As a remark, the condition $f(0)\neq 0$ is not needed, but the solution becomes somewhat messier.

We will prove that the only solution is $f \equiv 2$. Let $x=y=0$ in the given equation to obtain $2f(f(0))=f(0)^2$. Let $y=0$ in the given equation to obtain $f(f(x))=f(0)f(x)-f(f(0))$. Therefore, the given equation rewrites as $(f(x)+f(y))f(0)=f(0)^2+f(x+y)f(xy), \,\,\, (1)$. Now take $x=1,y=-1$ in this equation to obtain that $f(1)f(0)+f(-1)f(0)=f(0)^2+f(0)f(-1),$ and since $f(0) \neq 0$ we conclude that $f(1)=f(0)$. Thus, taking $y=1$ in $(1)$, we obtain $f(x)f(0)+f(0)^2=f(0)^2+f(x)f(x+1),$ and so $f(x)=0$ or $f(x+1)=f(0)$ for all $x$. The pith of the problem is the following Claim: Claim: $f(x) \neq 0$ for all $x$. Proof: Assume that $f(u)=0$ for some $u$. Then, $f(0)=f(f(u))=f(0)f(u)-f(f(0))=-f(f(0))=-\dfrac{f(0)^2}{2},$ and so we must have $f(0)=-2$. Note that this implies that $f(f(x))=-2f(x)-2$, and so $f(-2)=f(f(0))=f(f(1))=-2f(1)-2=2,$ hence $f(2)=f(f(-2))=-2f(-2)-2=-6,$ which is a contradiction since $f(1) \neq 0$ and $-6=f(2) \neq f(0)=-2$ $\blacksquare$ Now, from the Claim we conclude that $f(x+1)=f(0)$ for all $x$, and so $f$ is constant which easily implies that $f \equiv -2$, which evidently works.

Groupsolved with Strudan_Borisov, Pitagar, Maths_Girl, Iliyas Noman. Here's the solution: Let $P(x,y)$ denote the assertion of $(x,y)$ into the functional equation. $P(x,0)$ gives \[f(f(x)) = f(x)f(0)-\frac{1}{2}f(0)^2\quad (1)\]whence we can rewrite the original FE as \[f(x)f(0)+f(y)f(0)-f(0)^2 = f(x+y)f(xy)\]Define the function $g\colon \mathbb{R}\to\mathbb{R}$ as $g(x) = \frac{f(x)}{f(0)}$ which is real as $f(0)\neq 0$. The last equation then becomes \[g(x)+g(y)-1=g(x+y)g(xy)\]Denote by $Q(x,y)$ the assertion of $(x,y)$ in this variant of the FE. Then $Q(x,-x)$ gives: \[g(x)+g(-x)-1=g(0)g(-x^2)=g(-x^2)\]Plugging in $x = 1$ implies that $g(1) = 1\Longrightarrow f(1) = f(0)$. Now $Q(x,1)$ gives \[g(x) = g(x+1)g(x)\quad (\star)\]Case 1. $0\not\in \text{Im}(f)$. Then the starred equation implies that $g(x+1) = 1$ for all $x$, so $f(x) = f(0) = c$ for all $x\in\mathbb{R}$, so the FE becomes $2c = c^2\Longrightarrow c = 2$ and $f\equiv 2$ is indeed a solution. Case 2. $\exists x_{0}\in\mathbb{R}$ such that $f(x_{0}) = 0$. Plugging in $x = x_{0}-1$ into the starred equation implies that $f(x_{0}-1) = 0$ as well and continuing inductively, $f(x_{0}-n) = 0$ for all positive integers $n$. Now plugging in $Q(x_{0}, -n)$ gives \[g(x_{0})+g(-n)-1=g(x_{0}-n)g(nx_{0})\Longrightarrow g(-n) = 1\Longrightarrow f(-n) = f(0) \quad \forall n\in\mathbb{N}\]Also the starred equation implies that if $g(x)\neq 0$, then $g(x+1) = 1$, so as $f(1) = f(0) \neq 0$, we inductively get that $g(n) = 1\Longrightarrow f(n) = f(0)$ for all $n\in\mathbb{N}$ as well. Therefore $f$ is constant over $\mathbb{Z}$. Plugging $x_{0}$ into $(1)$ implies that: \[f(0) = 0 - \frac{1}{2}f(0)^2\Longrightarrow f(0) = -2\]Thus $f(n) = f(0) = -2$ for all integers $n$, but $P(x,y)$ for $x,y\in\mathbb{Z}$ derives a contradiction as \[-4 = f(-2)+f(-2) = f(f(x))+f(f(y)) = f(x+y)f(xy) = (-2)^2 = 4\]Finally, there are no solutions in this case, so the only function satisfying both conditions (the functional equation and $f(0) \neq 0$) is $f(x) = 2$ $\forall x\in\mathbb{R}$ which clearly works.

