similar: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=536873&p=3084849#p3084849

Different solution: Plugging in $x=0$ and $x=1$ respectively yields $f(1)=0$, $f(0)=-1$. Plugging in $y=-1$ yields $f(1-x)-f(x-1)=f(-1)f(x)$, which (after replacing $x$ with $1-x$) gives $f(1-x)+f(1+x)=0$. Hence $f(2)=1$. Plugging in $y=2$ into the original equation gives $f(1+2x)=f(2+x)+f(x)=f(x)-f(-x)=-f(-1)f(1+x)$. Replacing $y$ by $-y$ and adding the given and the resulting equation yields $f(x)(f(y)+f(-y))+f(x+y)+f(x-y)=0$ $(*)$. Setting $x=y+1$ and using the above gives $(f(y)+f(-y)-f(-1))f(1+y)=0$. Hence for all $x$ we either have $f(1+x)=0$ or $f(x)+f(-x)=f(-1)$. Suppose that $f(x)=0$ for some $x$ different from $1$. $(*)$ then gives $f(x+y)+f(x-y)=0$. For $y=x-1$ we get $f(2x-1)=0$ and iterating this we see that $f(1+k(x-1))=0$ for all $k\in\mathbb{Z}$. Due to $f(x+y)+f(x-y)=0$ whenever $f(x)=0$, we see that $f$ is periodic. However, $f$ cannot be periodic: Otherwise, say $f$ has period $d$. Let $y=d$. Then $f(1+xy)-f(x)=f(x)f(0)=-f(x)$ and hence $f(x)=0$ for all $x$, contradiction. Hence $f(x)+f(-x)=f(-1)$ for all $x$ (since the equation trivially holds for $x=1$). $(*)$ the becomes $f(x+y)+f(x-y)=-f(-1)f(x)$. Letting $y=0$ we see that $f(-1)=-2$ and hence $f(x+y)+f(x-y)=2f(x)$ for all real $x$ and $y$. This is easily seen to be equivalent to $f(x)+f(y)+1=f(x+y)$. Using this in the original equation yields $f(x)f(y)=f(xy)-f(x)-f(y)$ or $(f(x)+1)(f(y)+1)=f(xy)+1$. Let $g(x)=f(x)+1$. Then $g(x)+g(y)=g(x+y)$ and $g(x)g(y)=g(xy)$, whence $g(x)\ge0$ for $x\ge 0$, i.e. by a well-known result $g(x)=x$ and so $f(x)=x-1$.

My solution on this problem was 1. Get $f(x)+f(x+1)+f(x+2)=0$ or $f(x)-2f(x+1)+f(x+2)=0$, where the former will give a contradiction later. 2. put y, y+1, y+2 in the place of y. 3. Add or subtract them to use 1 and get rid of the product form on the right hand side. Notice that the second term of lhs also disappears. Now it's just a classical Cauchy-functional equation!

Well, I was asked to give some details. For step 1, you may follow Bugi's solution until where he finds the values of $c$. If $c=1$, $f(x)+f(x+1)+f(x+2)=0$, and if $c=-2$, $f(x)-2f(x+1)+f(x+2)=0$ Now $f(1+xy) - f(x+y) = f(x)f(y) $ $f(1+x(y+1)) - f(x+y+1) = f(x)f(y+1) $ $f(1+x(y+2)) - f(x+y+2) = f(x)f(y+2) $. If $c=1$, add them all and delete many terms to get something like $f(a) + f(\frac{a+b}{2}) + f(b) = 0$, thus $f$ is identically null in the rationals which is a contradiction. If $c=-2$, similarly you can get something like $f(a) - 2f(\frac{a+b}{2}) + f(b) = 0$ which is just Jensen's functional equation. Easy from here.

lyukhson wrote: Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfy the conditions \[f(1+xy)-f(x+y)=f(x)f(y) \quad \text{for all } x,y \in \mathbb{R},\] and $f(-1) \neq 0$. Let $P(x,y):f(1+xy)-f(x+y)=f(x)f(y) $. BY hands, it is easy to get $f(0)=-1,f(1)=0$. So, the guess is $f(x)=x-1\forall x\in\mathbb R$. Therefore, it is convenient to take $g(x)=f(x)+1$ and then show that $g(x)=x$ $\quad\forall x\in\mathbb R$(At first, I tried to show $g(x)=x-f(x)$ is constant using the identity $xy+1-(x+y)=(x-1)(y-1)=(1-x)(1-y)$ but no luck. Later I tried to find Cauchy from this). We already have $g(0)=0,g(1)=1$. Now, the equation becomes \[Q(x,y):g(xy+1)-g(x+y)=(g(x)-1)(g(y)-1)\] Let $a=g(-1)-1$. Then \[Q(x,-1):g(1-x)-g(x-1)=a(g(x)-1)\] Note that, we also have $g(2)=2$ after $Q(2,-1)$. \[Q(1-x,-1):g(x)-g(-x)=c(g(1-x)-1)\quad(*)\] Setting $x\to-x$ in $(*)$, $g(-x)-g(x)=c(g(1+x)-1)$. Therefore, adding them, \[g(1+x)+g(1-x)=2,g(x)+g(2-x)=2\quad\forall x\in\mathbb R\quad(\dagger)\] \[x+y=1,Q(x,y):g(xy+1)-g(1)=(g(x)-1)(g(y)-1)\] \[x+y=1,Q(2-x,2-y):g(xy+3)-g(3)=(g(2-x)-1)(g(2-y)-1)\] But $(g(2-x)-1)(g(2-y)-1)=(1-g(2-x))(1-g(2-x))$ and $g(x)-1=1-g(2-x)$. Therefore, \[g(xy+3_-g(xy+1)=g(3)-g(1)\] This suggests $xy+1\to t$ with $x+y=1$. Then we have $g(t+2)-g(t)=2$ when $t=x(1-x)+1$ or $x^2-x+(t-1)=0$ has real solutions. Taking discriminant gives $1-4(t-1)\geq0,t\leq\dfrac54$. Now, setting $t=1,g(3)=3$. To invoke this, just note $x>\dfrac54\implies -x<\dfrac54$. So putting this together with $(\dagger)$, we get \[g(x+2)-g(x)=2\quad\forall x\in\mathbb R\quad(\ddagger)\] $x\to-x$ in $(\dagger)$ infers $g(-x)+g(2+x)=2$ or $g(x)=-g(x)$ from $(\ddagger)$. \[Q(x,-y):g(1-xy)-g(x-y)=-(g(x)-1)(g(y)+1)\] \[Q(-x,y):g(1-xy)-g(-x+y)=-(g(x)+1)(g(y)-1)\] Thus, $g(1-xy)=1-g(x)g(y)$ and $g(1+xy)=1+g(x)g(y)$. So, from the assertion $Q$, $g(x+y)=g(x)+g(y)$ is additive, and also $g(xy+1)=g(1)+g(xy)=1+g(x)g(y)$ says $g$ is multiplicative. This is Cauchy! And $g(x)=g(1)x=x$.

PhelixD Due to $f(x+y)+f(x-y)=0$ whenever $f(x)=0$, we see that $f$ is periodic. I don't understand. Can you explain this?

Well, what if we relax the restriction $f(-1) \not= 0$?

Johann Peter Dirichlet wrote: Well, what if we relax the restriction $f(-1) \not= 0$? $f(x) = x^2 - 1$ becomes another solution. Maybe more. Haven't thought much yet.

Let $h(x) = f(x+1)$, so $a = h(-2) \neq 0$ and \[h(xy) = h(x-1)h(y-1) + h(x+y-1).\] Taking $y = -1$ gives $h(-x) = a h(x-1) + h(x-2)$, and replacing $x$ by $2-x$ gives $a \cdot \left( h(x-1) + h(1-x)\right) = 0$, i.e. $h$ is odd. Changing the signs of $x,y$ then gives \[h(xy) = h(x-1)h(y-1) + h(x+y-1) = h(x+1)h(y+1) - h(x+y+1).\] But $h(x) + ah(x-1) + h(x-2) = 0$, i.e. $h(x+1) = -ah(x) - h(x-1)$ for all $x$, so substituting above and simplifying gives $h(xy) = h(x)h(y)$ for all $x,y \in \mathbb{R}.$ Taking $y=0$ in the original equation easily gives $h(-1) = -1$, so $h(1) = 1$. Taking $y = 2-x$ gives $h(x(2-x)) = h(x(2-x) - 1) + 1$, i.e. $h(x+1) = h(x) + 1$ for all $x \le 0$. From $h(xy) = h(x+1)h(y+1) - h(x+y+1)$ we then get $h(x+y) = h(x) + h(y)$ for all $x, y \le 0$. But since $h$ is multiplicative and odd, $x$ and $h(x)$ have the same sign, so $h$ is bounded and additive on the negative reals $\implies h(x) = cx$ for all negative $x$ for some $c \in \mathbb{R}$, so by oddity $h(x) = x$ for all $x \in \mathbb{R}$, since $x = -1$ implies $c = 1$. Therefore $f(x) = h(x-1) = x-1$ for all $x \in \mathbb{R}$.

lyukhson wrote: Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfy the conditions \[f(1+xy)-f(x+y)=f(x)f(y) \quad \text{for all } x,y \in \mathbb{R},\] and $f(-1) \neq 0$. Solution on my face (In Vietnamese, sorry, but I think it is easy to understand): https://www.facebook.com/photo.php?fbid=10202582174572489&set=pcb.10202582182572689&type=1&theater

Excuse me, can anyone explain why we get additive and multiplicity then we can get g(x)=g(1)x for all x.

