Very nice. Let $P(x,y)$ denote the given proposition. Let $f(0)=c$. $P(x,0)$ gives $$f(f(x))=c(1-f(x))$$Putting this in the original equation, we get $$f(xy)=f(x)f(y)+c(1-f(x+y))$$Call this $Q(x,y)$. Now, if $c=0$, we get $f(f(x))=0$ and $f(xy)=f(x)f(y)$ for all $x,y$. Now if $f(t)=y \neq 0$ for some $t$, we get $f(xy)=0$ for all $x$, so $f$ is identically $0$, contradiction! Hence we get the solution $$\boxed{f(x)=0 \ \ \forall x \in \mathbb{R}}$$So consider the case when $c \neq 0$. $Q(x,1)$ gives $f(x)=f(x)f(1)+c(1-f(x+1))$ $$\implies f(x+1)=\frac{f(1)-1}{c}f(x)+1$$for all $x$. Let $a=\frac{f(1)-1}{c}$, so $f(x+1)=af(x)+1$. Then, $Q(x+1,y)$ gives \begin{align*} f(xy+y)&=f(y)(af(x)+1)+c(-af(x+y)) \\ &= f(y)+a(f(x)f(y)+c(1-f(x+y)))-ac \\ &= f(y)+af(xy)-ac \end{align*}For all $x,y$. If we let $z=xy$, then $(y,z)$ can take all values except $y=0$ and $z\neq 0$. Thus $f(y+z)=f(y)+af(z)-ac$ for all such $y,z$. If $a \neq 0$, then taking $y,z$ to be non-zero and switching them we get $f(y)=f(z)$. Thus $f$ is non-constant on non-zero values, say $d$. Then $P(x,y)$ for $x,y \neq 0$ with $x+y \neq 0$, we get $d-d^2=f(d)$. If $d \neq 0$, then $f(d)=d$ $\implies$ $-d^2=0$ $\implies$ $d=0$, contradiction! So $d=0$, bu then we get $f(0)=0$, contradiction! Therefore $a=1$, and we get $$f(y+z)=f(y)+f(z)-c$$for non-zero $y,z$, but this clearly holds when one of them is $0$ as well. Thus, if we define $g(x)=f(x)-c$, we get that $g$ is additive. Now putting $g$ back in $Q(x,y)$, we get $$g(xy)+c=(g(x)+c)(g(y)+c)+c(1-g(x+y)-c)$$$$\implies g(xy) =g(x)g(y)+c(g(x)+g(y)-g(x+y))$$$$\implies g(xy)=g(x)g(y)$$for all $x,y$. Thus $g$ is both additive and multiplicative, and it is well known that the only functions satisfying this are the zero function and identity. If $g$ is identically $0$, then $f$ is constant, but from $P(x,y)$ we see that the only constant function satisfying the equation is $f$ identically $0$, which is impossible since $f(0) \neq 0$. Thus $g$ is identity, and so $f(x)=x+c$ for all $x$. Putting this in $P(x,y)$, we get the value of $c$ as $-1$, so the only solution in this case is $$\boxed{f(x)=x-1 \ \ \forall x \in \mathbb{R}}$$$\blacksquare$

BMO Shortlist 2021 A6 wrote: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(xy) = f(x)f(y) + f(f(x + y))$$holds for all $x, y \in \mathbb{R}$. The answer is only $\boxed{f(x) = 0}$ and $\boxed{f(x) = x - 1}$ for all $x \in \mathbb{R}$, both of which works. We'll now prove that there are no other solutions to the above functional equation. Let $P(x,y)$ be the assertion of $x$ and $y$ to the given functional equation. $P(x,0)$ gives us $f(0)(1 - f(x)) = f(f(x))$. Thus, we can rewrite $P(x,y)$ as \[ Q(x,y) : f(xy) = f(x)f(y) + f(0) (1 - f(x+y)) \]Claim 01. $f(x) + \alpha f(x - 1) = 1$ for some constant $\alpha \in \mathbb{R}$, or $f(x) = 0$ for all $x \in \mathbb{R}$. Proof. $Q(x,0)$ and $Q(x-1,1)$ gives us \[ f(0)(1 - f(x)) = f(x - 1)(1 - f(1)) \]Suppose that $f(0) = 0$. If $f(1) = 1$, then we have $f(f(x)) = 0$. However, $f(f(1)) = 1 \not= 0$, a contradiction. Else, $f(x - 1) = 0$ for all $x \in \mathbb{R}$, which implies the result. Therefore, we must have $f(x) + \frac{1 - f(1)}{f(0)} \cdot f(x - 1) = 1$, which implies the conclusion by setting $\alpha = \frac{1 - f(1)}{f(0)}$. Claim 02. $\alpha = -1$. Proof. $Q \left( - \frac{1}{2}, 2 \right)$ gives us \begin{align*} f(-1) &= f \left( - \frac{1}{2} \right) f(2) + f(0) \left( 1 - f \left( \frac{3}{2} \right) \right) \\ \alpha f(-1) &= \alpha f \left( -\frac{1}{2} \right) f(2) + \alpha f(0) \left( 1 - f \left( \frac{3}{2} \right) \right) \\ 1 - f(0) &= \alpha f \left( -\frac{1}{2} \right) (1 - \alpha + \alpha^2 f(0) ) + \alpha^2 \left( 1 - \alpha f \left( - \frac{1}{2} \right) \right) f(0) \\ 1 + (\alpha^2 - \alpha) f \left( - \frac{1}{2} \right) &= (1 + \alpha^2) f(0) \end{align*}$Q \left( 2, \frac{1}{2} \right)$ gives us \begin{align*} f(1) &= f(2) f \left( \frac{1}{2} \right) + f(0) \left( 1 - f \left( \frac{5}{2} \right) \right) \\ 1 - \alpha f(0) &= (\alpha^2 f(0) - \alpha + 1) \left( 1 - \alpha f \left( - \frac{1}{2} \right) \right) + f(0) \left( \alpha - \alpha^2 + \alpha^3 f \left( - \frac{1}{2} \right) \right) \\ \alpha &= (\alpha^2 - \alpha) f \left( - \frac{1}{2} \right) + 2 \alpha f(0) \end{align*}Therefore, we conclude that \begin{align*} \alpha + 1 &= (\alpha^2 + 1)f(0) + 2 \alpha f(0) = (\alpha + 1)^2 f(0) \end{align*}which therefore implies $\alpha = -1$ or $f(0) = \frac{1}{\alpha + 1}$. Suppose $\alpha \not= -1$, then write $f(0) = \frac{1}{\alpha + 1} = c$ for some constant $c$. It is immediate to get $f(0) = f(-1) = f(1) = c$ and therefore $\alpha = \frac{1 - c}{c}$. Then, we have \begin{align*} f(f(0)) &= f(0)(1 - f(0)) \implies f(c) = c(1 - c) \\ f(c + 1) &= 1 - \alpha f(c) = 1 - \frac{1 - c}{c} \cdot c(1 - c) = 2c - c^2 \\ f(-c - 1) &= f(-1 \cdot (c + 1)) = c(2c - c^2) + c(1 - c + c^2) = c^2 + c \\ f(-c) &= 1 - \alpha f(-c - 1) = 1 - \frac{1 - c}{c} \cdot (c^2 + c) = c^2 \end{align*}However, \begin{align*} f(c)f(-c - 1) + c(1 - c) = f(-c(c + 1)) &= f(-c)f(c + 1) + c(1 - c) \\ c^2(1 - c^2) &= c^2(2 - c^2) \implies c = 0 \end{align*}However, $f(0) = \frac{1}{\alpha + 1} \not= 0$ for any $\alpha \in \mathbb{R}$, a contradiction. Claim 03. $f(0) = -1$. Proof. As $\alpha = -1$, we have \begin{align*} f(f(x + 1)) &= f(f(x) + 1) = f(f(x)) + 1 = f(0)(1 - f(x)) + 1 \\ f(f(x + 1)) &= f(0)(1 - f(x + 1)) = -f(0)f(x) \end{align*}which therefore forces $f(0) = -1$. Therefore, we just need to find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(0) = -1$, $f(x + 1) = f(x) + 1$, and \[ f(xy) = f(x)f(y) + f(x+y) - 1 \]Note that \begin{align*} f(xy + x) &= f(x)f(y + 1) + f(x + y + 1) - 1 &= f(x)f(y) + f(x) + f(x+y) = f(xy) + f(x) + 1 \end{align*}which