2020 OIM Problems/Problem 5

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

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2020 OIM Problems/Problem 5

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