@YaWNeeT http://www.artofproblemsolving.com/Forum/viewtopic.php?f=39&t=469012

Is there a way to find all solutions to the equation given $f(-1) = 0$ is permitted?

Assuming I have made no error, here is a solution to the complete functional equation without the $f(-1) \not= 0$ stipulation. As always, let \[f(1 + xy) - f(x + y) = f(x)f(y)\]be denoted by $P(x, y)$.

IMO ShortList 2012 A5 wrote: Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfy the conditions \[f(1+xy)-f(x+y)=f(x)f(y) \quad \text{for all } x,y \in \mathbb{R},\]and $f(-1) \neq 0$. In the same spirit as 2005 A4

EDIT: Fixed now. Thanks @ankoganit, I was quite sleepy last night typing it, oops.

anantmudgal09 wrote: Case 2. $f(-1)=-2$ .... we get that $f$ satisfies Jensen's functional equation and $f$ is linear... Don't you need something like continuity/monotonicity to conclude $f$ is linear from Jensen? Edit: The above post has been fixed.

I have found a possibly different idea for this problem. As earlier, we deal with the case $f(-1)=1$; so now take $f(-1)=-2$. Notice that $$f(x)-f(-x)=2f(x+1),$$for all $x \in \mathbb{R}$ since $f(x+2)=-f(-x)$. Add $P(x, y)$ and $P(x, -y)$ to see that $$f(x+y)+f(x-y)=-f(x) \cdot \left(f(y)+f(-y)\right)$$holds for all $x, y \in \mathbb{R}$. First, we consider Lemma: $f(x_0)=0 \implies x_0=1$. (Proof) Put $x \mapsto x_0$ in previous identity; then $f(a)=-f(b)$ whenever $a+b=2x_0$. Consequently, we have $f(2-a)=f(b)$ so $f$ is periodic with period $\ell=2(1-x_0)$. Fix $z=(1+xy)$ and compare $P(x, y)$ with $P(x, y+\ell)$; we get $f(z)=f(z+x\ell)$ for all $x, z \in \mathbb{R}$. As $f$ is non-constant, $\ell=0$, proving the lemma. $\blacksquare$ Next, we consider Lemma: $f(y)+f(-y)=-2$ for all $y \in \mathbb{R}$. (Proof) For $y=0$, it's clear so say $y \ne 0$. Hence $f(y+1) \ne 0$. Plug $x \mapsto y+1$ to conclude that $$\frac{-f(2y+1)}{f(y+1)}=f(y)+f(-y),$$for all $y \in \mathbb{R}$. Note that $$P(2, y) \implies f(2y+1)=f(y)-f(-y),$$and combine with $$f(y)-f(-y)=2f(y+1)$$to see that $f(y)+f(-y)=-2$; just as we claimed. $\blacksquare$ Hence, $f(x+y)+f(x-y)=2f(x)$; so the function $g=(1+f)$ is additive. Apply additivity on $P(x, y)$ to see that $g$ is also multiplicative; hence $g$ is the identity function. The conclusion $\boxed{f(x)=x-1}$ for all $x \in \mathbb{R}$ holds, just as we wished. $\blacksquare$ Comment: This solution is more natural to arrive than the earlier version. The crucial $y \mapsto -y$ is motivated by the fact that we already know $f(1+t)=-f(1-t)$ and there is this glaring $f(1+xy)$ term involved. If only we could change it a little bit, to fit in that form, we may arrive at some good result. It may also be noted that everything else that follows is quite straight forward, except maybe forcing the $\ell=0$ part.

Only $x\mapsto x-1$ works. It's very easy to check that this works. Let $P(x,y)$ denote the problem's equation. We begin with some observations. $P(-1,1)$ gives $f(1)=0$. $P(-1,0)$ gives $f(0)=-1$. $P(1-a,-1)$ gives $f(a)-f(-a)=f(1-a)f(-1)$. Replacing $a\mapsto -a$ and adding to it gives $f(1-a)+f(1+a)=0$ or $\boxed{f(2-a)=-f(a)}$ for all $a$. Plug in $a=0$ in above gives $f(2)=1$. Part I: Solving the equation in $\mathbb{Z}$. From $P(x,-1)$ and the above observation, we get $$-f(x+1)-f(x-1) = f(-1)f(x)$$or $f(x+1)=tf(x)-f(x-1)$ when $t=f(3)=-f(-1)$. This means \begin{align*} f(4) &= tf(3)-f(2) = t^2-1 \\ f(5) &= tf(4)-f(3) = t^3-2t \end{align*}However, $P(2,2)$ gives $f(5)-f(4)=f(2)^2=1$. Solving for $t$ gives $t=2,-1$. If $t=2$, then $\boxed{f(x+1)+f(x-1)=2f(x)}$ for all $x\in\mathbb{R}$. A simple induction gives $f(x)=x-1$ for all $x\in\mathbb{Z}$. Otherwise, we rule out $t=-1$. We have \begin{align*} f(x)+f(x-1)+f(x-2) &= 0 \\ f(x-1)+f(x-2)+f(x-3) &= 0. \end{align*}Subtracting yields $f(x)=f(x-3)$. Hence using $P(x,3)$ gives $f(3x+1)=0$ for all $x$, which is a contradiction. Part II: Solving over $\mathbb{R}$ The key idea is the following manipulation. Claim: [Main step] $f$ satisfies Jensen's Equation $f(x)+f(y)=2f\left(\tfrac{x+y}{2}\right)$ Proof: For any $a,b\in\mathbb{R}$, we have $$\begin{array}{rlrlc} P(a,b) &\implies & f(1+ab) - f(a+b) &= f(a)f(b) &\quad (1)\\[4pt] P(a,b-1)&\implies & f(1+a(b-1)) - f(a+b-1) &= f(a)f(b-1) &\quad (2)\\[4pt] P(a,b-2)&\implies & f(1+a(b-2)) - f(a+b-2) &= f(a)f(b-2) &\quad (3) \end{array}$$Keeping in mind that $f(x+1)+f(x-1)=2f(x)$, we add $(1)$ and $(3)$ and subtract with two times $(2)$. This gives $$f(1+ab) + f(1+ab-2a) = 2f(1+ab-a).$$Now pick a suitable $b$ such that $1+ab=x$. This gives $f(x)+f(x-2a)=2f(x-a)$ for any $a\ne 0$. This equation is trivially true when $a=0$ hence we are done. $\blacksquare$ Let $g(x)=f(x)+1$. Then $g$ is an additive function. To finish, we need a weak bound on $f$. Claim: $f(a)\leq 1$ for every $a\leq 2$. Proof: $P(x,2-x)$ gives $f(1+2x-x^2) = 1+f(x)f(2-x)=1-f(x)^2$, which is equivalent to the desired result. $\blacksquare$ Therefore $g(a)\leq 2$ for any $a\leq 2$. To finish, recall that any non-linear additive function must have a dense graph on $\mathbb{R}^2$ i.e. any disk contains point in form $(x,g(x))$ in it. This easily gives the result.

