Find all functions $f: \mathbb R \to \mathbb R$ such that for any $x,y \in \mathbb R$, the multiset $\{(f(xf(y)+1),f(yf(x)-1)\}$ is identical to the multiset $\{xf(f(y))+1,yf(f(x))-1\}$. Note: The multiset $\{a,b\}$ is identical to the multiset $\{c,d\}$ if and only if $a=c,b=d$ or $a=d,b=c$.
Problem
Source: 2022 China TST, Test 4 P3
Tags: functional equation, algebra
01.05.2022 04:33
NOTE: THE PREVIOUS SOLUTION I POSTED HAS A SIGN ERROR, hopefully this one is correct. Note 2: This solves for f(1)=1 only
04.07.2022 19:39
Case $f(1)=1$ is almost trivial but does the above deal with the case $f(1)=-1$ ? Let $M(x,y)$ denote the assertion that $\{f(xf(y)+1),f(yf(x)-1)\}=\{xf(f(y))+1,yf(f(x))-1\}$ ,that the multisets are equal but sometimes I will use just the fact that the sums are equal so let $S(x,y)$ denote $ f(xf(y)+1)+f(yf(x)-1)=xf(f(y))+y(f(f(x))$ $M(0,0)$ gives that $\{f(1),f(-1)\}=\{1,-1\}$ so $f(f(1))=1$ $M(x-1,1)$ gives $\{f(xf(1)+1),f(f(x)-1)\}=\{x,f(f(x))-1\}$ and this implies that $f$ is surjective.
Case $f(1)=1$ For $y\neq 0$ , $M(-\frac{2}{f(y)},y)$ gives $\{-1,f(yf(-\frac{2}{f(y)})-1)\}=\{-\frac{2f(f(y))}{f(y)}+1,yf(f(-\frac{2}{f(y)}))-1\}$ for which in view of the previous claim the only possible case is $-1=-\frac{2f(f(y))}{f(y)}+1\Rightarrow f(f(y))=f(y)$ and due to surjectivity of $f$ we get $f(x)=x$. For the rest we assume $f(1)=-1$ which is the hard case. To begin with $M(x,\frac{1}{f(x)})$ : $\{f(xf(\frac{1}{f(x)})+1),0)=\{xf(f(\frac{1}{f(x)}))+1,\frac{f(f(x))}{f(x)}-1\}$ $M(x,\frac{1}{f(f(x))})$: $\{ f(xf(\frac{1}{f(f(x))})+1),f(\frac{f(x)}{f(f(x))}-1)\}=\{xf(f(\frac{1}{f(f(x))}))+1,0\}$ $M(-\frac{1}{f(x)},x)$: $\{ 0,f(xf(-\frac{1}{f(x)})-1)\}=\{-\frac{f(f(x))}{f(x)}+1,xf(f(-\frac{1}{f(x)}))-1\}$ $M(-\frac{1}{f(f(x))},x)$:$\{f(-\frac{f(x)}{f(f(x))}+1),f(xf(-\frac{1}{f(f(x))})-1)\}=\{0,xf(f(-\frac{1}{f(f(x))}))-1\}$ These give that either (claim 1 is being used) Claim 2:$f(f(x))=f(x)$ or all of these hold: $xf(f(\frac{1}{f(x)}))+1=xf(\frac{1}{f(f(x))})+1=xf(f(-\frac{1}{f(x)}))-1=xf(-\frac{1}{f(f(x))})+1=0$
Injectivity with claim 2 show that either $f(f(x))=f(x)$ or $f(\frac{1}{f(x)})=\frac{1}{f(f(x)}$ and due to surjectivity this can be restated as $f(x)=x$ or $f(\frac{1}{x})=\frac{1}{f(x)}$ . If we suppose that $f(x)=x$ then the previous statement can be applied to $\frac{1}{x}$ to show that $f(\frac{1}{x})=\frac{1}{x}$ so $$f\left( \frac{1}{x}\right) =\frac{1}{f(x)}$$for every $x\neq 0$.
So $f(f(x))=f(x)$ cannot hold unless $x=0$ and claim 2 gives $xf(\frac{1}{f(f(x))})+1=0\Rightarrow $
Now consider $M(-1,x)$: $$\{f(-f(x)+1),f(x-1)\}=\{-f(f(x))+1,-x-1\}$$and $M(f(f(x)),\frac{x}{f(x)})$: (see claim 6) $$\{ f(-f(x)+1),f(-\frac{x^2}{f(x)}-1)\}=\{-\frac{f(f(x))^2}{x}+1,-x-1\}$$If $f(-f(x)+1)$ weren't $-x-1$ then $f(-f(x)+1)=-f(f(x))+1=-\frac{f(f(x))^2}{x}+1\Rightarrow f(f(x))=x$ using $f^3(x)=-x$ you get $f(x)=-x$ then $f(x+1)=f(-f(x)+1)=-x+1$ but $M(-1,x)$ also gives $f(x-1)=-x-1$ so $f(f(x+1))=x+1$ which then gives $f(x+1)=-x-1$ a contradiction. Thus $f(-f(x)+1)=-x-1$ always holds and we are almost done. Put $f(f(x))$ in place of $x$ to get $f(x+1)=-f(f(x))-1$ then $f(x+2)=-f(f(x+1))-1=-f(-f(f(x))-1)-1=f(f(f(x))+1)-1=-f^4(x)-1-1=f(x)-2$ using claim 7 we also get $f(x+1)=f(x)-1$ and then $-f(f(x))=f(x)\Rightarrow f(x)=-x$.