FabrizioFelen wrote: Find all funtions $f:\mathbb R\to\mathbb R$ such that: $$f(xy-1)+f(x)f(y)=2xy-1$$for all $x,y\in \mathbb{R}$. Let $P(x,y)$ be the assertion $f(xy-1)+f(x)f(y)=2xy-1$ If $f(0)\ne 0$, $P(x,0)$ $\implies$ $f(x)=-\frac{f(-1)+1}{f(0)}$ constant, which is never a solution. So $f(0)=0$ $P(0,0)$ $\implies$ $f(-1)=-1$ Let $x\ne 0$ : $P(x,\frac 1x)$ $\implies$ $f(\frac 1x)=\frac 1{f(x)}$ $P(x+1,\frac 1x)$ $\implies$ $(f(x+1)+1)f(\frac 1x)=\frac{x+2}x$ and so $f(x+1)=\frac{x+2}xf(x)-1$ $P(2,1)$ $\implies$ $f(1)(f(2)+1)=3$ and so $f(1)\ne 0$ $P(x+1,1)$ $\implies$ $f(x+1)=\frac{2x+1-f(x)}{f(1)}$ and so (two expressions for $f(x+1)$) : $\frac{x+2}xf(x)-1=\frac{2x+1-f(x)}{f(1)}$ And so assertion $Q(x)$ : $f(x)\frac{x(f(1)+1)+2f(1)}x=2x+1+f(1)$ If $f(1)=-1$, this implies $f(x)=-x^2$ $\forall x\ne 0$, still true when $x=0$ And so $\boxed{\text{S1 : }f(x)=-x^2\text{ }\forall x}$ which indeed is a solution If $f(1)\ne -1$, $Q(\frac{-2f(1)}{f(1)+1})$ $\implies$ $f(1)=1$ And $Q(x)$ becomes $f(x)=x$ $\forall x\notin\{0,-1\}$ We already know that $f(0)=0$ and $f(-1)=-1$ And so $\boxed{\text{S2 : }f(x)=x\text{ }\forall x}$ which indeed is a solution

$f(0) = 0$ Plug in $x = 0 $ so $f(-1) = -1$ Plug in $x-y-1$ so $f^2(1) = 1$. Here we have $2$ cases $\blacksquare $ If $f(1) = 1$. Plug in $y=1$ so $f(x-1)+f(x) = 2x-1$. Then $f(xy-1) = 2xy-1 - f(xy)$ $(1)$ Then $f(x)f(y) - f(xy) = 0$ So we have : $f(x)f(y) = f(xy)$. But we also have : $f(x)+f(x+1) = 2x+1$ Plug in $ x= y$ to $(1)$ then $f(x-1)f(x+1) +f^2(x) = 2x^2 - 1$ $\implies (2x-1-f(x))(2x+1-f(x))+f^2(x) = 2x^2-1$ $\implies (f(x)-x)^2 = 0$ so $f(x) = x$ $\blacksquare $ If $f(1) = -1$ . Plug in $y=1$ so $f(x-1)-f(x) = 2x-1$. Do similary with case $1$ Then $f(x)f(y) = -f(xy)$ And : $-f(x-1)f(x+1) +f^2(x) = 2x^2-1$ Do similary with case $1$ we have $4x^2-1+2f(x) = 2x^2-1$ or $f(x) = -x^2$

Nice solution PCO.This is also Azerbaijan BMO 2016 TST.