Let $P(x,y)$ denote the given FE. We claim the only solution is $f(x)=x-1$. Clearly $f$ cannot be constant, as that would imply $f\equiv 0$, and thus $f(-1)=0$. Now suppose $f$ is a solution. We see that \[P(x,0)\implies f(1)-f(x) = f(x)f(0)\implies f(x)[f(0)+1]=f(1),\]and since $f$ is not constant, we must have $\boxed{f(0)=-1}$ and ${f(1)=0}$. Now, \[P(x,-1)\implies f(1-x) - f(x-1) = f(x)f(-1)\implies\boxed{f(u)-f(-u)=f(1-u)f(-1)}.\]Thus, \[f(1-u)f(-1) = -f(1+u)f(-1),\]so since $f(-1)\ne 0$, we have that $f(1-u)=-f(1+u)$, or $\boxed{f(u)=-f(2-u)}$. Now, using $f(1-x^2)+f(1+x^2)=0$, we see that \[P(x,x)+P(x,-x)\implies \boxed{-f(2x)+1 = f(x)[f(x)+f(-x)] = f(x)[f(x)-f(x+2)]}.\]We also have \[P(x,2)\implies \boxed{f(2x+1) = f(x)+f(x+2)}.\]Now, let $a=f(-1)$. We will now compute $f(13)$ in two ways and $f(17)$ in two ways to narrow down the possibilities for $a$. We have \begin{align*} f(0) &= -1 \\ f(1) &= 0 \\ f(2) &= -f(0) = 1 \\ f(3) &= -f(-1) = -a \\ f(4) &= -f(-2) = f(-1)[f(1)+f(-1)] - 1 = a^2-1\\ f(5) &= f(2)+f(4) = a^2 \\ f(6) &= 1-f(3)[f(3)-f(5)] = 1+a[-a-a^2] = 1-a^2-a^3\\ f(7) &= f(3)+f(5) = a^2-a \\ f(8) &= 1-f(4)[f(4)-f(6)] = 1-(a^2-1)[2a^2+a^3-2] = -a^5-2a^4+a^3+4a^2-1 \\ f(9) &= f(4)+f(6) = -a^3 \\ f(10) &= 1-f(5)[f(5)-f(7)] = 1-a^2[a]=1-a^3. \end{align*}Now, \[P(3,4)\implies f(13) = f(7)+f(3)f(4) = a^2-a + a-a^3 = a^2-a^3,\]and \[f(13) = f(6)+f(8) = -a^5 - 2a^4+3a^2,\]so \[a^5+2a^4-a^3-2a^2=0\implies a^2(a+2)(a^2-1)=0\implies a\in\{-2,1,-1\}.\]Now, \[f(17) = f(8)+f(10) = f(8)+1-a^3\]and \[P(4,4)\implies f(17) = f(8) + f(4)^2 = f(8)+(a^2-1)^2,\]so \[1-a^3 = (a^2-1)^2 \implies (1-a)(1+a+a^2) = (1-a)^2(1+a)^2.\]This equation is false for $a=-1$, so we have $a\in\{-2,1\}$. We now have two cases based on the value of $a$. Case 1: Suppose $a=1$. It is easy to see by induction that $f(x)\equiv x-1\pmod{3}$ and $f(x)\in\{-1,0,1\}$ for $x\in\mathbb{Z}$. We have $f(2x) = f(x-1/2)+f(x+3/2)$ from $P(x-1/2,2)$. Now, \[P(2x,3)\implies f(1+6x) - f(2x+3) = -f(2x)\]and \[P(x,6)\implies f(1+6x) - f(x+6) = -f(x),\]so \[f(2x+3)-f(x+6) = f(2x)-f(x),\]so \[f(x+1)+f(x+3)-f(x+6) = f(x-1/2)+f(x+3/2)-f(x).\]Let $a_n=f(n/2+x)$ for integers $n$ and some fixed $x$. This implies that \[a_{n+2}+a_{n+6}-a_{n+12} = a_{n-1}+a_{n+3}-a_n.\]This recursion has characteristic polynomial \[P(\lambda):=\lambda^{13}-\lambda^7 + \lambda^4 - \lambda^3-\lambda+1.\]Now, \[P(x,4)\implies f(1+4x)=f(x+4)\implies f(2x)+f(2x+2) = f(x+4)\], so \[f(x-1/2)+f(x+3/2)+f(x+1/2)+f(x+5/2)=f(x+4),\]so we now get the recursion \[a_{n+8} = a_{n-1} + a_{n+3}+a_{n+1}+a_{n+5}.\]This has characteristic polynomial \[Q(\lambda) = \lambda^9 - \lambda^6-\lambda^4-\lambda^2-1.\]It is easy to check that $\gcd(P,Q)=\lambda^2+\lambda+1$, so $a_n$ is a linear combination of $\omega^n$ and $\omega^{-n}$, where $\omega$ is a primitive third root of unity. Thus, \[a_n = \alpha\cdot \omega^n + \beta\cdot \omega^{-n},\]where $\alpha,\beta\in\mathbb{C}$. Since $a_n$ is real, we see that $\beta=\bar{\alpha}$. This implies that $a_{n+3}=a_n$, so \[f(x+3/2)=f(x).\]Now, \[P(x+3/2,y)-P(x,y)\implies P(xy+1+3y/2) = f(xy+1).\]For $y\ne 0$, we can vary $x$ to make $xy+1$ hit anything, so $f(z+3y/2) = f(z)$ for all nonzero $y$, so $f$ is a constant, which is the desired contradiction. Case 2: Thus, we must have $a=-2$. An easy induction then implies that $f(x)=x-1$ for all $x\in\mathbb{Z}$. Now, \[P(2x,3)\implies f(1+6x) - f(2x+3) = 2f(2x)\]and \[P(x,6)\implies f(1+6x) - f(x+6) = 5f(x),\]so \[f(2x+3)-f(x+6) = 5f(x)-2f(2x),\]so \[f(x+1)+f(x+3)-f(x+6) = 5f(x)-2f(x-1/2)-2f(x+3/2).\]Let $a_n=f(n/2+x)$ for integers $n$ and some fixed $x$. This implies that \[a_{n+2}+a_{n+6}-a_{n+12} = 5a_n-2a_{n-1}-2a_{n+3}.\]This recursion has characteristic polynomial \[P(\lambda):=\lambda^{13}-\lambda^7 - 2\lambda^4 - \lambda^3+5\lambda-2.\]Now, \[P(x,4)\implies f(1+4x)=f(x+4)+3f(x)\implies f(2x)+f(2x+2) = f(x+4)+3f(x)\], so \[f(x-1/2)+f(x+3/2)+f(x+1/2)+f(x+5/2)=f(x+4)+3f(x),\]so we now get the recursion \[a_{n+8} = a_{n-1} + a_{n+3}+a_{n+1}+a_{n+5}-3a_n.\]This has characteristic polynomial \[Q(\lambda) = \lambda^9 - \lambda^6-\lambda^4-\lambda^2+3\lambda-1.\]It is easy to see that $\gcd(P,Q)=(\lambda-1)^2$, so $a_n$ is a linear combination of $1$ and $n$, so \[a_n=cn+d.\]Thus, fixing $z$, we see that there is some $c$ such that $f(x+1/2)=f(x)+c/2$ for all $x\in\frac{1}{2}\mathbb{N}+z$. Now, \[1-f(x-1/2)-f(x+3/2) = f(x)^2-f(x)f(x+2).\]Replacing $x$ with $x+1$ and tracking the difference, we see that \[-2c = (f(x)+c)^2-f(x)^2 -(f(x)+c)(f(x+2)+c)+f(x)f(x+2) = 2cf(x)-c(f(x)+f(x+2)),\]so assuming $c\ne 0$, we see that $f(x+2)-f(x)=2$, so $c=1$. Thus, $c\in\{0,1\}$ always. However, if $c=0$, the equation \[1-f(x-1/2)-f(x+3/2) = f(x)^2-f(x)f(x+2)\]becomes \[1-2f(x) = f(x)^2-f(x)^2,\]so $f(x)=1/2$. Thus, $c=0\implies f(z)=1/2$. But if we replace $z$ with $-z$, then $f(x)+f(2-x)=0$ implies that the $f(x+1)=f(x)+c$ is also true for $z$ replaced with $-z$, so $f(-z)=1/2$ also, so $f(2-z)=1/2$, which is clearly a contradiction as $f(z)+f(2-z)=0$. Thus, $c=1$ always. Therefore, $f(x+1)=f(x)+1$ for all $x$. Now \[P(x+1,y)-P(x,y)\implies f(xy+1+y) - f(xy+1) = f(y)+1.\]As $y\ne 0$ is fixed, $xy+1$ reaches anything, so we actually have \[f(a+b)=f(a)+f(b)-1\]for all $a,b$ (the case $b=0$ is trivially true). It suffices now to show that $f$ is bounded on some interval, as Cauchy then proves that $f(x)\equiv x-1$. Note \[f(-x) = -2+f(2-x) = -2-f(x),\]so \[P(x,-x)\implies f(1-x^2) = -1+f(x)f(-x) = -1-f(x)(f(x)+2)=-(f(x)+1)^2,\]so $f(1-x^2)\le 0$, so $f$ is bounded on an interval, so $f(x)\equiv x-1$ for all $x$.