Solved with GavinLoveMath. We claim the only solution is $\boxed{f\equiv 2}$, which works. Now we prove it's the only one. Let $P(x,y)$ denote the given assertion. $P(0,0): 2f(f(0)) = f(0)^2$. $P(x,0): f(f(x)) + f(f(0)) = f(x)f(0)$, so $f(f(x)) = f(x)f(0) - f(f(0))$. $P(x,1): f(f(x)) + f(f(1)) = f(x+1)f(x)$. Substituting $f(f(x))$ with $f(x)f(0) - f(f(0))$ gives $f(x)f(0) + f(f(1)) - f(f(0)) = f(x+1)f(x)$. Setting $x = -1$ here gives that $f(f(1)) - f(f(0)) =0 $. Therefore, $f(x)f(0) = f(x+1)f(x)$ for all $x$, so for each $x$, either $f(x)= 0$, or $f(x+1) = f(0)$. Now we prove that $f$ cannot have any zeroes. Suppose it did. Let $a$ denote a zero of $f$. $P(a,0): f(f(0)) + f(0) = 0\implies f(f(0)) = -f(0)$. In fact, putting this into $P(0,0)$ gives $-2f(0) = f(0)^2$, so $f(0) = -2$. $P(a,1): f(0) + f(f(1)) =0 \implies f(f(1)) = -f(0) = 2$. $P(0,1): -2f(0) = f(0)f(1)\implies f(1) =-2$. Since $f(f(0)) = -f(0)$, we have $f(-2) = 2$. $P(1,1): 2f(f(1)) = f(1)f(2)$, so $f(1)f(2) = 4\implies f(2) = -2$. Hence we have \[ f(f(x)) = f(x)f(0) - f(f(0)) = -2f(x) - 2\]for all reals $x$. Now setting $x = -2$ here gives $f(2) = -2 \cdot f(-2) - 2 = -6$, absurd since $f(2)= -2$. Therefore, there are no zeroes of $f$. The rest is simple because it implies $f(x+1) = f(0)$ for all $x$, which means $f$ is constant. Checking, we see $f\equiv 2$ is the only solution.

Solution: Substitute x = 0 = y, we get that 2(f(f(0))) = f(0)^2. Then substitute y = 0, and this results in f(f(x)) = -f(f(0)) + f(x)*f(0). Substitute the new expression for f(f(x)) into the expression, we get that f(0)(f(x) + f(y)) = f(x+y)f(xy)+ f(0)^2. Plug that x = -1 and y = 1. We have that f(0) = f(1) always true after we substitute.Then just make y = 1, we have that f(x+1)*f(x)=f(0)*f(x) for all x. Then f(x+1) is always constant and equal to f(0). Plug f(x) = c into the equation we get that f(x)= 2. So the functions are f(x)=2

First let f(0)=a≠0 P(0,0) ==> f(a)=a^2/2 P(x,0) ==> f(f(x))=f(x)a-a^2/2 ==> f(x)a+f(y)a-a^2=f(x+y)f(xy) Now let g(x)=f(x)/a and g(0)=1 ==> g(x)+ g(y)-1=g(x+y) g(xy) G(1,-1) g(1)= g(0)=1 G(x,1/x) g(x)+ g(1/x)-1=g(x+y) Also now let h(x)= g(x)-1 ==> h(x)+h(1/x)=h(x+1/x) we can found h(x)=xc for some c and g(x)=xc+1 and because g(1)=1 ==> c=0 and g(x)=1 and also f(x)=a for all x ===> 2a=a^2and because a≠0==> a=2 and f(x)=2 it done .

