For $x=y=1$ we have $f(f(1)^2-1)=0$ let $a=f(1)^2-1$ then: $P(a,y)$ gives :$f(-1)=ay-1$ and because $y=free$ we have $a=0$ which means $f(1)=+-1$ If $f(1)=-1$ then $P(1,1)$ gives:$f(0)=0$ $P(0,0)$gives $f(-1)=-1$ $P(1,-1)$ gives $f(0)=-2$ contradiction If $f(1)=1$ then[ $P(1,y+1)$ gives $f(y)=y$ which is a solution.]=wrong

Let $P(x,y): f(f(x)f(y)-1)=xy-1$ $P(x,1): f(f(x)f(1)-1)=x-1$ so $f$ is bijective. It is clear that $f$ isn't constant. $P(x,0): f(f(x)f(0)-1)=-1$ so $f(x)f(0)=constant \Rightarrow f(0)=0$ $P(1,1): f(f(1)^2-1)=0\Leftrightarrow f(1)=\pm 1$ Let $f(1)=-1$ $P(x,1): f(-f(x)-1)=x-1$ so $f(-f(-xy)-1)=xy-1$ and since $f$ is $1-1$ we get that $-f(xy)=f(x)f(y)$ $-f(1)=f(-1)f(-1)\Leftrightarrow f(-1)=1$ $P(x,-1): f(f(x)-1)=-x-1$ and $-f(-x)=f(x)f(-1)\Leftrightarrow f(-x)-f(x)$ $P(-x,-1): f(f(-x)-1)=x-1\Leftrightarrow f(f(x)+1)=1-x=-(x-2)-1=f(f(x-2)-1)$ so $f(x-2)=f(x)+2$ $f(x)f(y-2)=-f(x(y-2)\Leftrightarrow...\Leftrightarrow f(x+y)=f(x)+f(y)$ Also $f(x^2)=-f(x)^2$ so we have $f(x)=Cx$....and we get a contradiction in $P(x,y)$ Therefore we must have that $f(1)=1$ $P(xy,1) :f(f(xy)+1)=xy+1=f(f(x)f(y)+1$ so $f(x)f(y)=f(xy)\Rightarrow f(x)\geq 0,\forall x>0$ Also $f(-1)f(-1)=f(1)=1$ so $f(-1)=-1$ (since $f(-1)\neq 1=f(1)$) So $f(x)f(-1)=f(-x)\Leftrightarrow f(-x)=-f(x)$ Also $f(f(x)+1)=x-1,\forall x\in R$ $P(-x,1): f(f(-x)-1)=-x-1\Leftrightarrow f(f(x)+1)=x+1=(x+2)-1=f(f(x+2)-1)$ so $f(x)+1=f(x+2)-1\Leftrightarrow f(x+2)=f(x)+2\Rightarrow f(2)=2$ $f(x)f(2)=f(2x)\Leftrightarrow f(2x)=2f(x)$ $f(x)f(y+2)=f(x(y+2))\Leftrightarrow f(x)f(y)+2f(x)=f(xy+2x)$ $\Leftrightarrow f(xy)+f(2x)=f(xy+2x)\Leftrightarrow f(x)+f(y)=f(x+y)$ Since $f(x)>0,\forall x>0$ we have that $f(x)=f(1)x=x$ So $f(x)=x$ the only solution. Edit: i have a small mistake in the case $f(1)=-1$, I'm trying to fix it Edit 2: I think that now is correct

P2nisic wrote: If $f(1)=1$ then $P(1,y+1)$ gives $f(y)=y$ which is a solution. I don't understand this step

bsf714 wrote: Find all functions $f:\mathbb R\to\mathbb R$ so that the following relation holds for all $x, y\in\mathbb R$. $$f(f(x)f(y)-1) = xy - 1$$ My solution in the contest. It is clear that $f(x)$ is surjective so take $a$ such that $f(a)=0$. $P(x,a):f(-1)=ax-1$ $P(0,a):f(-1)=-1$ So $f(0)=0$ and $f(-1)=-1$ From surjectivity take a $b$ such that $f(b)=1$. $P(b,b):b^2=1\rightarrow f(1)=1$ or $f(-1)=1$. But since $f(-1)=-1$ we have that only $f(1)=1$ counts. $P(x,1): f(f(x)-1)=x-1$. From this it is easy to get that $f(x)$ is injective. From $P(f(x)-1,1)$, $f(x-2)=f(x)-2$ $f(x+2)=f(x)+2...............(1)$ Using $P(x,1)$ in $P(x,y)$ and injectivity we have, $f(f(x)f(y)-1)=xy-1=f(f(xy)-1)$ $f(x)f(y)=f(xy)...............(2)$ So the function is multyplicative. From $(1)$ we have $f(2)=2$ and using this in $(2)$ with $y=2$ we get that $f(2x)=2f(x)$. Placing $y\rightarrow y+2$ and using $(1)$ and $(2)$ we get, $f(xy)+2f(x)=f(xy+2x)$ $f(x)+f(y)=f(x+y)$ Hence the function is additive and since it is multyplicative as well we get $f(x)=x$ as the only solution.