Does this work? The only solution is $\boxed{f\equiv x-1}.$ It is easy to check this works, so it remains to show it is the only ones. $\textbf{Claim: }$ $f(x)=-f(2-x)$ for all $x.$ $\emph{Proof: }$ $P(x,-1)$ and $P(2-x,-1)$ yield $$f(1-x)-f(x-1)=f(x)f(-1),$$$$f(x-1)-f(1-x)=f(2-x)f(-1).$$Therefore, since $f(-1)\ne 0,$ we have $f(x)=-f(2-x).$ $\blacksquare$ $\textbf{Claim: }$ $f$ is bounded on $(0,\infty).$ $\emph{Proof: }$ Let $x$ be a positive real. The assertion $P(1+\sqrt{x},1-\sqrt{x})$ gives $$f(2-x)-f(2)=f(1+\sqrt{x})f(1-\sqrt{x}).$$Since $f(2-x)=-f(x)$ and $f(1-\sqrt{x})=-f(1+\sqrt{x}),$ we have $$f(x)=f(1+\sqrt{x})^2-f(2)\ge -f(2),$$as needed. $\blacksquare$ Now, from $P(2-x,2-y)$ and $P(x,y),$ we have \begin{align*} f(5-2x-2y+xy) &= f(2-x)f(2-y)+f(4-x-y)\\ &= f(x)f(y)-f(x+y-2)\\ &= f(1+xy)-f(x+y)-f(x+y-2).\\ \end{align*}Fix $x+y=c\in (0,2).$ Then, as $x$ and $y$ vary, $1+xy$ can take any real value less than $1.$ Hence, for all $a<1,$ we have $$f(a)-f(a+(4-2c))=f(c)+f(c-2).$$As $c$ varies, this is equivalent to Jensen's functional equation, so since $f$ is bounded on $(0,\infty),$ we know $f$ is linear. Now we can easily check that the claimed solutions are the only ones.

The only function that work is $f(x)\equiv x-1$. This clearly works. Let $P(x,y)$ denote the assertion that $f(1+xy)-f(x+y)=f(x)f(y)$. $P(1,y)$ means $f(1+y)-f(1+y)=f(1)f(y)$. Since $f(-1)\ne 0$, $f(1)=0$. $P(0,-1)$ yields $0=(f(0)+1)f(-1)=0$ so $f(0)=-1$. Let $k=f(-1)$. Claim: $k=2$. Proof of Claim: $P(-1,y)$ yields $f(1-y)-f(y-1)=kf(y)$ $P(-1,2-y)$ yields $f(y-1)-f(1-y)=f(-1)f(2-y)$ so $$f(y)=-f(2-y)$$ (Call equation 1) Note $f(y-1)=-f(3-y)$ so $-f(3-y)-f(1-y)-cf(2-y)=0$, or $f(x+1)+cf(x)+f(x-1)=0$ Let $g(x)=f(x+1)$. P(2,x) on $g$ means $g(2y)=g(y+1)+g(y-1)$ so $g(2x)=cg(x)$. Backsolving with $g(x+1)+cg(x)+g(x-1)=0$, we get $g(2)=-c, g(3)=c^2-1, g(4)=-c^3+2c$. We have $-c^2=-c^3+2c$, so $-c^2+c-2=0$, $c\in \{1,-2\}$. If $c=1$, then $g(x-1)+g(x)+g(x+1)=0$ so $g(x)=1 \forall x\in 3\mathbb{Z}+1, 0\forall 3\mid x$ and $-1$ otherwise. We also have $g(x)=g(3+x)$ and $g(x)=-g(-x)$ $P(3x+1,\frac{3y+2}{3x+1})$ yields $-1-g(\frac{3y+2}{3x+1})=0$ so $g(\frac{3y+2}{3x+1})=-1\forall x,y$. Similarly, $P(3x+1,\frac{3y}{3x+1})$ yields $P(\frac{3y}{3x+1})=0$, $P(3x+1, \frac{3y+1}{3x+1})$ yields $P(\frac{3y+1}{3x+1})=0$ However, $P(3x,\frac{3y+1}{3x}): 1-g(\frac{3y+1}{3x})=-g(\frac{3y+1}{3x})$, contradiction. Thus our claim is proven. Hence $c=-2$, $g(x+1)+g(x-1)=2g(x)=g(2x)$. It can be proven via induction that $g(x)=x\forall x\in\mathbb{Z}$ Note $P(x,y),P(-x,y)$ yields $$f(1+xy)-f(x+y)=f(x)f(y)$$ $$f(1-xy)-f(y-x)=f(-x)f(y)$$ Hence $-f(x+y)-f(y-x)=((f(x)+f(-x))f(y)$. For all integers $x$, $f(x+y)+f(y-x)=2f(y)$, implying $f$ is linear. Hence $f(x)\equiv x-1.$

Let $P(x,y)$ be the given assertion. $P(1,-1) \implies f(0)-f(0)=f(1)f(-1) \implies f(1)=0$ $P(-1,0) \implies f(1)-f(-1)=f(-1)f(0) \implies f(0)=-1$ Assume $f(-1)=c$. Then $P(x,-1) \implies f(1-x)-f(x-1)=cf(x)$ $x \rightarrow x+1$ gives $f(-x)=f(x)+cf(x+1)$ $x \rightarrow x+1$ gives $f(-x-1)=f(x+1)+cf(x+2)$ $P(-x,-1) \implies f(x+1)=f(-x-1)+cf(-x)=(f(x+1)+cf(x+2))+c(f(x)+cf(x+1))$ $\implies f(x)+f(x+2)+cf(x+1)=0$ $x=1 \implies f(3)=-c$ $x=2 \implies f(4)=c^2-1$ $x=3 \implies f(5)=-c^3+2c$ Also $P(2,2) \implies f(5)=c^2$ Thus $-c^3+2c=c^2 \implies c\in\{1,-2\}$ Case 1:($c=1$) Then we get $f(x-1)=f(x+2)=f(-x)$ by taking $x \rightarrow x-1$ Also $f(-x)=f(x)+f(x+1)=-f(x+2)$ $\implies f(x+2)=-f(x+2)$ $\implies f(x+2)=0$ for all reals $x$, which is a contradiction as $f(-1)=1$. $\blacksquare$ Case 2:($c=-2$) So $f(x)+f(x+2)=2f(x+1)$ As proved by @anantmudgal09, we will prove that $f(y)+f(-y)=-2$. So $P(x,y)+P(x,-y) => f(x+y)+f(x-y)=-f(x)(f(y)+f(-y))=2f(x)$ Thus $f$ is a linear function. Checking leads to $f(x)=x-1$ being the only solution.

Here's an equivalent FE, which I'll solve because of simplicity: Equivalent Problem wrote: Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfy the conditions \[f(1+xy)-f(x+y)=(f(x)-1)(f(y)-1) \quad \text{for all } x,y \in \mathbb{R},\]and $f(-1)-1 \neq 0$. Let $P(x,y)$ denote the given assertion. $P(x,-1)\implies f(1-x)-f(x-1)=(f(x)-1)(f(-1)-1).$ Noting that $f(-1)-1\neq 0,$ we get $f(0)=0, f(1)=1,$ and $f(2)=2.$ Moreover comparing $P(x-1,-1)$ and $P(-x-1,-1)$ yields $f(x)+f(2-x)=2. \qquad (*)$ Observe that $\forall x\leq 5/4$, we can write $x$ in the form $ab+1.$ Let $a+b=1,$ then combining $P(a,b)$ and $P(2-a,2-b)$ implies $f(ab+3)-f(ab+1)=f(3)-f(1) \implies f(x+2)-f(x)=f(3)-f(1).$ Plugging $x=0$ yields $f(3)=3,$ it follows that $f(x+2)-f(x)=2.$ Now if $x>5/4$ then $-x<5/4,$ implies $f(2-x)-f(-x)=2.$ But $(*)\implies f(x)=2-f(2-x)$ and $f(x+2)=-f(-x)+2.$ Thus $f(x+2)-f(x)=f(2-x)-f(-x)=2~~\forall x\in R.$ Setting $x=-x$ in $(*)$ we get $f(-x)=-f(x).$ Now comparing $P(-x,y)$ and $P(x,-y)$ yields $f(1-xy)=1-f(x)f(y)\implies f(1+xy)=1+f(x)f(y),$ clearly $f$ is additive. Also we get multiplicative from $f(1+xy)=1+f(x)f(y)=1+f(xy)$ hence $f(x)\geq 0\forall x\geq 0.$ The well known solution is $f(x)=x$ for all real $x$ which fits. Footnote: Notice that the actual problem is just a corollary of this one. So if the function of the original one is $\phi(x)$ then $f(x)=\phi(x)+1.$ Thus the only solution of that is $\phi\equiv\text{Id}-1,$ which works indeed.