Good problem (my $50^{th}$ post). Let $f(0)=c$. $P(0,0)$ gives $f(c)=c^2/2$. $P(x,0)$ gives $f(f(x))+c^2/2=cf(x)$ if $f(x)=t$ we have $f(t)=ct-c^2/2$, then if there is $z$ such that $f(z)=k$, then $f(k)=ck-c^2/2$ $(*)$. Then we have $f(f(x))+f(f(y))=cf(x)+cf(y)-c^2=f(x+y)f(xy)$, then $P(-1,1)$ gives $cf(1)+cf(-1)-c^2=cf(-1)$, which means that $f(1)=c$. Here we need this claim. Claim: there is a no $r$ such that $f(r)=0$. Suppose that $f(r)=0$, then according to $(*)$ we have $c=(-c^2)/2$, then $c=-2$ (since $c$ is not zero). And according to $(*)$ we have $f(c)=c^2/2$, then $f(-2)=2$. Then $f(-2)=2$, then $P(0,1)$ gives $f(1)=-2$, then $P(-2,1)$ gives $f(-1)=-2$, then $P(-1,-1)$ gives that $f(-2)=-2$, which is contradiction to $f(-2)=2$. Claim is proved (yeah I know, my proof is ugly). Then $P(x,1)$ gives that $cf(x)=f(x+1)f(x)$, since $f(x)$ is not zero, we have $f(x+1)=c$, checking it in $P(x,y)$ gives that $c=2$. Then answer is $f(x)\equiv 2$. Comment: Oh, it is BMO 2022 A4 (I saw that it is Azerbaijan BMO TST problem), I have seen it now, i was thinking that it is normal FE(when I was solving:)).

The only solution is $f(x)=2$ for every $x \in \mathbb{R}.$ Let's prove it: Let $P(x,y)$ be the assertion. $P(x,0)$ gives $$f(f(x))+f(f(0))=f(x)f(0) (*).$$$P(x,-x)$ gives $$f(f(x))+f(f(-x))=f(0)f(-x^{2}) (**).$$If we let $x=-1,$ from $(*)$ and $(**)$ we get $f(f(0))=f(f(1)).$ Since $f(f(1))=f(f(0))$ and $f(0)\neq 0$, from $P(1,y)$ you get $$f(f(1))+f(f(y))=f(y)f(y+1)=f(0)f(y).$$Now let's prove $f(y)$ can't be equal to 0: If $f(y)=0$ then from $(*)$ we have $f(0)=-f(f(0)).$ From $P(0,0)$ we have $-2f(0)=f(0)^{2}\implies f(0)=-2\implies f(f(0))=2.$ From $(*)$ we have $$f(f(x))=-2f(x)-2\implies f(x)=-2x-2.$$Put it to first equation and we have $4(x+y+1)=4(x+y+1)(xy+1)$ meaning it would only work for $x=0$, $y=0$ or $x+y=-1$ leading to a contradiction meaning $f(x)\neq0.$ $f(0)=f(y+1)\implies f(y+1)=c$ where $c$ is a real number. Now the first equation becomes $2c=c^{2}$ and since $c\neq0$ we have $f(x)=2$ as the only solution.

Captain_Baran wrote: ...$f(f(1))+f(f(y))=f(y)f(y+1)=f(0)f(y)$ $f(0)=f(y+1)\implies f(y+1)=c$... @Captain_Baran I think you must prove that $f(y)$ is not zero

oVlad wrote: As a remark, the condition $f(0)\neq 0$ is not needed, but the solution becomes somewhat messier. I think the remark is needed. I haven't characterized all of the solutions yet, but you can show that for example: $$f(x) = \left\{\begin{matrix}0,&x\ne 2\\1,& x=2\end{matrix}\right.$$ Works.

tadpoleloop wrote: I think the remark is needed. I can assure you that $f(0)\neq 0$ is not needed. The solution you found works and you can indeed characterize all of the other solutions in one line.

oVlad wrote: tadpoleloop wrote: I think the remark is needed. I can assure you that $f(0)\neq 0$ is not needed. The solution you found works and you can indeed characterize all of the other solutions in one line. I found a solution where $f(0)=0$ so the condition was needed. Your remark implies that $f(0)\ne 0$ regardless, but that isn't the case.

