of the functional equation: \[f(f(f(z)))=f(z+2)\] implying \[f(z+2)=f(z)+2\] In particular, using $f(0)=1$ and $f(1)=2$ from above, we can use induction to show that for all integers $n$, we have \[f(n)=n+1\]

We return to our original functional equation. Consider $F(-1,y)$: \[f(-f(-1-y))+yf(-1)=-1+y+f(1)\] But we just found that $f(-1)=0$, so \[\Rightarrow f(-f(-1-y))=y+1\] Let $z=-y-1$; then, since the domain of $y$ is all reals, the domain for $z$ is also all reals; therefore, for all real $z$, \[f(-f(z))=-z\] If we apply our function to both sides, we get \[f(f(-f(z)))=f(-z)\] But from before, $f(f(z))=z+2$, so \[\Rightarrow -f(z)+2=f(-z)\] \[\Rightarrow f(z)+f(-z)=2\]

Now we define a new function $g(x)=f(x)-1$. Then, from the condition we just found, $g(x)+g(-x)=0$, implying that $g(x)$ is odd. Substituting into the original functional equation: \[g(xg(x-y)+x)+1+yg(x)+y=x+y+g(x^2)+1\] \[\Rightarrow g(xg(x-y)+x)+yg(x)=x+g(x^2)\] Clearly $g(0)=0$, so consider $G(x,y)$: \[g(x)+xg(x)=x+g(x^2)\] But if we consider $G(-x,-y)$: \[g(-x)-xg(-x)=-x+g(x^2)\] Subtracting the second equation from the first: \[g(x)-g(-x)+xg(x)+xg(-x)=2x\] But now we utilize the odd condition of $g(-x)=-g(x)$: \[\Rightarrow g(x)+g(x)+xg(x)-xg(x)=2x\] \[\Rightarrow 2g(x)=2x\] \[\Rightarrow g(x)=x\] Therefore, $f(x)=g(x)+1=x+1$, so we are done.

~ eevee9406

See also

OIM Problems and Solutions