bsf714 wrote: Find all functions $f:\mathbb R\to\mathbb R$ so that the following relation holds for all $x, y\in\mathbb R$. $$f(f(x)f(y)-1) = xy - 1$$ $P(x,1)\implies f(f(x)f(1)-1)=x-1\implies$ bijective. Let $r$ be the root. Then, if $r\neq 0$, $P(x,r)\implies \frac{f(-1)+1}{r}=f(x)\implies f$ is constant, which fails. So, $f(0)=0$ and $P(x,0)\implies f(-1)=-1$ Claim: $f$ is odd. proof: Let $a,b\neq 0$. Then, $f(a),f(b),f(-a),f(-b)\neq 0$. Comparing $P(a,b)$ with $P(-a,-b)$ together with injectivity gives $$f(a)f(b)=f(-a)f(-b)$$And comparing $P(-a,b)$ with $P(a,-b)$ together with injectivity gives $$f(-a)f(b)=f(a)f(-b)$$So, $$\frac{f(a)}{f(-a)}=\frac{f(-b)}{f(b)}=\frac{f(b)}{f(-b)}$$Hence, $f(b)=f(-b)$ or $-f(b)=f(-b)\implies$. by $f(f(x)f(1)-1)=x-1, f(b)=f(-b)\implies b=0$. and so, $-f(b)=f(-b)$ and $f$ is odd $\blacksquare$ So, $f(1)=f(-(-1))=-f(-1)=-(-1)=1$ and $P(x,1)$ is actually $$f(f(x)-1)=x-1$$Also $P(-x,1)\implies f(f(-x)-1)=-x-1\implies f(-f(x)-1)=-x-1$ or $$f(f(x)+1)=x+1$$Comparing $f(f(x)-1)=x-1$ with $f(f(x)+1)=x+1$, we deduce $f(x+2)=f(x)+2$. Claim: $f$ is completely multiplicative. proof: $f(f(x)f(y)-1) = xy - 1=f(f(xy)-1)\implies f(x)f(y)=f(xy)$ $\blacksquare$ Claim: $f$ is additive. proof: $f(1)=1=2-1=f(f(2)-1)\implies f(2)=2$. So, by multiplicativity, $f(2x)=2f(x)$ $f(x)f(y+2)=f(x(y+2))\implies 2f(x)+f(y)f(x)=f(xy+2x)\implies f(2x)+f(xy)=f(xy+2x)$ $\blacksquare$. Since $f(x^2)=f(x)^2$, $f(x)\geq 0$ for $x\geq 0$, which together with additivity implies that $f(x)=xf(1)$ and plugging will give $f(x)=x, \forall x\in \mathbb{R}$, which works.

Let $P(x,y) : f(f(x)f(y)-1)=xy-1$. Step 1. $f$ is bijective. It is clear that $f$ is surjective. If $f(a)=f(b)$ Compare $P(x,a)$ and $P(x,b) \rightarrow xa-1=f(f(x)f(a)-1)=f(f(x)f(b)-1)=xb-1$ for every real number $x$, which means $a=b$. Therefore, $f$ is injective. Step 2. $f$ is multiplicative. Compare $P(x,y)$ and $P(xy,1) \rightarrow f(f(x)f(y)-1)=xy-1=f(f(xy)f(1)-1)$. $f$ is injective, so $f(x)f(y)=f(xy)f(1) \cdots (*)$. Put $y=0$ in $(*)$, and we get $f(x)f(0)=f(0)f(1)$ for all real number $x$. If $f(0) \neq 0$, then $\forall_x , f(x)=f(1)$, which can't be true. $\therefore f(0)=0$. $P(0,0) \rightarrow f(-1)=-1$. Put $x=y=-1$ in $(*)$, then $f(-1)^2=f(1)^2$, then $f(1)=1$ because f is injective. Therefore, $f(xy)=f(x)f(y)$. Step 3. f is additive. $P(t,1) : f(f(t)-1)=t-1$ $P(f(t)-1,1) : f(f(f(t)-1)-1)=f(t)-1-1 \Leftrightarrow f(t-2)=f(t)-2$ $f(0)=0 \rightarrow f(2)=2$ Then, $f(xy+2x)=f(x(y+2))=f(x)f(y+2)=f(x)(f(y)+2)=f(x)f(y)+2f(x)=f(xy)+f(2x)$ Thus, $\forall_{a, b} , f(a+b)=f(a)+f(b)$ Step 4. $\forall_x , f(x)=x$ $a \geq 0 \rightarrow f(a)=f(\sqrt{a})^2 \geq 0$ $\therefore , \forall_x , f(x)=kx$ $P(x,y) \rightarrow k=1 \rightarrow f(x)=x$