oVlad wrote: Find all functions $f : \mathbb{R} \to\mathbb{R}$ such that $f(0)\neq 0$ and \[f(f(x)) + f(f(y)) = f(x + y)f(xy),\]for all $x, y \in\mathbb{R}$. oVlad wrote: As a remark, the condition $f(0)\neq 0$ is not needed, but the solution becomes somewhat messier. We will characterize all of the functions that satisfy the assertion $f(f(x)) + f(f(y)) = f(x + y)f(xy), $ with $f(0)=0$ Clearly $\boxed{f(x) = 0 \quad\forall x}$ is a solution, so we will assume that $f$ is not constant throughout. Claim: $f$ is a solution to the given assertion if and only if $\exists S\ne\emptyset$ (support), $Z = \mathbb R - S$ (zeroes) with properties 1) $S \subseteq (0,4)$ 2) $x<2\sqrt{y}\quad \forall x,y\in S$ and an arbitrary function $g:\mathbb R \to Z-\{0\}$ such that $$\boxed{f(x) = \left\{\begin{matrix} g(x)&x\in S\\0&x\in Z\end{matrix}\right.}$$

For #1 (f(0)$\neq$0): Let $P(x,y)$ be the assertation $f(f(x)) + f(f(y)) = f(x + y)f(xy)$ $P(0,0) \Rightarrow 2f(f(0))=f(0)^2$ $P(x,0) \Rightarrow f(f(x))=f(x)f(0)-f(f(0))$ $P(x,y) \Rightarrow (f(x)+f(y))f(0)-f(0)^2=f(x+y)f(xy)$ $P(1,-1) \Rightarrow (f(1)+f(-1))f(0)-f(0)^2=f(0)f(-1) \Rightarrow f(1)=f(0)$ $P(x,1) \Rightarrow (f(x)+f(1))f(0)-f(0)^2=f(x+1)f(x) \Rightarrow$ $f(x)f(0)=f(x+1)f(x) \Rightarrow f(x)=0 \text{ or } f(x+1)=f(0), \forall x \in \mathbb R$ Let $a \in \mathbb R$ be such that $f(a)\neq 0 ,f(a)\neq f(0) \Rightarrow$ $P(a,1) \Rightarrow f(a+1)=f(0); P(a-1,1) \Rightarrow f(a-1)=0; P(a-2,1) \Rightarrow f(a-2)=0$ $P(a,-1) \Rightarrow f(a)f(0)+f(-1)f(0)=f(0)^2 \Rightarrow$ $f(a)=f(0)-f(-1) \Rightarrow f(0)\neq f(-1)\Rightarrow P(-2,1) \Rightarrow f(-2)=0$ $P(a,-2) \Rightarrow f(a)f(0)-f(0)^2=0 \Rightarrow f(0)=f(a)=f(0)-f(-1) \Rightarrow$ $f(-1)=0 \Rightarrow f(a)=f(0) \Rightarrow Contradiction$ This means that $f(x)=0$ or $f(x)=f(0)$, $\forall x \in \mathbb R$ Let there exist $u$ and $v$ $\in \mathbb R$ such that $f(u)=0, f(v)=0$; $P(u,v) \Rightarrow -f(0)^2=f(u+v)f(uv) \Rightarrow$ Both when $f(u+v)=0$ or $f(uv)=0$ and $f(u+v)=f(0), f(uv)=f(0)$ we get a $contradiction$ $\Rightarrow$ There exist at most one $w \in \mathbb R$ for which $f(w)=0$; $P(w,2) \Rightarrow f(w)f(0)=f(w+2)f(2w)=f(0)^2 \Rightarrow f(w)=f(0)\Rightarrow Contradiction$ We got that $f(x)=f(0) \Rightarrow 2f(0)=f(0)^2 \Rightarrow f(0)=2 \Rightarrow$ $\boxed{f(x)=2 \; \forall x \in \mathbb R}$ is the only solution.

Let $f(0)=c$. $1.P(x,-x)$ $\implies f(f(x))+f(f(-x))=f(0)f(-x²)$,$2.P(x,0) \implies f(f(x))+f(c)=cf(x)$. Add $1$ and $2$ but this time take $f(f(-x)),f(c)$ and $2f(f(x))$ to get: $c(f(-x^2)-f(x)-f(-x)+c)=0$. So,we have $f(-x^2)=f(x)+f(-x)-c$.$P(1)$ implies $f(1)=c$,and we have two cases: $1. f(x)=0$ $2.f(x)=c$ and $c$ ≠ $0$.Assume there is an $f(u)=0$ and $c$ ≠0,and this will cause a contradiction.So $f(x)=2$ for all $x$.