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfy the conditions \[f(1+xy)-f(x+y)=f(x)f(y) \quad \text{for all } x,y \in \mathbb{R},\]and $f(-1) \neq 0$. The only solution is $\boxed{f(x)=x-1}$, which works. Let $P(x,y)$ denote the given assertion. $P(1,-1): f(1)f(-1)=0\implies f(1)=0$. $P(0,0): f(0)^2 + f(0) =0\implies f(0)\in \{-1,0\}$. If $f(0)=0$, then $P(-1,0): -f(-1)=0\implies f(-1)=0$, contradiction. So $f(0)=-1$. Let $f(-1)=c$. $P(x+1,-1): f(-x) - f(x) = cf(x+1)$. $P(x+2,-1): f(-x-1) - f(x+1) = cf(x+2)$. $P(-x,-1): f(x+1) - f(-x-1) = cf(-x)$. So $-cf(-x) = cf(x+2)\implies -f(-x) = f(x+2)$. Thus, $f(3) = -f(-1) = -c$ and $f(2) = -f(0) = 1$. Since $f(-2) - f(2) = cf(3) = -c^2$, we have $f(-2) = 1-c^2$. $P(2,-2): f(-3) +1 = f(2)f(-2) = 1-c^2\implies f(-3) = -c^2$. So $f(5) = -f(-3) = c^2$. $P(2,2): f(5) - f(4) = 1\implies f(4) = c^2 -1 $. $P(4,-1): f(-3) - f(3) = c(c^2-1)\implies c-c^2 = c^3 - c\implies c(c-1)(c+2)=0$. Since $c\ne 0$, we have $c\in \{-2,1\}$. Case 1: $c=1$. $P(x+1,-1): f(-x) - f(x) = f(x+1)$ and $P(x+2,-1): f(-x-1) - f(x+1) = f(x+2)$. So\begin{align*} -f(-x) \\ = f(x+2) \\ = f(-x-1) - f(x+1) \\ = f(-x-1) - f(-x) + f(x) \\ \end{align*}This gives $f(-x-1) = -f(x)$. Since $-f(-x) = f(x+2)$ for all $x\in \mathbb{R}$, we have $-f(x) = f(2-x)$. So $f(-1-x) = f(2-x)$. Setting $x\to -x-1$ gives $f(x) = f(x+3)\forall x\in \mathbb{R}$. $P(x,3): f(3x+1) - f(x) = -f(x)\implies f(3x+1) =0$. Setting $x=\frac{-2}{3}$ gives $f(-1)=0$, contradiction. Case 2: $c=-2$. By $P(x+1,-1): f(-x) - f(x) = 2f(x+1)$. Since $-f(-x) = f(x+2)$ for all $x$, we have\[-f(1-x) = -f(-(x-1)) = f(1+x)\implies f(1+x) + f(1-x) = 0.\]Claim: If $f(x) =0$, then $x=1$. Proof: Suppose $k\ne 1$ satisfied $f(k ) =0$. $P(k,x): f(1+kx) = f(x+k)$. $P(k,-x): f(1-kx) = f(k-x)$. So $f(k+x) = -f(k-x)$ Now, $-f(k-x) = f(x-k+2)$, so $f(x+k) = f(x+(2-k))$. Setting $x\to x-k$ gives $f(x) = f(x+(2-2k))$. Since $k\ne 1$, we have $f(x) = f(x+d) $ for some $d\ne 0$. $P(d,x): f(dx+1) - f(x) = f(d) f(x) = -f(x)\implies f(dx+1)=0$. Since $dx+1$ can take any real value, we have $f\equiv 0$, which is not possible since $f(-1)\ne 0$. $\square$ Claim: $f(x)+f(-x) = -2$. Proof: This is true if $x=0$. Henceforth assume $x\ne 0$. $P(x+1,-x): f(1-x^2 - x ) = f(x+1) f(-x) $. Since $f(1-x^2 - x) = -f(x^2 + x + 1)$, we have $f(x^2 + x + 1) =- f(x+1)f(-x)$. $P(2,x): f(2x+1) - f(x+2) = f(x)\implies f(2x+1) = f(x) - f(-x) = -2f(x+1)$. $P(x,x+1): f(x^2 + x + 1) - f(2x+1) = f(x)f(x+1)$, so\[f(x+1)(f(x)+f(-x)) = -f(2x+1) = -2f(x+1)\]Since $x\ne 0$, we have $f(x+1)\ne 0$, so dividing both sides by $f(x+1)$ gives $f(x)+f(-x) = -2$. $\square$ $P(-x,y): f(1-xy) - f(y-x) = f(-x)f(y)$. Adding this to the original FE gives\[(f(1+xy) + f(1-xy)) - (f(y+x) + f(y-x)) = (f(x)+f(-x))f(y),\]so\[-(f(y+x) + f(y-x)) = (f(x)+f(-x)) f(y) = -2f(y), \]so\[f(y+x) + f(y-x) = 2f(y)\]Setting $g(x) = f(x) +1 $ gives $g(y+x) + g(y-x) = 2g(y)$. Let $Q(x,y)$ be this assertion. $Q(x,x)$ gives $g(2x) = 2g(x)$ because $g(0) =0$. So $g(x+y) + g(y-x) = g(2y)$. For any reals $x,y$, $Q\left(\frac{x-y}{2}, \frac{x+y}{2} \right): g(x) + g(y) = g(x+y)$ The original FE becomes\[g(xy+1) -g(x+y) = (g(x)-1)(g(y)-1) = g(x)g(y) - g(x)-g(y)+ 1.\]Notice that $g(1) = f(1)+1 = 1$. Since $g(x+y) = g(x)+g(y)$ and $g(xy+1) = g(xy) + g(1) = g(xy) + 1$, we have $g(xy) = g(x)g(y)$. So $g$ is additive and multiplicative, either $g\equiv 0$ or $g(x) = x$. The former isn't true as $g(1)=1$. So we must have $g(x) =x \implies f(x)=x-1$.

