1)take x=y=1 (f(1))^2-2f(1)+1=0 f(1)=1 2)take y=a f(x)+f(a/x)=2f(ax) 3)take y=1 f(a/x)=f(x) 4)so; f(a/x)=f(x)=f(ax) 5)take a=1 f(1/x)=f(x) 6)y=1/x f(x)f(1/x)+f(a/x)f(ax)=2 2(f(x))^2=2 f(x) can't be -1 because f(1)=1 so f(x)=1

I think $a$ is fixed, so you can't take $a=1$. You can reach $f^{2}(x)=1$ anyway: $x=y=1$ $\to$ $f(1)=1$ (using $f(a)=1$). $y=1$ $\to$ $f(x)=f(a/x)$ (using $f(1)=1$, $f(a)=1$). $y=a/x$ $\to$ $f^{2}(x)=1$ (using $f(x)=f(a/x)$, $f(a)=1$). But from here, for each separate $x$ there's $f(x)=\pm 1$. So $f(x)=1$ if $x\in P$ and $f(x)=-1$ if $x\not\in P$. There's still some work to do to prove that we must have $P\equiv\mathbb{R}$. Maybe it helps that $f(xy)=f(x)f(y)$ (using $f(x)=f(a/x)$ in the functional equation).

I think this is all a which satisfies the condition and 1 is.

lordWings wrote: Maybe it helps that $f(xy)=f(x)f(y)$ (using $f(x)=f(a/x)$ in the functional equation). Take $x=y$ then $f(x^{2})=f^{2}(x)$, we have $f(x)=f^{2}(x^{\frac{1}{2}})\geq 0$. So $f(x)=1$?? Correct me if I am wrong...

You're right, ychjae. The most simple solution has only four steps! $x=y=1$ $\to$ $f(1)=1$ (using $f(a)=1$). $y=1$ $\to$ $f(x)=f(a/x)$ (using $f(1)=1$, $f(a)=1$). $y=a/x$ $\to$ $f^{2}(x)=1$ (using $f(x)=f(a/x)$, $f(a)=1$). $x=y$ $\to$ $f(x^{2})=1$ (using $f(x)=f(a/x)$, $f^{2}(x)=1$). Therefore, $f(x)=1$ for every $x>0$.

Let $P(x,y)$ be the given assertion, \[P\left(x,\frac{a}{x}\right)\implies f(x)f\left(\frac{a}{x}\right)=1 \]Hence, we have \[P(x,x)\implies f(x)^2+\frac{1}{f(x)^2}=2f(x^2)\geq 2 \implies f(x)\geq 1 \ \forall x\in \mathbb{R^+}\]by AM-GM. However, \[f(x)f\left(\frac{a}{x}\right)=1\]so, this means \[f(x)\equiv 1. \square\]

No way I solved a fneqn Please contact westskigamer@gmail.com if there is an error with my solution for cash bounties by 3/18/2025.

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