Let $P(x,y)$ be the assertion in the given equation. Then, $$P(0,0) \Rightarrow f(f(0)) + f(f(0)) = f(0)^2.$$Let $f(0) = \alpha$; then, we have $f(\alpha) = \frac{\alpha^2}{2}.$ $$P(x,0) \Rightarrow f(f(x)) + f(\alpha) = \alpha f(x) \Rightarrow f(f(x)) = \alpha f(x) - \frac{\alpha^2}{2}. $$Then our initial equation can be written as $\alpha (f(x) + f(y)) =\alpha^2 + f(x+y)f(xy).$ Now setting $P(1,-1)$ gives $\alpha (f(1) + f(-1)) = \alpha^2 + \alpha f(-1)$, and as $\alpha \ne 0$ we get $f(1) = \alpha.$ $$P(1,1) \Rightarrow 2\alpha f(1) = \alpha^2 + \alpha f(2), f(2) = \alpha$$We can continue this way and by induction we get $P(1,r-1) \Rightarrow f(r) = \alpha$ for all $r \in \mathbb{R}^{+}$. Now take $P(1,-2)$ we will similarly get $f(-1) = \alpha$. Now again by induction we can see that $P(1,r-1)$ implies $f(r) = \alpha$ for all non-positive $r$. Since we now have $f(r) = \alpha$ for all real $r$'s we can write $f(\alpha) = \frac{\alpha^2}{2} = \alpha \Rightarrow \alpha = 2$. So the final answer is $\boxed{f(x) = 2}$ for all $x \in \mathbb{R}$.

$$ P(0;0) \Longrightarrow 2f(f(0)) = f^2(0) $$$$ P(x; 0) \Longrightarrow f(f(x)) + \frac{1}{2} f^2(0) = f(x) \cdot f(0)$$$$ f(f(x)) = f(0) (f(x) - \frac{1}{2} f(0)) $$ $$f(0) (f(x) + f(y) - f(0)) = f(x+y) \cdot f(xy)$$ $$ P(x; -x) \Longrightarrow f(0) (f(x) + f(-x) - f(0)) = f(0) \cdot f(-x^2) $$$$ f(x) + f(-x) = f(-x^2) + f(0) $$$$ P(1) \Longrightarrow f(1) + f(-1) = f(-1) + f(0) $$$$ f(1) = f(0) $$ $$ P(x; 1) \Longrightarrow f(0) \cdot f(x) = f(x+1) \cdot f(x) $$$$ f(x+1) = f(0) = const = 2$$

As usual let $P(x;y)$ denote the given assertion. Substituting $0$ for $y$ we get $f(f(x))=f(0)f(x)-f(f(0))$ and from $P(0;0)$ we can substitute $f(f(0))$ with $\frac{f(0)^2}{2}$. Now replacing we get that $P(x;y)$ with $f(0)(f(x)+f(y))-f(0)^2=f(x+y)f(xy)$ and substituting 1 and -1 we get that $f(0)=f(1)$. And $P(x;1)$ Implies that either $f(0)=f(x+1)$ or $f(x)=0$ And now assuming that 0 is in the image of $f$ we get that $f(f(0))+f(0)=0$ meaning $f(0)=-2$ this easily implies $f(x)=-2$ for any positive integer input $f$ is -2 and we easily get from here that $f$ is 0 for negative inputs by proving that $f(f^{-1}(0)-n)$ is 0, and now just substituting any integers we get that 4=-4 and thus finally we got that $f$ is constant, thus $f \equiv 2$.