implies $f(y + z) = f(y) + f(z) + 1$ for any $y,z \in \mathbb{R}$. Thus, we have \[ f(xy) = f(x)f(y) + f(x) + f(y) = (f(x) + 1)(f(y) + 1) - 1 \]Writing $g(x) = f(x) + 1$, we get that $g$ is multiplicative and additive, and therefore $g(x) = x$ for all $x \in \mathbb{R}$, which implies $f(x) = x - 1$. Motivational Remark. Claim 01 is straightforward. To prove Claim 02, we want to try to use the "linearity" property of Claim 01 and somehow force it using multiplicative condition $f(x)f(y)$ on the problem to maybe cancel things out. Because we already used $y = 1$ to prove the linearity structure in Claim 01, using $y = 1$ again at $Q(x,y)$ might be useless. To force multiplicativity, we might consider plugging $y$ such that $|y| \ge 2$ -- apparently $y = 2$ is a good choice. Now, by picking $x = -\frac{1}{2}$ and $x= 2$ -- we could simplify all $\frac{1}{2} \mathbb{Z}$ in forms of $f \left( - \frac{1}{2} \right)$ and $f(0)$ from our first claim. Now, we get that either $f(0) = f(1)$ or $\alpha = -1$. The former case gives us a weird structure: $f(\mathbb{Z})$ being constant -- and apparently this has to be a solution for $Q(x,y)$. Therefore, we must somehow use $f(f(x)) = f(0)(1 - f(x))$ to eliminate this case. Playing with its structure for a while, we can get that as long as $x + y, z + w \in \mathbb{Z}$, then the condition forces $f(x)f(y) = f(z)f(w)$ if $xy = zw$

Let $P(x,y)$ denote the proposition that $f(xy) = f(x)f(y) + f(f(x + y))$ $\forall x, y \in \mathbb{R}$ $$P(x,0): f(0) = f(x)f(0) + f(f(x))$$If $f(0) = 0, f(f(x)) = 0$ $\forall x \in \mathbb{R} \implies f(xy) = f(x)f(y)$ from the original equation. Replace $x$ with $f(x) \implies f(f(x)y) = 0$ $\forall x, y \in \mathbb{R} $ . If $f$ is not identically zero, then selecting $x$ such that $f(x) \neq 0$ and varying $y$ still results in $f$ being identically zero. Contradiction, hence one solution that works is$$\boxed{f(x)=0 \ \ \forall x \in \mathbb{R}}$$Now suppose $f(0) \neq 0 \implies f(f(x)) = f(0)[1-f(x)] \textcircled{X} \implies f(xy) = f(x)f(y) + f(0)[1-f(x+y)]$ $$ \textcircled{1} P(1,1): f(1) = f(1)^2 + f(0)[1-f(2)] $$$$ \textcircled{2} P(1,2): f(2) = f(1)f(2) + f(0)[1-f(3)] $$$$ \textcircled{3} P(1,3): f(3) = f(1)f(3) + f(0)[1-f(4)] $$$$ \textcircled{4} P(1,4): f(4) = f(1)f(4) + f(0)[1-f(5)] $$$$ \textcircled{5} P(1,5): f(5) = f(1)f(5) + f(0)[1-f(6)] $$$$ \textcircled{6} P(2,2): f(4) = f(2)^2 + f(0)[1-f(4)] $$$$ \textcircled{7} P(2,3): f(6) = f(2)f(3) + f(0)[1-f(5)] $$With $7$ distinct equations and $7$ unknowns, we solve by substitution and find that $f(0) = -1, f(1) = 0.$ $f(0) = -1$ and $ \textcircled{X} \implies f(f(x)) = f(x) -1 \implies f(xy) = f(x)f(y) + f(x+y)-1\textcircled{A}$ $f(1) = 0$ and $\textcircled{A}$ $\implies f(x+1) = f(x) + 1$ $P(x+1, -1): f(-x-1) = -f(x)-3.