We claim the only solution is $\boxed{f(x) = x-1}$. This works since: \[f(1+xy) - f(x+y) = xy - x - y + 1 = (x-1)(y-1) = f(x)f(y). \]Let $P(x, y)$ denote the FE. We begin with some initial observations. $P(1, -1): 0 = f(1)f(-1) \Rightarrow \boxed{f(1) = 0}$. $P(0, -1): -f(-1) = f(0)f(-1) \Rightarrow \boxed{f(0) = -1}$. $P(-1, 2): f(-1) - f(1) = f(-1)f(2) \Rightarrow \boxed{f(2) = 1}$. \[P(2, y-1): f(2y-1) - f(y+1) = f(2)f(y-1)\]\[\Rightarrow \boxed{f(2y-1) = f(y-1) + f(y+1)} \tag{1}\]\[P(-1, y): \boxed{f(1-y) - f(-1 + y) = f(-1)f(y)} \tag{2}\]\[[2](1-x): \boxed{f(x) - f(-x) = f(-1)f(1-x)} \tag{3}\]\[[3](-x): f(-x) - f(x) = f(-1)f(1+x) \Rightarrow f(-1)f(1-x) = -f(-1)f(1+x)\]\[\Rightarrow \boxed{f(1-x) = -f(1+x)} \tag{4}\]\[[2](y), [4](y): -f(1+y) - f(y-1) = f(-1)f(y)\]\[[1](x): \boxed{f(2x-1) = -f(-1)f(x)} \tag{5}\]\[[4](1-x): \boxed{f(x) = -f(2-x)} \tag{6}\] Claim 1: $f$ cannot be periodic. Proof: Assume FTSOC $f$ is periodic with period $p$. Then $P(p, x): f(1 + px) - f(x + p) = f(p)f(x) \Rightarrow f(1 + px) = f(0)f(x) + f(x) = 0$, but now $1 + px = -1$ has a solution in $x$ since $p \neq 0$ means $f(-1) = 0$, contradiction. $\square$ Now we determine $f(-1)$. $P(2, -2): f(-3) - f(0) = f(2)f(-2) \Rightarrow f(-3) = f(-2) - 1$. $[3](3): f(3) - f(-3) = f(-1)f(-2)$ and $[4](2): f(3) = -f(-1)$. Thus, $-f(-1) - f(-2) + 1 = f(-1)f(-2)$. Now $[3](2)$ gives $f(2) - f(-2) = f(-1)f(-1) \Rightarrow 1 - f(-2) = f(-1)^2$. Hence $-f(-1) + f(-1)^2 - 1 = f(-1)(1 - f(-1)^2) \Rightarrow f(-1)(f(-1)^2 + f(-1) - 2) = 0$. Solving we get $f(-1) = -2$ or $f(-1) = 1$. Claim 2: $f(-1) = -2$. Proof: Assume FTSOC $f(-1) = 1$, then $f(-2) = 0$. $P(-2, -y): f(1 + 2y) + f(-y-2) = 0 \Rightarrow f(2y+1) = -f(-y-2)$. $[5](y+1): -f(y+1) = f(2y+1) = -f(-y-2) \Rightarrow f(y+1) = f(-y-2)$. $[6](-y-2): f(y+1) = -f(4+y) \Rightarrow f(x) = -f(x+3) \Rightarrow f(x) = f(x+6)$. But now $f$ is periodic, contradiction by Claim 1. $\square$ We continue with some more observations: \[\boxed{f(1-y) - f(y-1) = -2f(y)} \tag{2}\]\[\boxed{f(x) - f(-x) = -2f(1-x)} \tag{3}\]\[\boxed{f(2x-1) = 2f(x)} \tag{5}\]\[\boxed{2f(y) = f(y-1) + f(y+1)} \tag{7}\]\[P(x, x): f(1+x^2) = f(x)^2 + f(2x)\]\[[7](x^2): -f(1 - x^2) = f(x)^2 + f(2x)\]\[P(x, -x): f(1-x^2) = f(x)f(-x) - 1 \Rightarrow 1 - f(x)f(-x) = f(x)^2 + f(2x)\]\[\therefore \boxed{f(2x) = 1 - f(x)f(-x) - f(x)^2} \tag{8}\]\[[8](-x): f(-2x) = 1 - f(-x)f(x) - f(-x)^2 \Rightarrow f(2x) - f(-2x) = f(-x)^2 - f(x)^2\]\[[3](2x), [3](x): -2f(1-2x) = (f(-x) + f(x))(2f(1-x))\]\[\Rightarrow f(1-2x) = -f(1-x)(f(x) + f(-x))\]\[[5](1-x): 2f(1-x) = -f(1-x)(f(x) + f(-x))\]\[\Rightarrow \boxed{f(1-x) = 0 \textrm{ or } f(x) + f(-x) = -2} \tag{9}\] Claim 3: $f(x) = 0 \Rightarrow x = 1$. Proof: Assume FTSOC there exists $f(x_0) = 0$ where $x_0 \neq 1$. Note that $x_0 \neq 0, 1$. \[P(x_0, y): \boxed{f(x_0y + 1) = f(y+x_0)} \tag{*}\]\[[8](x_0): f(2x_0) = 1\]Furthermore, note that if $z_0$ is a zero of $f$, then by $(*)$, for any $k \in \mathbb{Z}_{\ge 0}$: \[\boxed{f(y + z_0 + 1) = f(z_0y + z_0 + 1) = ... = f(z_0^ky + z_0 + 1)} \tag{**}\] Case 1: $f(1-x_0) = 0$. \begin{align*} [4](x_0): f(1 + x_0) &= 0 \\ [7](x_0 + 1): f(x_0) + f(x_0 + 2) &= 2f(x_0 + 1) \\ \Rightarrow f(x_0 + 2) &= 0 \\ [6](x_0 + 2): f(-x_0) &= 0 \\ [**](x_0, 2): f(y + x_0 + 1) &= f(x_0^2y + x_0 + 1) \\ [**](-x_0, 2): f(y - x_0 + 1) &= f(x_0^2y - x_0 + 1) \\ \Rightarrow f((y + \frac{2}{x_0}) - x_0 + 1) &= f(x_0^2 + x_0 + 1) \\ \Rightarrow f(y + x_0 + 1) &= f(y + \frac{2}{x_0} - x_0 + 1) \\ \Rightarrow f(x) &= f(x + (\frac{2}{x_0} - 2x_0)) \end{align*}Since $x_0 \neq 1$, $\frac{2}{x_0} - 2x_0 \neq 0$ hence $f$ is periodic, contradiction by Claim 1. Case 2: By $[9](x_0)$, otherwise we must have $f(-x_0) = -2$. \begin{align*} [3](x_0): f(1 - x_0) &= -1 \\ [2](x_0): f(x_0 - 1) &= -1 \\ [4](x_0): f(1 + x_0) &= 1 \\ P(x_0-1, x_0+1): f(x_0^2) - f(2x_0) &= f(x_0-1)f(x_0+1) \\ \therefore f(x_0^2) &= 1 + (-1)(1) = 0 \end{align*}We repeat a similar argument to above: \begin{align*} [**](x_0, 2): f(y + x_0 + 1) &= f(x_0^2y + x_0 + 1) \\ [**](x_0^2, 1): f(y + x_0^2 + 1) &= f(x_0^2y + x_0^2 + 1) \\ \Rightarrow f((y - 1 + \frac{1}{x_0}) + x_0^2 + 1) &= f(x_0^2 + x_0 + 1) \\ \Rightarrow f(y + \frac{1}{x_0} + x_0^2) &= f(y + x_0 + 1) \\ \Rightarrow f(x) &= f(x + \frac{1}{x_0} + x_0^2 - x_0 - 1) \end{align*}Once again, $\frac{1}{x_0} + x_0^2 - x_0 - 1 = 0 \Rightarrow x_0^3 - x_0^2 - x_0 + 1 = 0 \Rightarrow (x_0 + 1)(x_0 - 1)^2 = 0 \Rightarrow x_0 = 1$ or $x_0 = -1$. $x_0 \neq 1, -1$, so $f$ is periodic, contradiction. $\square$ Thus, by $(9)$, we get $f(x) + f(-x) = -2$ unless $x = 1$, which is true anyway. Thus, \[\boxed{f(x) + f(-x) = -2} \tag{10}\]\[[3](-x): \boxed{f(x+1) = f(x) + 1} \tag{11}\]\[P(2, x): f(1 + 2x) - f(x+2) = f(x)f(2) \Rightarrow \boxed{f(2x+1) = 2f(x) + 2} \tag{12}\] Now consider the function $g(x) = f(x+1)$. Thus, we get: \[\boxed{g(x+1) = g(x) + 1} \tag{I}\]\[P(x, y): g(xy) - g(x + y - 1) = g(x-1)g(y-1)\]\[[11]: \boxed{g(xy) - g(x)g(y) = g(x+y) - g(x) - g(y)} \tag{II}\]\[[12]: \boxed{g(2x) = 2g(x)} \tag{III}\]\[[III], [II](x, x): \boxed{g(x^2) = g(x)^2} \tag{IV}\] Claim 4: Suppose function $h: \mathbb{R} \to \mathbb{R}$ satisfies $h(x+1) = h(x) + 1$ and $h(x^2) = h(x)^2$. Then $h \equiv x$. Proof: $h(1) = h(0) + 1$, $h(1) = h(1)^2$ and $h(0) = h(0)^2$ readily imply $h(0) = 0, h(1) = 1$. Now by non-negativity of squares, $h(x) \ge 0$ for all $x \ge 0$. $h(x^2) = h(-x)^2 \Rightarrow h(x) = |h(-x)|$. Assume FTSOC $h(x) = x + \epsilon > 0$. By appending some large $L \in \mathbb{N}$, we get $h(x + L) = x + L + \epsilon$ and can deal with $x > 0$ only. Then $h(x^{2n}) = (x + \epsilon)^{2n} > x^{2n} + 2n\epsilon > x^{2n} + N$ for arbitrarily large $N$. Now let $M$ be smallest $M \in \mathbb{N}$ whereby $h(x^{2n} - M) < 0$. We have $x^{2n} + N - M < 0 \Rightarrow -N > x^{2n} - M$, so $x_{2n} - M$ negative. Thus, $h(M - x^{2n}) = |h(x^{2n} - M)|$. But now notice that by minimality of $M$, $-1 \le h(x^{2n} - M) < 0$, hence $0 < h(M - x^{2n}) \le 1$. We can pick $M - x^{2n} > 2$ by unboundedness of $N$, thus $h(M - x^{2n} - 2) \le -1$ yet $M - x^{2n} - 2 > 0$, contradiction. Similarly, if FTSOC $h(x) = x - \epsilon < x$, then after shifting to sufficiently large $x$ such that $x > 2 + \epsilon$, $h(x^{2n}) = (x - \epsilon)^{2n} = (x^{2n})(1 - \frac{\epsilon}{x})^{2n}$. For arbitarily large $n$, we can get $(1 - \frac{\epsilon}{x})^{2n} < \frac12$ hence $h(x^{2n}) < \frac12 x^{2n}$. Since $x > 2$ and $n$ arbitrarily large, we get $h(x^{2n}) < x^{2n} - N$ for arbitrarily large $N$ with fractional part equal to $x^{2n}$. Hence, $h(N) = h(x^{2n}) - (x^{2n} - N) < 0$, yet $N > 0$ contradiction. Thus, $h(x) = x$ necessarily for all $x$. $\square$ Thus, $f(x+1) = x \Rightarrow f(x) = x-1$ holds for all $x$. $\blacksquare$