\[f(f(x))+f(f(y))=f(x+y)f(xy)\]$P(1,f(1))\implies 2f(f(1))=f(f(1)+1)f(f(1))\iff f(f(1))(f(f(1)+1)-2)=0$ If $f(f(1))=0,$ then $P(f(1),1)$ would give $f(0)=f(f(1))=0$ which is impossible. Thus, $f(f(1)+1)=2$. $P(f(1),1)\implies f(f(f(1)))+f(f(1))=2f(f(1))\iff f(f(f(1)))=f(f(1))$ Comparing $P(f(1),y)$ with $P(1,y)$ gives \[f(1+y)f(y)=f(f(1))+f(f(y))=f(f(f(1)))+f(f(y))=f(f(1)+y)f(f(1)y)\]Plugging $y=0$ gives $f(0)f(1)=f(f(1))f(0)\iff f(1)=f(f(1))$ since $f(0)\neq 0$. If we replace $y=1,$ then we get $f(1)f(2)=2f(f(1))\iff f(1)(f(2)-2)=0$. If $f(1)=0, \ P(f(1),1)\implies 2f(0)=0$ which is not true thus, we have $f(2)=2$. $P(2,0)\implies 2+f(f(0))=2f(0)\iff f(f(0))=2f(0)-2$ $P(0,0)\implies 4f(0)-4=f(0)^2\iff (f(0)-2)^2=0\iff f(0)=2$ $P(x,0)\implies f(f(x))=2f(x)-2$ and plugging $x=1$ yields $f(1)+2=f(f(1))+2=2f(1)$ hence $f(1)=2$ $P(x,1)\implies 2f(x)=f(f(x))+2=f(x+1)f(x)\iff f(x)(f(x+1)-2)=0$ If $f(a)=0$ where $a\neq 0,$ then $x=1,y=a$ gives $4=0$ which is impossible. Thus, $f(x)\neq 0$. This gives that $f(x+1)=2$. So $\boxed{f\equiv 2}$ is the only solution.$\blacksquare$

The only solution is $f(x)=2 \ \forall \ x \in \mathbb{R}$. Clearly this choice of $f$ works, now we will prove that this is the only one. Let $P(x,y)$ be the assertion given in the problem. From $P(x,0)$ we have $$f(f(x))+f(f(0))=f(x)f(0) \Rightarrow f(f(x))=f(x)f(0)-f(f(0))$$. In particular, for $x=0$ the equation above yields $2f(f(0))=f(0)^2$. Substitute $f(f(x))=f(x)f(0)-f(f(0))$ to $P(x,y)$ and also use $2f(f(0))=f(0)^2$ to get $$f(x)f(0)+f(y)f(0)-2f(f(0))=f(x+y)f(xy) \Rightarrow f(0)(f(x)+f(y))=f(x+y)f(xy)+f(0)^2$$Let $Q(x,y)$ be the assertion of $f(0)(f(x)+f(y))=f(x+y)f(xy)+f(0)^2$. From $Q(1,-1)$, we have $$f(0)(f(1)+f(-1))=f(0)f(-1)+f(0)^2 \Rightarrow f(0)f(1)=f(0)^2$$. Since $f(0)\neq 0$, we have $f(1)=f(0)$. Next, from $Q(x,1)$ and using $f(1)=f(0)$ we have $$f(0)(f(x)+f(1))=f(x+1)f(x)+f(0)^2 \Rightarrow f(0)f(x)=f(x+1)f(x)$$Let $f(0)f(x)=f(x+1)f(x)$ be equation $(1)$. Claim. $f(x) \neq 0\ \forall \ x \in \mathbb{R}$ Proof. Suppose there exist $x_0 \in \mathbb{R}$ such that $f(x_0)=0$. Then, plugging $x=x_{0}-1$ to equation $(1)$ yields $f(x_0 - 1)=0$ since $f(0) \neq 0$. By Induction, we can show that $f(x_0 - n)=0 \ \forall \ n \in \mathbb{N}$. From $Q(x_0,-n)$ with $n \in \mathbb{N}$ and using $f(x_0 - n)=0 \ \forall \ n \in \mathbb{N}$, we have $$f(0)(f(x_0)+f(-n))=f(x_0 - n)f(-x_{0}n)+f(0)^2 \Rightarrow f(0)f(-n)=f(0)^2$$. So $f(-n)=0 \ \forall \ n \in \mathbb{N}$. But $Q(-x,-y)$ with $x,y \in \mathbb{N}$ yields $f(0)^2=0$, a contradiction. As desired. So, equation (1) yields $f(x+1)=f(0)$. Hence, $f$ is constant. Easy to check that only $f(x)=2 \ \forall \ x \in \mathbb{R}$ satisfies. We are done .