$ Replace $x$ with $-x: f(x) = -f(-x)-2$ Now compare $P(x,y)$ vs $P(-x,-y): f(x)f(y) + f(x+y) -1 = [f(x)+2][f(y)+2]-f(x+y)-3$ Expanding and simplifying, $f(x+y) = f(x)+f(y)+1 \textcircled{B} $ Let $g(x) = f(x) +1 \implies g(x+y) = g(x) + g(y).$ This is Cauchy's equation, and we now prove that $g$ is bounded over a range. From $\textcircled{A}$ and $\textcircled{B},$ $$f(xy) = f(x)f(y) + f(x)+f(y) \implies f(xy)+1 = [f(x)+1][f(y)+1] \implies g(xy) = g(x)g(y)$$Then, replace $y$ with $x: g(x^2) = g(x)^2 \implies g(x) \geq 0$ $\forall x \geq 0 \implies g(x) = cx$ for some constant$c$. Substituting, $c=1$ and the other solution, which works, is $$\boxed{f(x)=x-1 \ \ \forall x \in \mathbb{R}}$$

Let $P(x,y)$ denote the assertion of $x$ and $y$ into the functional equation. \[P(x,0):\quad f(0)=f(x)f(0)+f(f(x))\Longrightarrow f(f(x))=f(0)-f(0)f(x)\quad (1)\]Now we use $(1)$ to rewrite the equation: $f(xy)=f(x)f(y)+f(0)-f(0)f(x+y)$. Let's denote $f(0)=a$ and $f(1)=b$. Also, let's denote \[P(x,1):\quad f(x)=f(1)f(x)+f(0)-f(0)f(x+1)=bf(x)+a-af(x+1)\quad (2)\]$\textbf{Claim 1:}$ If $a=0$, then $f\equiv 0$. $\textbf{Proof:}$ Using $(1)$ we have $f(f(x))=f(0)-f(0)f(x)=0$. Now if we look at $P(f(x),y)$ we get: \[f(f(x)y)=f(f(x))f(y)+f(0)-f(0)f(f(x)+y)\Longrightarrow f(f(x)y)=0\]If we assume that there exists a $t\in\mathbb{R}$, such that $f(t)\neq 0$, then substituting $x=t$ into $f(f(x)y)=0$ and letting $y$ be any real number we get that $f\equiv 0$, so we reach a contradiction. This however also implies that $f(x)=0\quad \forall x\in\mathbb{R}$, so we've proven the Claim. Note that obviously $\boxed{f\equiv 0}$ is obviously a solution, so we can assume that $a\neq 0$ from now on. Now we can calculate $f(1),f(2),f(3),f(4)$ by substituting $x=1,2,3$ into $(2)$:

If $a\neq -1$, then $f(4)=\frac{f(2)^2+a}{1+a}$ and we have:

$\underline{\textbf{Case 5:}}$ $a-b+1=0$. We implement the same strategy as above. We can transform $(2)$: \[f(x)=bf(x)+a-af(x+1)=(a+1)f(x)+a-af(x+1)\Longrightarrow f(x+1)=f(x)+1\]Note that $f(a)=f(f(0))=a-af(0)=a-a^2$. By induction, $(2)$ implies that $f(x+n)=f(x)+n\quad \forall n\in\mathbb{Z}$. Now $\forall m,n\in \mathbb{Z}$ we have that: \[f(mn)=f(m)f(n)+f(f(m+n))\Longrightarrow f(0)+mn=(f(0)+m)(f(0)+n)+f(a+m+n)\]\[\Longrightarrow a+xy=(a+x)(a+y)+f(a+x+y)=a^2+ax+ay+xy+a-a^2+x+y\]\[\Longrightarrow (a+1)(x+y)=0\Longrightarrow a=-1\]Now the functional equation becomes $f(xy)=f(x)f(y)+f(x+y)-1$. We proceed with the following claims: $\newline$ $\textbf{Claim 2:}$ If $f(z)=0$, then $z=1$. $\textbf{Proof:}$ Assume the contrary, that is, $z\neq 1$. Then $P\left(z,\frac{z}{z-1}\right)$ implies that: \[f\left(z\cdot \frac{z}{z-1}\right)=f(z)f\left(\frac{z}{z-1}\right)+f\left(z+\frac{z}{z-1}\right)-1\]\[\Longrightarrow f\left(\frac{z^2}{z-1}\right)=f\left(\frac{z^2}{z-1}\right)-1\]since $f(z)=0$. This is obviously a contradiction, so we've proven the Claim. $\textbf{Claim 3:}$ $f$ is injective $\textbf{Proof:}$ If $c\neq d$ and $f(c)=f(d)$, then $f(c-k)=f(c)-k=f(d)-k=f(d-k)$ for any integer $k$ using $(2)$. Choose such a $k$ so that $c-k\leq 1$. Then there exist real numbers $x_{1}$ and $x_{2}$ such that $x_{1}x_{2}+1=c-k$, $x_{1}+x_{2}=d-k$ (that's because the quadratic $x^2-(d-k)x+(c-k-1)$ has a non-negative discriminant because $c-k-1\leq 0$). However, now we have: \[P(x_{1},x_{2}):\quad f(x_{1}x_{2})=f(x_{1})f(x_{2})+f(x_{1}+x_{2})-1\]\[\Longrightarrow f(c-k-1)=f(x_{1})f(x_{2})+f(d-k)-1\Longrightarrow f(x_{1})f(x_{2})=0\]By Claim 2 it follows that at least one from the numbers $x_{1}$ and $x_{2}$ is equal to $1$ (WLOG $x_{1}=1$), but then $c-k=1\times x_{2}+1=1+x_{2}=d-k\Longrightarrow c=d$, contradiction with the assumption. Now if we notice that combining $(1)$ and $(2)$ we have that $f(f(x))=f(x)-1=f(x-1)$, but we showed that $f$ is injective, so $\boxed{f(x)=x-1\quad\forall x\in\mathbb{R}}$. It's easy to see that this is indeed also a solution: \[f(xy)=xy-1=(x-1)(y-1)+(x+y-2)=f(x)f(y)+f(f(x+y))\]Finally, all solutions are $\boxed{f(x)=0\quad \forall x\in\mathbb{R}}$ and $\boxed{f(x)=x-1\quad \forall x\in \mathbb{R}}$

This one was proposed by me. The story behind this problem is quite interesting, I think Basically, I was in 9th grade when I wrote down this random functional equation, and I was unable to solve it then. I tried it a couple of weeks, after which I gave up on it and decided I need more practice with functional equations. One year passed, then I suddenly remembered about the problem, after randomly seeing one of the sheets containing my failed attempts to solve it. It took me 5 hours to take it to the end and bam, I had the problem from A to Z. 3 more years passed, and I wasn't anymore in highschool, so I decided that, maybe, I could do something with this functional equation

Let $P(x,y)$ denote the assertion $f(xy)=f(x)f(y)+f(f(x+y))$. Note that $f\equiv 0$ is the only constant solution. Claim 1. $f(0)=0\implies f\equiv 0$ Proof. If $f(0)=0$, then $P(x,0)$ implies $f(f(x))=0$. If $f(1)\neq 0$, then $P(f(1),\tfrac 1{f(1)})$ yields contradiction. So now $P(x,1)$ implies $f(x)\equiv 0$. $\blacksquare$ Claim 2. $f(1)\neq 1$ Proof. If $f(1)=1$, then $P(1,-1)$ implies $f(f(0))=0$ and $P(0,f(0))$ implies $f(0)=0$. So $f\equiv 0$, a contradiction. $\blacksquare$ Now assume $f(0)\neq 0$ and $f$ is non constant. $P(x,0)$ implies $f(f(x))=f(0)-f(0)f(x)$ which compared with $P(x-1,1)$ implies $f(x)-1=kf(x-1)$ where $k=\tfrac{f(1)-1}{f(0)}$. Claim 2. function $f(x)-f(0)$ is additive Proof. Multiply $P(\tfrac xy,y)$ by $k$ and compare with $P(\tfrac xy+1,y)$ to get $f(x+y)=kf(x)+f(y)-kf(0)$. Symmetry shows $k=1$ since $f$ is non constant. So indeed $f(x)-f(0)$ is additive. $\blacksquare$ $P(x,x)$ easily implies $f(x^2)-f(0)=(f(0)-f(x))^2\geqslant 0$, and so $f(x)-f(0)$ is linear. Checking we get $f(x)=x-1$, which works.