$(0,y)$ : $f(-1)+f(0)f(y)=-1$ $f(0)\ne0 \implies$ $f(x)=-\frac{f(-1)+1}{f(0)}$ which means function $f(x)$ is constant function. (contradiction) So, $f(0)=0, f(-1)=-1$ $(x,1/x)$ $\implies$ $f(x)f(1/x)=1$ $(x,-1)$ $\implies$ $f(-x-1)-f(x)=-2x-1$ $....(*)$ and one more substitution here, ($x$ $\implies$ $-xy$) We get $f(xy-1)-f(-xy)=2xy-1=f(xy-1)+f(x)f(y)$ (from original condition) then, $-f(-xy)=f(x)f(y)$ $....(**)$ Put,$(1,1)$ : $f^2(1)=1$ yields two case CASE1 $f(1)=1$ $y\implies -\frac{1}{x}$ at $(**)$ and multiply $f(1/x)$ Then, $-f(\frac{1}{x})=f(-\frac{1}{x})$ which means $-f(x)=f(-x)$ Apply this to $(*)$ : $f(x+1)+f(x)=2x+1$ from $f(0)=0, f(1)=1$ we easily can succeed induction $f(k)=k \in\mathbb N$ and odd function $f$, So, $f(k)=k \in\mathbb Z$ Expansion to $\mathbb R$ By comparing two substitutions, $(x,y+\frac{t}{x})$, $(x+\frac{t}{y},y)$ We get $f(x)f(y+\frac{t}{x})=f(y)f(x+\frac{t}{y})$ ....(A) Then, there exist $x\notin\mathbb Z$, $y\notin\mathbb Z$ but $t\in\mathbb R\implies x+\frac{t}{y}, y+\frac{t}{x}\in\mathbb Z$ Such $t$ must be exist by checking, $t=xy\alpha,\alpha\in\mathbb R \quad x=\frac{x'}{1+\alpha},y=\frac{y'}{1+\alpha},\quad x', y'\in\mathbb Z$ Not to be too complicative check this : $ x+\frac{t}{y}=x(1+\alpha),y+\frac{t}{x}=y(1+\alpha)$ , $\quad (t,x,y)$ is derived this idea. So, $f(x+\frac{t}{y})=x(1+\alpha) \quad and \quad f(y+\frac{t}{x})=y(1+\alpha)$ $\quad where (x,y,t,\alpha) are on R$ Put these to (A), then, $f(x)y(1+\alpha)=f(y)x(1+\alpha)$ $\implies$ $f(x)y=f(y)x$ ($x,y$ themselves are on $\mathbb R$) Now, independently take $y$ from $\mathbb Z$ and divide both side by $y$. Finally, we get $f(x)=x\in\mathbb R$ CASE2 $f(1)=-1$ Same way, induction, expansion $f(x)=-x^2$ Complement : $f(\alpha)=0$ , such $\alpha$ is only a value 0. If there is exception, contradiction comes from putting that value to $f(x)f(\frac{1}{x})=1$ So, we can freely divide both side except 0.

Pluto1708 wrote:

This was a partial solution given by Pluto1708 in another thread, could someone please continue on this idea? I am having trouble solving the quadratic (and then finding $f(1)$)

Answer: $f(x)=x$ and $f(x)=-x^2$ for all $x\in \mathbb{R}.$ Proof: It's easy to see that these are indeed solutions. Let $P(x,y)$ denote the given assertion, we have \[P(0,x): f(-1)+f(0)f(x)=-1.\]If $f(0)\ne 0$, then $f\equiv \textrm{const.}$ which obviously isn't a solution to our FE. Therefore, $f(0)=0$ and so $f(-1)=-1$ and \[P(1,1): f(1)^2=1 \implies f(1)=\pm 1.\]Consider two cases: Case 1: $f(1)=-1$. We know $P\left(x, \frac{1}{x}\right)\implies f(x)f\left(\frac{1}{x}\right)=1$ for all $x\ne 0$. So, when $x\ne 0$, \[P\left(x+1,\frac{1}{x}\right): f(x+1)=\frac{x+2}{x}f(x)-1.\]Then, as $f(x)=\frac{x+1}{x-1}f(x-1)-1\implies f(x-1)=\frac{x-1}{x+1}\left(f(x)+1\right)$ when $x\ne 1,$ \[P(x,1): \frac{x-1}{x+1}\left(f(x)+1\right)-f(x)=2x-1 \implies f(x)=-x^2\]together with fact that $f(1)=-1,$ $\boxed{f(x)\equiv -x^2}.$ Case 2: $f(1)=1$. First, we know $P\left(x, \frac{1}{x}\right)\implies f(x)f\left(\frac{1}{x}\right)=1$ for all $x\ne 0$. Then, \[P(x+1,1): f(x)+f(x+1)=2x+1\]and \[P(x+2,1): f(x+1)+f(x+2)=2x+3\]imply $f(x+2)=f(x)+2$. From this, we find that $f(2)=f(0)+2=2$ and $f\left(\frac{1}{2}\right)=\frac{1}{2}.$ Also, comparing \[P(x,1): f(x-1)+f(x)=2x-1\]and \[P\left(2x,\frac{1}{2}\right): f(x-1)+\frac{1}{2}f(2x)=2x-1\]we get $f(2x)=2f(x)$. Thus, \begin{align*} P\left(x+2,\frac{1}{2}\right)&: f\left(\frac{x}{2}\right)+\frac{1}{2}f(x+2)=x+1 \\ & \implies \frac{1}{2}f(x)+\frac{1}{2}(f(x)+2)=x+1 \\ & \implies \boxed{f(x)\equiv x}. \end{align*}