I claim the answer is only $f(x) = 2$, which obviously works. Substitute $y = 0$, then we get $f(f(x)) = f(x)f(0) - f(f(0))$, we can plug this in to get $(f(x) + f(y))f(0) = f(x+y)f(xy) + 2f(f(0))$, then note $f(f(0)) = f(0)^2 - f(f(0))$, so we can write this as $(f(x) + f(y))f(0) = f(x + y)f(xy) + f(0)^2$. Plugging in $x, -x$ gives $f(x) + f(-x) = f(-x^2) + f(0)$, $x = 1$ then gives $f(1) = f(0)$. Now we also have $f(x)f(0) + f(1)f(0) = f(x + 1)f(x) + f(0)^2$, which gives $f(0) = f(x + 1)$ OR $f(x) = 0$. We now divide into cases. If $f(x) = 0 $ is never true, then $f(x + 1) = f(0)$, so $f(x) = c$, clearly $c^2 = 2c$ so $c = 2$. Now we consider the other case where $f$ does have a root. Now we first prove that $f(x) = f(0)$ over all integers. Clearly, since $f(1) \neq 0$, we can utilize $f(1+ 1) = f(0)$, and then repeat this inductively to get $f(n) = 0$ for nonnegative $n$. Now assume some negative $a$ satisfies $f(a) \neq f(0)$, then $f(a - 1) = 0$, so we have $P(a - 1, a -1)$ gives $2f(0) = f(2a - 2) f((a-1)^2)$, then inductively note that $f(k) = 0$ implies $f(k) \neq f(0)$ implies $f(k - 1) = 0$, so we have $2f(0) = 0 $, contradiction so $f(n) = f(0)$ for all integers $n$. Now to finish off, consider $f(a) = 0$, set $a = x + y$, then $f(x) + f(y) = f(0)$, then let $x$ vary over all integers we get $f(a + n) = 0 $ for all integers $n$. Now consider $f(a) = 0$, then $f(\frac a2) + f(\frac a2) = f(0)$, so $f(\frac a2) \neq f(0)$ so $f(\frac a2 - 1) = 0$, however this contradicts what we got before in the previous sentence, so $f$ cannot have a root and this case is dead.

$P(x,y) : f(f(x))+f(f(y))=f(x+y)f(xy)$ The answer is $f\equiv 2$. From $P(0,0)$ we get that $2f(f(0))=f(0)^{2}$ and from $P(x,0)$ we get \begin{align*} f(f(x))&=f(x)f(0)-f(f(0))\\ &=f(x)f(0)-\frac{f(0)^{2}}{2} \ \ \dots \textbf{(1)} \end{align*} Combining these two in the original equation we get: $$Q(x,y):f(0)(f(x)+f(y))=f(0)^{2} + f(x+y)f(xy)$$ $Q(1,-1) \Rightarrow f(0)(f(1)+f(-1))= f(0)^{2} + f(0)f(-1)$ so $f(1)=f(0)$. $$Q(x,1) \Rightarrow f(0)f(x) = f(x)f(x+1), \forall x\in\mathbb R \dots \textbf{(2)}$$ Claim: $f(x) \neq 0$ for all $x\in \mathbb R$ Proof: assume FTSOC that there exists $x$ such that $f(x) = 0$. Plug in $x$ into $\textbf{(1)}$ to get $f(0) = -\frac{f(0)^{2}}{2}$. Hence $f(0) = -2 = f(1)$. Plug in $x=0$ into $\textbf{(1)}$ to get $f(-2)=2$. Now, from $Q(-1,-1)$ we get $-4f(-1)=4+f(-2)f(1)=0$. So $f(-1)=0.$ Finally, from $Q(1,-2)$ we get $0=4$, a contradiction. Now going back to $\textbf{(2)}$ we have that $f(0)=f(x+1)$ for every $x\in\mathbb R$. So $f$ is constant and substituting back in the original equation we get $f\equiv2$.