As usual let $P(x;y)$ denote the given assertion. Setting $y=0$ we know that $f(f(x))=f(0)-f(0)f(x)$ If $f(0)=0$ the condition rewrites as $f(xy)=f(x)f(y)$ and $f(f(x))=0$ and assuming $f$ has a nonzero value we get that there is a nonzero number $\alpha$ such that $f(\alpha)=0$, and together with multiplicativity this gives $f \equiv 0$. It is easy to see that the only constant solution is $f \equiv 0$. Assume from now on that the function is not constant. Now setting $y=1$ we now know that $f(x)-f(1)f(x)=f(0)-f(0)f(x+1)$. Thus $f(x+1)=f(x)\frac{f(1)-1}{f(0)}+1$. And now let $\frac{f(1)-1}{f(0)} =a$. If $a \neq 1$ we can easily see that for natural inputs the function has the form $f(n)=ka^n+c$ by and substituting this in $P(x;y)$ we get that as long as $a \neq 1$ the function is constant. So now assume $a=1$ thus $f(x+1)=f(x)+1$. Now from $P(x+1;y)-P(x;y)$ we get that $f(x)+f(y)+1=f(x+y)$. (in other words $f+1$ is Cauchy) Implying $f(0)=-1$ and by substituting $f(f(x+y))$ to $f(x)+f(y)$ in the initial relation we also get that $f+1$ is multiplicative and thus positive over positives and thus $f+1$ is either $0$ or $x$ and substituting we get that the only working solutions are $f(x)=0$ and $f(x)=x$

The only solutions are $\boxed{f\equiv 0}$ and $\boxed{f(x) = x - 1}$. These work. Now we show that they are the only solutions. Let $P(x,y)$ be the given assertion. $P(x,0): f(0) = f(x) f(0) + f(f(x))$, so $f(f(x)) = f(0) - f(x) f(0) = f(0)(1 - f(x))$. Now, this implies that \[ f(xy) = f(x) f(y) + f(0) - f(x+y) f(0)\]Claim: If $f(0) = 0$, then $f$ is constant. Proof: $P(x,0)$ gives $f(f(x)) = 0$, so $P(x,y)$ implies $f(xy)= f(x)f(y)$. Now if $f(k) = 0$ for some $k\ne 0$, then $P(x,k): f(xk) = 0$, so $f\equiv 0$. Otherwise, if $f$ is injective at $0$, then $f(x) = 0$ for all $x$, absurd. $\square$ Henceforth assume $f(0) \ne 0$. Claim: There exists a real number $c$ with $f(c) \le \frac{c^2}{4}$. Proof: Suppose otherwise. Then $f(x) \ge 0$ for all reals $x$. Additionally, $f(f(x)) = f(0) (1 - f(x))$ must be greater than or equal to $0$, implying that since $f(0) > 0$, $f(x) \le 1$. Taking $c>2$ gives $f(c) \le 1 < \frac{c^2}{4}$ $\square$ Claim: $0$ is in the image of $f$. Proof: Now, with the $c$ in the claim above, let $f(c) = x(c-x)$ for some real number $x$ (this is possible as $f(c) \le \frac{c^2}{4}$) $P(x, c-x)$ gives that $f(x) f(c-x) = 0$, so $0\in \{f(x), f(c-x)\}$. $\square$ Claim: $f(1) = 0$ Proof: Let $f(k) = 0$ for some real number $k$. $P(k,1): f(f(k+1)) = 0$, so $f(0) (1 - f(k+1)) = 0\implies f(k+1) = 1$. $P(k+1,0): f(1) = 0$. $\square$ Now $P(1,1)$ gives that $f(f(2)) = 0$, so $f(0) (1 - f(2)) = 0\implies f(2) = 1$. $P(x,1): f(f(x+1)) = f(x)$, so $f(0) - f(x+1) f(0) = f(x)$. Claim: $f(0) \in \{-1,1\}$. Proof: Suppose $f(0) \ne -1$. Hence $f(0)(1 - f(3)) = f(2) = 1$, so $f(3) = 