Posting for storage purposes

f(xy-1)+f(x)f(y)=2xy-1 First look P(0,0); x=y=0 f(-1)+f(0)×f(0)=-1 Assume f(0)=0, and prove it And also if f(0)=0, f(-1)=-1 Let's look P(x,0) f(-1)+f(x)*f(0)=-1 In that case, find f(x); f(x)=-(1+f(-1))/f(0) In that time f(x) must be constant, but that's not correct. So, f(0)=0, we proved. Look P(x, 1/x); f(0)+f(x)*f(1/x)=1 Use this claim; f(0)=0 We'll get f(x)*f(1/x)=1 f²(1)=1 We have two cases: i) f(1)=1 For this f(x)=x. For all real number x ii) f(1)= -1 f(x)= -x² For all real number x

We claim that the two solutions are $f(x) \equiv x$ and $f(x) \equiv -x^2$. Plugging these in, they work. We will show that no other solutions exist. Plug in $(x, 0)$ to get \[f(-1) + f(x)f(0) = -1.\]Suppose that $f(0) \ne 0$. Then, $f$ must be a constant function, which is a contradiction. Thus, $f(0) = 0$. Plug in $(0, 0)$ to get \[f(-1) + f(0)^2 = -1 \implies f(-1) = -1.\]Plug in $(1, 1)$ to get \[f(0) + f(1)^2 = 1 \implies f(1) = \pm 1.\]Plug in $\bigl(x, \frac{1}{x}\bigr)$ to get \[f(x)f\bigl(\tfrac{1}{x}\bigr) = 1. \qquad (1)\]Notice that $(x + 1) \cdot \frac{1}{x} - 1 = \frac{1}{x}$. Thus, plug in $\bigl(x + 1, \frac{1}{x}\bigr)$ to get \[f\bigl(\tfrac{1}{x}\bigr) + f(x + 1)f\bigl(\tfrac{1}{x}\bigr) = 1 + \tfrac{2}{x}.\]Multiplying both sides by $xf(x)$ and substituting $(1)$ gets us \[(x + 2)f(x) - xf(x + 1) = x.\]Plug in $(x + 1, 1)$ to get \[f(x) + f(x + 1)f(1) = 2x + 1.\]Since $f(1) = \pm 1$, we have two cases. We have two equations and two variables ($f(x)$ and $f(x + 1)$). Solving, we get that \[f(x) = x\text { or }f(x) = -x^2,\]as desired.

It's easy to see that $f(0)=0,f(-1)=-1,f(1)=1,-1$ $P(x,\frac{1}{x}) \implies f(\frac{1}{x}) = \frac{1}{f(x)}$ $$P(x+1,\frac{1}{x}) \implies 1+f(x+1)=f(x)[\frac{2}{x} +1]...(1)$$ Then, $$P(x+1,1) \implies f(x) + f(x+1)f(1)=2x+1$$ If $f(1)=1$ then $f(x+1)=2x+1-f(x)$ Plugging in $(1)$ gives $\boxed{f(x)=x}$ If $f(1)=-1$ then $f(x+1)=f(x)-2x-1$ Plugging in $(1)$ gives $\boxed{f(x)=-x^2}$