Hertz wrote: $$ P(0;0) \Longrightarrow 2f(f(0)) = f^2(0) $$$$ P(x; 0) \Longrightarrow f(f(x)) + \frac{1}{2} f^2(0) = f(x) \cdot f(0)$$$$ f(f(x)) = f(0) (f(x) - \frac{1}{2} f(0)) $$ $$f(0) (f(x) + f(y) - f(0)) = f(x+y) \cdot f(xy)$$ $$ P(x; -x) \Longrightarrow f(0) (f(x) + f(-x) - f(0)) = f(0) \cdot f(-x^2) $$$$ f(x) + f(-x) = f(-x^2) + f(0) $$$$ P(1) \Longrightarrow f(1) + f(-1) = f(-1) + f(0) $$$$ f(1) = f(0) $$ $$ P(x; 1) \Longrightarrow f(0) \cdot f(x) = f(x+1) \cdot f(x) $$$$ f(x+1) = f(0) = const = 2$$ I think you need to prove that $f(x)$ is never zero.

my eyes hurt from using aops too much i wanna rest We will denote the assertion in the problem by $P(x,y)$. $P(x,0)$ yields that $f(f(x))+f(f(0))=f(x)f(0)$. In particular, $P(0,0)$ tells us that $f(f(0))=\frac{f(0)^2}{2}$. Suppose $f(0)=c\neq 0$. The previous two statements imply that $f(f(x))=cf(x)-\frac{c}{2}$. Substituting this back into $P(x,y)$, we have $c(f(x)+f(y))-c^2=f(x+y)f(xy)$. We will replace $P(x,y)$ from now on with this statement. $P(1,-1)$ tells us that $c(f(1)+f(-1))-c^2=f(-1)c$, expanding, $cf(1)=c^2$, or $c=f(1)$. Thus, $f(0)=f(1)$. Now, using $P(x,1)$, we have $c(f(x)+c)-c^2=f(x)f(x+1)$, or simplifying, $cf(x)=f(x)f(x+1)$. Key Claim: $f$ is never $0$. Proof: Suppose for the sake of contradiction that $f(\alpha)=0$. Looking at the RHS of $P(x,y)$, if it is every $0$, then $c(f(x)+f(y))=c^2$, or $f(x)+f(y)=c$. We see that we will achieve this whenever $x+y=\alpha$. Now, the particular case of $x=1$ yields $f(1)+f(\alpha-1)=c$, but as $f(1)=f(0)=c$, $f(\alpha-1)=0$. Now, comparing $f(x)+f(\alpha-x)=c$ with the analogous expression for $\alpha-1$, which is that $f(x)+f(\alpha-x-1)=c$, we have $f(\alpha-x)=f(\alpha-x-1)$, or taking $x\to x+\alpha$, $f(x)=f(x-1)$ for all $x$, or that $f$ is periodic. Now compare $P(x+1, y-1)$ with $P(x,y)$, The LHS on each actually remains the same as $f(y)=f(y-1)$ and $f(x)=f(x+1)$ by periodicity. Now, comparing the RHS, and cancelling $f(x+u)=f(x+1+y-1)$, we get $f(xy)=f(xy+y-x-1)=f(xy+y-x)$, where the final euqality is by periodicity. However, by symmetry, we may swap around $x$ and $y$. Thus, for all real $x$ and $y$, we have $f(xy+x-y)=f(xy+y-x)$. Lemma: For any $u,v\geq 0$, we may find suitable reals $x$, $y$ such that $xy+x-y=u$ and $xy-x+y=v$. This is equivalent to solving the system of equations for $x,y$, $\frac{u+v}{2}=xy$ and $\frac{u-v}{2}=x-y$. Let $\frac{u+v}{2}=d$ and $\frac{u-v}{2}=c$. Then, $y+c=x$, so we just want to solve $y(y+c)=d$. The discriminant of this quadratic is $c^2+4d\geq 0$ as $c^2\geq 0$ and from $u,v\geq 0$, we have $4d\geq 0$. Thus, $f(u)=f(v)$ for any non-negative reals $u,v$, implying that $f$ must be constant over the nonnegative reals. However, by periodicitiy, $f$ must consequently be constant for all real. However, this implies that $f(0)=f(\alpha)=0$, a blatant contradiction. As $f$ never attains $0$, we are able to cancel the $f(x)$, and get $f(x+1)=c$. This implies that $f$ is constant. Indeed, checking constant solutions, $2c=c^2$, so $c=0,2$. But $c=0$ is impossible as $f(0)\neq 0$, so $f(x)=2$ is the only solution.