1 - \frac{1}{f(0)}$. $P(2,2): f(4) = 1 + f(0) - f(4) f(0)$, so $f(4)(1 + f(0)) = 1 + f(0)$. Then $f(4) = 1$. $P(3,1): f(3) = 0$. But this implies that $1 - \frac{1}{f(0)} = 0$, so $f(0) = 1$. $\square$ Case 1: $f(0) = 1$. Then $P(x,1)$ gives that $f(x) + f(x+1) = 1$, implying that $f(-1) = 0$. $P(x, -1): f(-x) = f(f(x - 1))$. However, $P(-x,1)$ gives that $f(-x) = f(f(1 - x))$, implying that $f(f(x-1)) = f(f(1-x))$, so $f(f(x)) = f(f(-x))$ for each $x$. But then, since $f(f(x)) = f(0) - f(0) f(x) = 1 - f(x)$, we have $1 - f(x) = 1 - f(-x)$, so $f$ is even. Now comparing $P(x,y)$ and $P(x,-y)$ gives that $f(f(x+y)) = f(f(x-y))$, implying that $f(f(x))$ is constant over all reals $x$. Since $f(0) = 1$ and $f(1) = 0$, we have $f(f(x)) = 0$ for all reals $x$. $P(x,y)$ implies that $f(xy) = f(x)f(y)$, so if $f(c) = 0$ for some $c$ ($c\ne 0$), then $P(c,0)$ gives that $f(0) = 0$, contradiction. Case 2: $f(0) = -1$ $P(x,1): f(x+1) - 1 = f(x)$. We also have \[ f(xy) = f(x)f(y) + f(x+y) - 1\]$P(x, y+1) - P(x,y): f(xy + x) - f(xy) = f(x) + 1$. For each $x\ne 0$, setting $y\to \frac yx$ gives that $f(x+y) = f(x) + f(y) + 1$ for all $x,y$ with $x\ne 0$, so it must be true for all $x,y$ (as it is obvious when $x$ or $y$ equals zero). Hence if $g(x)=f(x)+1$, we get $g(x+y)=g(x)+g(y)$. Now from the simplified $P(x,y)$, we find \[ f(xy)=f(x)f(y)+f(x)+f(y)\]Therefore $g(xy)=g(x)g(y)$. Now $g$ is multiplicative and additive, so it is linear and $g(1)=1$, $g(x)=x$ so $f(x)=x-1$.

Wow I actually unironically like this problem for once sometimes functional equations are actually OP So we claim (as everyone else did) that the solutions are: \begin{align*} f(x)\equiv\boxed{0},\boxed{x-1} \end{align*}Let $P(x,y)$ denote the given assertion as always.

So now we begin by proving some major facts about $f$. Step 1: Eliminating the $f(f(x+y))$ term $P(x,0)$ will immediately give you $f(0)=f(0)f(x)+f(f(x))$, so we actually have $f(f(x))=f(0)(1-f(x))$. Hence $P(x,y)$ reduces down to: \begin{align*} f(xy)&=f(x)f(y)+f(0)(1-f(x+y)) \end{align*}We’ll be using this as our main statement from now on and forget there used to be an $f(f(x+y))$ term in the equation. Remark 1: I almost got badly reverse-trolled because of trauma from BMOSL 2022/A4, thinking that the $f(f(x+y))$ term was actually going to be extremely useful somehow in getting a really short solution; it was actually Step 1.25: If $f(0)=0$, then $f\equiv0$ Well then $f(f(x))=0$ from Step 1, so we can eliminate the double $f$ and get $f(xy)=f(x)f(y)$. Now consider $P(f(x),y)$ for some $f(x)\neq0$. Then $f(f(x)y)=f(f(x))f(y)=0$, so by varying $y$, we get that for all $x\in\mathbb{R}$, $f(x)=0$, contradiction and end of proof. Step 2: Obtaining $f(x)=f(x+1)-1$ for all $x\in\mathbb{R}$ or another condition So this is the more interesting part. We are going to pull