This is so long why. Denote the assertion with $P(x, y)$. Claim: We have that $f(0) = -1$, $f(1) = 0$, and $f(2) = 1$. Proof. By $P(1, 1)$ it follows that $f(1)^2 = 0$ and thus $f(1) = 0$. Then, by $P(x, 0)$ it follows that \[ -f(x) = f(x)f(0) \]Since $f$ is not uniformly $0$ it follows that $f(0) = -1$. By $P(x, -1)$ it follows that \[ f(1 - x) - f(x - 1) = f(x)f(-1) \]Then, by $P(2-x, -1)$, since $f(-1) \ne 0$ it follows that $f(x) = -f(2-x)$. As such, $f(0) = -f(2)$ so $f(2) = 1$. $\blacksquare$ Define $g(x) = f(x + 1)$ so $g$ is odd and the identity on $-1, 0, 1$. The expression becomes \[ g(xy) - g(x + y - 1) = g(x - 1)g(y - 1) \]which we denote with $Q(x, y)$. Claim: $g$ is the identity on the integers Proof. Thus, by $Q(x, -1)$ \[ g(-x) - g(x - 2) = g(x - 1)g(-2) \]or \[ g(x) = g(x-1)g(2) - g(x - 2) \]Since $g(4) - g(3) = 1$ and $g(4) = g(3)g(2) - g(2)$ and $g(3) = g(2)g(2) - 1$ it follows that \[ g(2)^2 - g(2) - 2 = 0 \]so either $g(2) = 2$ or $g(2) = -1$. First suppose that $g(2) = -1$. Then, \[ g(x) = \begin{cases} -1 & x \equiv -1 \pmod{3} \\ 0 & x \equiv 0 \pmod{3} \\ 1 & x \equiv 1 \pmod{3} \\ \end{cases} \]As such, by $Q\left(x, \frac12\right)$, \[ g\left(\frac{x}{2}\right) - g\left(\frac{x - 1}{2}\right) = g(x-1)g\left(-\frac{1}{2}\right) \]so \[ g\left(\frac{x}{2}\right) - g\left(\frac{x}{2} - 1\right) = (g(x-2) + g(x-1))g\left(-\frac{1}{2}\right) \]However, this leads to a contradiction. Thus, $g$ is the identity on the integers. $\blacksquare$ As such, $g(x) + g(x-2) = 2g(x-1)$ so $g(x), g(x+1), \dots$ is an arithmetic progression. Furthermore, since $g(x(x - 2)) + 1 = g(x - 1)^2 \ge 0$, so $g(x) \ge -1$ for $x \ge -1$. Claim: $g$ is the identity and multiplicative on the rationals Proof. Thus, by $Q(x, n)$ for integer $n$, \[ g(nx) - g(x) - (n-1)(g(x) - g(x-1)) = (n-1)g(x-1) \]or \[ g(nx) = ng(x) \]$\blacksquare$ Furthermore, by $Q(x, 1-x)$ \[ g(x(x-1)) = g(x-1)g(x) \]As such, by $Q(x, x-1)$ \[ g(x)g(x-1) - g(2x - 2) = g(x-1)g(x-2) \]or \[ g(x-1)(g(x) - g(x-2)) = 2g(x - 1) \]so either $g(x-1) = 0$ or $g(x) - g(x-2) = 2$. As such, each arithmetic progression either has common difference $1$ or is only $0$. Thus, $g(x) \ge 0$ for $x \ge 0$ and $g(x) \le 0$ for $x \le 0$. Thus, $g(xy) \ge g(x + y - 1)$ holds whenever $x - 1$ and $y - 1$ are the same sign. Thus, $g(x) \ge g(y)$ for $x \ge y \ge 2$ which by multiplicity means that it holds for all $x$ and $y$. As such, $g(x) = x$ by density of rationals and $f(x) = x + 1$.

Let $g(x)=f(x)-1$, with $g(1+xy)-g(x+y)=(g(x)-1)(g(y)-1)$ Let $P(x,y)$ denote that assertion. $P(1,-1)$ gives $(g(-1)-1)(g(1)-1)=0$ but since $g(-1)\neq -1$, $g(1)=1$. $P(0,-1)$ gives $1-g(-1)=(g(-1)-1)(g(0)-1)$ so $g(0)=0$. $P(x,-1)+P(2-x,-1)$ gives $(g(-1)-1)(g(x)-1+g(2-x)-1)=0$ so $g(x)+g(2-x)=2$. Now, $P(x,1-x)-P(2-x,1+x)$ gives \[g(1+x-x^2)+1=(g(x)-1)(g(1-x)-1)=(1-g(2-x))(1-g(1+x))=g(3+x-x^2)+g(3)\]We may deduce that for all $y\le \tfrac54$, $g(2+y)-g(y)=g(3)-1$. Note that $g(2)-g(0)=g(3)-1$ and $g(2)+g(0)=2$. Thus, $g(2)=2$ and so $g(3)=3$. Then $g(x)+g(2-x)=2$ extends this result to $g(2+x)-g(x)=2$ for all $x$. Now, $g(2+x)+g(-x)=2$ so $g(-x)=-g(x)$. Now, $P(x,y)+P(-x,-y)$ gives $g(1+xy)=1+g(x)g(y)$. Plugging into $P$, $g(x+y)=g(x)+g(y)$. Thus, $g(x)^2=g(1+x^2)-1=g(x^2)$ so $g(x)$ is bounded on the positives and satisfies Cauchy, so it's linear. Clearly it must be $g(x)=x$, and we can extend to negatives by odd.

Solved with $\textbf{AlperenInan}$ Let $P(x,y)$ be the denotation of given equation. From $P(x,1)$ we obtain $f(1)=0$ $P(-1,0) \Rightarrow f(0)=-1$ $P(2,-1) \Rightarrow f(2)f(-1)=f(-1)$ since $f(-1)\neq0$ we obtain $f(2)=1$ From compairing $P(x,-1)$ and $P(2-x,-1)$ we obtain $f(x)=-f(2-x)$ Let $g(x)=f(x+1)$. If we substitute the function $g$ on the first given equation we obtain that $P(x,y)$ is equivalent to equation $$g(x+y-1)+g(x-1)g(y-1)=g(xy)$$Since $g(x)=-g(-x)$ from $P(-x,-1)$ we obtain the equation $$g(x+2)+g(x)=g(2)g(x+1)$$From the values we have found from $f$ it's obvious that if $g(2)=c$ then $g(3)=c^2-1$ , $g(4)=c^3-2c$ , $g(5)=c^4-3c^2+1$ And with $P(-2,-2)$ we obtain $c(c-2)(c+1)=0$. Since $c\neq0$ if $g(2)=-1$ then $P(-x,-1)$ is equivilant to the equation $g(x)+g(x+1)+g(x+2)=0$. From that it is obvious that $g(x)=g(x+3)$ thus $g$ is periodic. From compairing $P(-x,-y)$ and $P(-x-3,-y)$ we obtain $g(xy+3y)=g(xy)$. Thus from substituting $x=\frac{1}{y}$ we obtain $g$ is constant. Which leads us to contradiction. So $g(2)=2$ and $g(-2)=-2$.After consider we did the same solution of #28 and $g(n)=n \forall n \in \mathbb{Z}$. That means $f(n)=n-1 \forall n \in \mathbb{Z}$.

This might be one of the few FEs where the domain and co-domain are $\mathbb{R}$ for a reason. The following generalization has a solution that I believe is complete, but it took me a lot of time and energy to even just solve it in my head; writing them down takes even more. There are some tricks you can do to solve all cases, including the $f(-1) = 0$ case when working with the function's domain being $\mathbb{R}$. None of them are available when you generalize the domain. Quote: Let $R$ be a commutative ring and $S$ be a field. Find all functions $f : R \to S$ such that for any $x, y \in R$, \[ f(1 + xy) - f(x + y) = f(x) f(y). \]

The solution is attached as a PDF file.

