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Source: 2025 Turkey TST P3

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swynca

#1 h Mar 18, 2025, 6:24 AM • 3 Y

Y by AnSoLiN, dangerousliri, sami1618

Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x,y \in \mathbb{R}-\{0\}$ ,
$f(x) \neq 0 \text{ and } \frac{f(x)}{f(y)} + \frac{f(y)}{f(x)} - f \left( \frac{x}{y}-\frac{y}{x} \right) =2$

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swynca

#2 h Mar 18, 2025, 8:27 AM • 6 Y

Y by MS_asdfgzxcvb, Wikiliks, AnSoLiN, sami1618, PerfectPlayer, bin_sherlo

Answer
Solution

This post has been edited 1 time. Last edited by swynca, Mar 18, 2025, 9:40 AM
Reason: typo

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66 posts

#3 h Mar 18, 2025, 11:44 AM • 3 Y

Y by swynca, sami1618, PerfectPlayer

let $P(x,y)$ be the given equation.

Claim 1. $f(x)=0\iff x=0$
Proof. By $P(x,x)$ we get $f(0)=0$ . And for any $x\neq 0$ , we have $f(x)\neq 0$ by the problem statement.

Claim 2. $f(x)=f(y)\iff x^2=y^2$
Proof. If $f(x)=f(y)$ , we get $f\left(\frac{x}{y}-\frac{y}{x}\right)=0$ and by Claim 1, we get $\frac{x}{y}-\frac{y}{x}=0$ , which means $x^2=y^2$ . On the other hand, by $P(x,y)$ and $P(y,x)$ , we get $f(x)=f(-x)$ .

Lemma. $a+\frac{1}{a}=b+\frac{1}{b}\implies (a-b)\left(1-\frac{1}{ab}\right)=0\implies a=b\text{ or }a=\frac{1}{b}$
Claim 3. There is no $x,y$ such that $f(y)<0<f(x)$
Proof. Partition the set $\mathbb{R}-\{0\}$ into two groups $A=\{x:f(x)>0\}$ and $B=\{x:f(x)<0\}$ . Let $x\sim y$ denote $x$ and $y$ are in the same group. If $x\sim y$ , we get $f\left(\frac{x}{y}-\frac{y}{x}\right)\geq 0$ because $\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}\geq 2$ , and on the other hand if $x\not\sim y$ , we get $f\left(\frac{x}{y}-\frac{y}{x}\right)\leq -4$ because $\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}\leq -2$ . These observations mean $x\sim y\implies xz\sim yz$ . Then, for any $u\neq 0$ , depending on whether or not $1\sim u$ , we have either $x\sim xu\sim xu^2$ or $x\not\sim xu\not\sim xu^2$ , both of which give $x\sim xu^2$ . That is, multiplying $x$ with a positive number does not change its group. With Claim 2, we get either $A$ or $B$ is empty.

Claim 4. For all $x,y\in\mathbb{R}$ , we have $f(xy)f(1)=f(x)f(y)$
Proof. If $x=0$ or $y=0$ , it is true. So let's assume $x,y\in\mathbb{R}-\{0\}$ . By $P(1,y)$ and $P(x,xy)$ , we get $\frac{f(1)}{f(y)}+\frac{f(y)}{f(1)}-f\left(\frac{1}{y}-y\right)=2=\frac{f(x)}{f(xy)}+\frac{f(xy)}{f(x)}-f\left(\frac{1}{y}-y\right)$ With the lemma, this means $f(xy)f(1)=f(x)f(y)\text{ or }f(xy)f(y)=f(x)f(1)$ Assume there exists $x,y$ such that $f(xy)f(1)\neq f(x)f(y)$ . Then we must have $f(xy)f(y)=f(x)f(1)$ and $f(xy)f(x)=f(y)f(1)$ . Multiplying these, we get $f(x)^2=f(y)^2$ and $f(xy)^2=f(1)^2$ . By Claim 3, we get $f(x)=f(y)$ and $f(xy)=f(1)$ , and by Claim 2, we get $x^2=y^2$ and $(xy)^2=1$ . That means we have cases $(x,y)=(1,1),(1,-1),(-1,1),(-1,-1)$ , all of which already satisfies $f(xy)f(1)=f(x)f(y)$ , hence the result follows.

Now that we get a nice multiplicative property, we will use it in the main equation. For all $x,y\in\mathbb{R}-\{0\}$ ,
$f\left(\frac{x}{y}-\frac{y}{x}\right)=f\left(\frac{x^2-y^2}{xy}\right)=\frac{f(x^2-y^2)f(1)}{f(xy)}=\frac{f(\sqrt{|x^2-y^2|})^2f(1)}{f(x)f(y)}$ so $\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}-\frac{f(\sqrt{|x^2-y^2|})^2f(1)}{f(x)f(y)}=2\implies f(x)^2+f(y)^2-2f(x)f(y)-f(1)f(\sqrt{|x^2-y^2|})^2=0$ Let $R(x,y)$ be the last equation. Let's look at $R(x,z)$ and $R(y,z)$ where $z=\sqrt{x^2+y^2}$ .
$f(x)^2+f(z)^2-2f(x)f(z)-f(1)f(y)^2=0$ $f(y)^2+f(z)^2-2f(y)f(z)-f(1)f(x)^2=0$ Subtracting one from other, we get
$(f(1)+1)(f(x)-f(y))(f(x)+f(y))=2f(z)(f(x)-f(y))$ If $x^2\neq y^2$ , then by Claim 2, we have $f(x)\neq f(y)$ , so we get $f(z)=k(f(x)+f(y))$ where $k=\frac{f(1)+1}{2}$ . Rewrite this in $R(x,z)$ (or $R(y,z)$ ) and we get $(k^2-f(1))f(x)^2+(k^2-f(1))f(y)^2+(2k^2-f(1)-1)f(x)f(y)=0$ If $f(1)\neq 1$ , then $k^2\neq f(1)$ and take any different $x_1,x_2,x_3>0$ . $x_1,x_2,x_3$ all satisfy this quadratic equation if we fix a $y\neq 0$ , which is not possible. So $f(1)=1=k$ and $f(z)=f(x)+f(y)$ . We get a nice additive property, time to combine this with multiplicativity. Let $g(x)=f(\sqrt{|x|})$ . Then we have $g(x+y)=g(x)+g(y)$ and $g(xy)g(1)=g(x)g(y)$ , which means $g(x)=g(1)x$ and $f(x)=cx^2$ . Putting this into main equation, we get $c=1$ and the only solution $\boxed{f(x)=x^2}$

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#4 h Yesterday at 8:38 AM • 2 Y

Y by egxa, sami1618

Answer is $f(x)=x^2$
Let $P(x,y)$ be the given assertion. Similar to above solutions we can get $f(x)>0$ whenever $x\neq 0$ , $f(0)=0$ , and $f(x)=f(y) \iff |x|=|y|$
unnecessarily long proof for kinda multiplicativity
The rest of the solution comes from $\textbf{Selim Doğan:}$
bash time
final finish

This post has been edited 2 times. Last edited by PerfectPlayer, Yesterday at 8:39 PM

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