out another analysis-flavoured argument on this part First in interest of readability later let $f(n)=a_n$ for integers $n\in\mathbb{Z}$; the next part is going to be very funny. Consider $P(x,1)$; then note: \begin{align*} f(x)&=f(x)f(1)+f(0)(1-f(x+1))\\ f(x+1)&=f(x)\frac{f(1)-1}{f(0)}+1\\ f(x+1)&=kf(x)+1 \end{align*}So let’s determine the value of $k$. Notice that we actually must have $a_n=k^na_0+\frac{k^n-1}{k-1}$ if $k\neq1$ (let’s assume so for now) by induction. Then we must have by $P(n,2)$: \begin{align*} a_{2n}&=a_na_2+a_0(1-a_{n+2})\\ k^{2n}a_0+\frac{k^{2n}-1}{k-1}&=\left(a_0k^n+\frac{k^n-1}{k-1}\right)\left(a_0k^2+\frac{k^2-1}{k-1}\right)+a_0\left(1-\frac{k^{n+2}-1}{k-1}-k^{n+2}a_0\right)\\ k^{2n}a_0+\frac{k^{2n}-1}{k-1}&=\frac{(k^n-1)(k^2-1)}{(k-1)^2}+a_0\frac{2k^{n+2}-k^n-k^2}{k-1}-a_0\frac{k^{n+2}-1}{k-1}+a_0\\ k^{2n}a_0+\frac{k^{2n}-1}{k-1}&=\frac{(k^n-1)(k^2-1)}{(k-1)^2}+a_0\frac{k^{n+2}-k^n-k^2+1}{k-1}+a_0\\ k^{2n}a_0+\frac{k^{2n}-1}{k-1}&=\frac{(k^n-1)(k^2-1)}{(k-1)^2}+a_0\frac{(k^n-1)(k^2-1)}{k-1}+a_0\\ (k^{n}+1)a_0+\frac{k^{n}+1}{k-1}&=\frac{k+1}{k-1}(1+(k-1)a_0) \end{align*}So let’s begin comparing; notice that if $k\neq0,1$ then the LHS is not constant, at all, whilst the RHS is stoically constant. So in fact we must seek for $k=0,1$. If $k=0$ then $a_0-1=1-a_0$, i.e. $a_0=1$. Further we then must have $f(x+1)=1$ for all real $x$, so $f\equiv1$; unfortunately this does not satisfy the original equation. So we must have $k=1$, and our desired objective is reached. Here comes the entire crux of the problem: Step 3: $f(1)=0$, and only $x=1$ satisfies $f(x)=0$. So with the original equation, we actually have $f(0)(1-f(x+1))=-f(0)f(x)$, so the original statement becomes \begin{align*} f(xy)=f(x)f(y)-f(0)f(x+y-1) \end{align*}It suffices to show that $f(x)=-1$ iff $x=0$. First notice that we have if $n\in\mathbb{Z}$: \begin{align*} f(f(n))&=f(0)(1-f(n))\\ f(n+f(0))&=f(0)(1-f(n))\\ f(f(0))+n&=f(0)(1-n-f(0))\\ f(0)(1-f(0))+n&=f(0)(1-f(0))-nf(0)\\ n(1+f(0))&=0 \end{align*}so $f(0)=-1$ and thus $f(1)=0$. We can thus rewrite $P(x,y)$ again as $f(xy)=f(x)f(y)+f(x+y-1)$. So assume $f(z)=0$ and $z\neq1$. Using $P(z,z(z-1)^{-1})$, we have $xy=x+y$, and so $f(x)f(y)-1=0$, or $f(x)f(y)=1$. However since $x=z$, $f(x)f(y)=0$ as well, contradiction. Step 4: $f$ injective Assume $f(a)=f(b)$. Choose positive integer $N>0$ so big that there are two real numbers $x,y$ with $x+y=a+N+1$ and $xy=b+N$, which is possible by thinking about the discriminant of $x^2-(a+N+1)x+(b+N)=0$. Now we must have $f(x)f(y)=0$, so WLOGging $f(x)=0$, we must have $a+N=b+N$, or $a=b$, as desired. Step 4.01: $f\equiv x-1$ So we use the final form of $f(f(x))=f(x-1)$ to deduce that $f(x)=x-1$, done. Remark: The end is essentially what you do in IMO 2017/2 actually.

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