We can actually solve the case where $f$ vanishes at $-1$. Let $g(x)=f(x)+1$ and let $\mathbb{P}(x,y):g(1+xy)-g(x+y)=(g(x)-1)(g(y)-1)$. Assume that $g(0)=0$. By $\mathbb{P}(x,-1)$ we get $g(1-x)=g(x-1)$ and hence, $g$ is even. Now, notice that by $\mathbb{P}(x, -x)$ we get $0 \leq (g(x)-1)^2=g(x^2-1)$. And, we also must have $$g(1+xy)-g(x+y)=(g(x)-1)(g(y)-1)=(g(-x)-1)(g(y)-1)=g(1-xy)-g(y-x)$$Substituting $y=x$ into this relation we get $g(x^2+1)-g(x^2-1)=g(2x)$. Then, if $xy>0$ we must have $g(2\sqrt{xy})=g(xy+1)-g(xy-1)=g(x+y)-g(x-y)$ Denote by $\mathbb{R}^{+}_{0}$ the set of nonnegative real numbers. Now, consider a function $h:\mathbb{R}^{+}_{0} \rightarrow \mathbb{R}^{+}_{0}$ such that $h(x^2)=g(x)$ holds for all $x \in \mathbb{R}$. Then, we must have $h((x-y)^2)+h(4xy)=h((x+y)^2)$. It follows that $h(x)+h(y)=h(x+y)$ for all $x,y \in \mathbb{R}^{+}_{0}$. Then, expanding the relation $h((1+xy)^2)-h((x+y)^2)=(h(x^2)-1)(h(y^2)-1)$ we get $h(x^2)h(y^2)=h((xy)^2)$ and hence $h$ is multiplicative. It follows that $h(x)=x$ or $h(x)=0$ and hence $g(x)=x^2$ or $g(0)=0$. Assume that $g(0) \neq 0$. Suppose that there is a real number such that $g(x) \neq 1$. Then, $\mathbb{P}(1,x)$ shows us that $0=(g(1)-1)(g(x)-1)$ which means $g(1)=1$. And now, by $\mathbb{P}(0,x)$ we get $1-g(x)=(g(0)-1)(g(x)-1)$ and hence $g(0)=0$ a contradiction. Therefore $g(x)=1$. Finally, checking our answers, we get that $f(x)=x^2-1$ and $f(x)=0$ are the only ones that work.

The answer is $f(x) = \boxed{x-1}$, which works. Denote the given assertion as $P(x,y)$ and let $f(-1)=c$. Notice that $P(-1,1)$ gives $f(1)=0$ and $P(-1,0)$ gives $f(0)=-1$. Also, comparing $P(x,-1)$ and $P(2-x,-1)$ gives $f(x) = -f(2-x)$. Plug in $P(-x,-1)$ to get \[f(x+1)-f(-x-1) = cf(-x)\]\[\implies f(x+1)+f(x+3) = -cf(x+2)\] With this, we can calculate every integer value of $f(x)$ in terms of $c$. Then, plugging in $P(2,2)$ yields \[(2c-c^3)-(c^2-1) = 1\] Factoring and solving yields $c=0, 1, -2$. The first case is impossible due to the given condition, so we are left with two cases. Case 1: $c=1$ In this case, we have $f(x)=f(x+3)$ for all $x$. Then, $P(-\tfrac{2}{3},3)$ yields \[f(-1) - f\left(\frac{7}{3} \right) = f(3) f\left(-\frac{2}{3} \right) \implies f(-1) = 0,\] which is a contradiction. Case 2: $c=-2$ The recursion becomes \[f(x-1)+f(x+1) = 2f(x),\] and in particular, $f(x) = x-1$ over the integers. Claim: $f$ satisfies Jensen's Equation; in other words, \[\frac{f(x)+f(y)}{2} = f \left(\frac{x+y}{2} \right).\] Proof: Plugging in $P(a,b)$, $P(a,b-1)$, and $P(a,b-2)$ yields \begin{align} f(1+ab)-f(a+b) &= f(a)f(b) \\ f(1+a(b-1)) -f(a+b-1) &= f(a)f(b-1) \\ f(1+a(b-2)) - f(a+b-2) &= f(a)f(b-2). \end{align} Taking twice (2) and subtracting it from the sum of (1) and (3) gives \[\frac{f(1+ab)+f(1+ab-2a)}{2} = f(1+ab-a),\] as desired. $\square$ Due to the claim, we know that $g(x) = f(x)-f(0)= f(x)+1$ is additive. Moreover, $P(1+\sqrt{x}, 1-\sqrt{x})$ yields \[f(2-x) - f(2) = f(1+\sqrt{x})f(1-\sqrt{x})\]\[\implies f(x) = f(1+\sqrt{x})^2 - f(2) \ge -f(2).\] This bound holds for $x \in (0,\infty)$, which means that $g$ is linear. Hence, $f$ is linear as well. Solving yields $f(x) = 0, x-1$, but the former is absurd.

The only answer is $f(x)=x-1$, which works. Taking $P(x,0)$ implies $f$ is constant, which implies $f\equiv 0$, which implies $f(-1)=0$, or $f(0)=-1$. Summing $P(-1,1-x)$ and $P(-1,1+x)$ gives $f(1-x)+f(1+x)=0$. Taking $P(1-x,1-y)$ and $P(1+x,1+y)$ implies $f(2-x-y+xy)-f(2-x-y)=f(2+x+y+xy)-f(2+x+y)$. Letting $x+y=a,xy=b$ we get $f(2+a)-f(2-a)=f(2+a+b)-f(2-a+b)$ for all $(a,b)$ with a solution in $x,y$, which in particular works for all negative $b$. Setting $b\le -2$, we get $f(a)-f(-a)=f(a+k)-f(-a+k)$ for all negative $k$. From $f(1-x)+f(1+x)=0$, we also get $f(-a)-f(a)=f(-a-b)-f(a-b)$ for negative $b$, or equivalently $f(a)-f(-a)=f(a+k)-f(-a+k)$ for all positive $k$. Combining these two, notice we can write $f(x+y)-f(y)=f(x)-f(0)$, or $f(x+y)=f(x)+f(y)+1$, or $f+1$ is additive. Finally, $P(1-x,1+x)$ gives $f(2-x^2)-f(2)=-f(1+x)^2\le 0$, so $f(x)\le f(2)$ for $x\le 2$, so $f$ is bounded on an interval. Since $f+1$ is bounded and additive, we have $f(x)=cx-1$ for some constant $c$. Checking, $c=1$, which gives the answer $f(x)=x-1$.

$(1, -1) \implies f(1) = 0, (0, -1) \implies -f(-1) = f(0)f(-1) \implies f(0) = -1$. $(x, -1)\implies f(1 - x) - f(x - 1) = f(x)f(-1), (2 - x, -1)\implies f(x - 1) - f(1 - x) = f(2 - x)f(-1).$ Thus, $f(x) = -f(2 - x).$ let $f(-1) = c.$ $(1 + x, -1) \implies f(-x) = f(x) + f(1 + x)c, f(-x - 1) = f(x + 1) + f(2 + x)c.$ From $(-x, -1),$ we have $f(1 + x) = f(-x - 1) + cf(-x) = f(x + 1) + cf(2 + x) + c(f(x) + f(1 + x)c),$ or $f(x) + f(1 + x)c + f(2 + x) = 0\implies f(2 + x) = -f(1 + x)c - f(x)$ $f(0) = -1, f(1) = 0,$ so we can deduce some more terms: $f(2) = 1, f(3) = -c, f(4) = c^2 - 1, f(5) = 2c-c^3.$ However, from the given, $(2, 2) \implies f(5) - f(4) = f(2)^2, 2c - c^3 - c^2 + 1 = 1,$ so $c^2 + c - 2 = 0, c = -2, 1.$ If $c = 1,$ we can deduce $f(3k) = -1, f(3k + 1) = 0, f(3k + 2) = 1$ for $k\in\mathbb Z.$ Also note that $f$ has period 3. Then, take $x = 3k, y = \frac1x$ to get $1 = f(3k + y) + f(x)f(y) = f(y) - f(y) = 0,$ a contradiction. Thus, $f(-1)=-2.$ By induction, $f\equiv n - 1$ for $n\in\mathbb N.$ \begin{align*} (x, y)\implies f(1 + xy) - f(x + y) & = f(x)f(y) \\ (x, y + 1) \implies f(1 + xy + x) - f(x + y + 1) & = f(x)f(y + 1) \\ (x, y + 2) \implies f(1 + xy + 2x) - f(x + y + 2) & = f(x)f(y + 2) \end{align*}Next, note that $2f(y) = f(y - 1) + f(y + 1).$ Thus, if we multiply the second equation by 2 and subtract the first and 3rd, one gets $f(1 + xy + 2x) + f(1 + xy) = 2f(1 + xy + x),$ which is just Jensen's, or $f(a) + f(b) = 2f\left(\frac{a + b}2\right)$. Finally, take $(1 - x, 1 + x)$ to get $f(2 - x^2) - 1 = f(1 - x)f(1 + x) = -f(1 - x)^2 < 0,$ so $f(x) < 1$ if $x < 2.$ Thus, by bounded cauchy, we deduce $f$ is linear. Hence, $f\equiv x - 1.$