I can see $f(x)=0$ and $f(x)=x-1$ are solutions, are they the only ones? @Tintarn thanks for pointing out $f(x)=1-x$ is also a solution, I wrote that down on my paper and promptly forgot about it.

champion999 wrote: I can see $f(x)=0$ and $f(x)=x-1$ are solutions, are they the only ones? that's what i got as well

champion999 wrote: I can see $f(x)=0$ and $f(x)=x-1$ are solutions, are they the only ones? Nope. $f(x)=1-x$ also works.

Where is pco when you need him :/

Tintarn wrote: champion999 wrote: I can see $f(x)=0$ and $f(x)=x-1$ are solutions, are they the only ones? Nope. $f(x)=1-x$ also works. Yeah, I also got these. If $f(x)$ is a solution to this problem, clearly $-f(x)$ is also.

Does the substitution $y=\frac{x}{x-1}, x\neq 1$ give anything useful?

The solutions are $f(x) = 0, x - 1, 1 - x$. It's easy to check that these all work. Let $P(x, y)$ denote the assertion given in the problem statement. 1. If $f(0) = 0$, then $P(x, 0)$ gives $f(x) = 0$ for all $x$. Now assume $f(0) \neq 0$. 2. $P(0, 0)$ gives $f(f(0)^2) = 0$. 3. Claim: If $f(c) = 0$ then $c = 1$. Proof: Suppose otherwise. Take $P(\frac{c}{c - 1}, c)$. This gives $f(0) = 0$, contradiction. 4. Thus $f(0) = \pm 1$ and $f(1) = 0$. Since $f$ is a solution iff $-f$ is a solution, assume $f(0) = -1$. 5. $P(x, 1)$ gives $f(x + 1) - 1 = f(x)$. 6. Claim: $f$ is injective. Proof: Suppose $f(a) = f(b)$ but $a\neq b$. By 5 we may replace $a, b$ with $a - n, b - n$, so assume $a < 1$. Then there exists $r, s$ such that $rs + 1 = a$ and $r + s = b$, which gives $f(f(r)f(s)) + f(b) = f(a - 1) = f(a) - 1 = f(b) - 1$, so $f(f(r)f(s)) = -1$, so by 5, $f(f(r)f(s) + 1) = 0$ and thus $f(r)f(s) + 1 = 1$. Thus $f(r)f(s) = 0$, so $r$ or $s$ is 1. However, that means that $a = b$, contradiction. 7. $P(x, -x)$ gives $f(f(x)f(-x)) - 1 = f(-x^2) = f(-x^2 + 1) - 1$, so $f(x)f(-x) = 1 - x^2$. 8. $P(x, 1 - x)$ gives $f(f(x)f(1 - x)) = f(x(1 - x))$ so $f(x)(1 + f(-x)) = f(x)f(1 - x) = x - x^2$. 9. Combining 7 and 8 gives $f(x) = x - 1$ for all $x$. Likewise, in the case $f(0) = 1$, we get $f(x) = 1 - x$.

Here's a fix to my (now deleted) fakesolve: Let $P(x,y)$ be the given assertion. First, $P(0,0)$ guarantees there is a value $z$ with $f(z)=0$. Now, suppose $f(z)=0$ and $z\neq 1$. Then take $y$ with $z+y=zy$. It follows $f(f(z)f(y))=0$, or $f(0)=0$. But then $P(x,0)$ gives $\boxed{f(x)\equiv 0}$ as a solution. Now, suppose $f(z)=0$ has only one solution, $z=1$. First, note that $P(0,0)$ gives $f(f(0)^2)=0$, so $f(0)^2=1$. WLOG $f(0)=-1$; if $f$ is a solution, then so is $-f$. Note that $P(x,1)$ gives $f(x+1)=1+f(x)$. Suppose $f(z)=-1$. Then note that $f(z+1)=0$, so $z+1=1$ and $z=0$, so we get that $f(z)=-1$ has a unique solution $z=0$. Suppose $f(x)=f(y)$. Then take an integer $N$ such that $a+b=x+N+1$, $ab=y+N$ has a real solution (which we can do for $N$ large). Note that $f(x+N)=f(y+N)$. Take $P(a,b)$. Then $f(f(a)f(b))+f(a+b)=f(ab)$, so $f(f(a)f(b))+f(x+N+1)=f(y+N)$, so $f(f(a)f(b))=-1$. Thus $f(a)f(b)=0$. WLOG $f(a)=0$; then $a=1$ so $b=x+N=y+N$, and $x=y$. Thus $f$ is injective. Finally, take $P(x,0)$; then $f(-f(x))+f(x)=-1$. Taking $P(-f(x), 0)$ gives $f(-f(x))+f(-f(-f(x)))=-1$, so $f(-f(-f(x)))=f(x)$, and thus $x=-f(-f(x))$. Since $f(-f(x))+f(x)=-1$, we get $f(x)=x-1$. Since we WLOGed $f(0)=-1$, we get two solutions: $\boxed{f(x)\equiv x-1}$ and $\boxed{f(x)\equiv 1-x}$.

Let P(xy) denote the FE. Note that if f is a solution then −f is also a solution. Now, P(22) imply that f(f(2)2)=0. Let t=f(2)2. Case 1: t=1 P(xt) implies f(0)+f(x+t)=f(xt), and we can find x so that x+t=xt. So f(0)=0. Now P(x0) implies f(x)=0 for all real x. Case 2: t=1 So f(2)=1 or −1. It suffices to deal with the f(2)=1 case since f being a solution implies −f is also a solution. P(x1) implies f(0)+f(x+1)=f(x), plugging in 1 here we get that f(0)=−1 which implies that f(x+1)=f(x)+1 and thus f(x+n)=f(x)+n by induction. P(x0) implies that f(−f(x))+f(x)=−1 and so f(−f(x))=−1−f(x). P(−f(x)0) implies that f(−f(−f(x)))=−1−f(−f(x)) which implies f(f(x))=f(x−1). Now assume that f(k)=0 but k=1 . So there exists r with k+r=kr. P(kr) implies f(0)=0 which is a contradiction. Note that this implies that f(k)=n for integer n is only possible when k=n+1. Now note that if there exists p and q so that f(f(p)f(q))=−1 then f(p)f(q)=0 then at least one of p and q equals to 1. So if there exists p and q so that f(pq)−f(p+q)=−1 or f(pq+1)−f(p+q)=0 or f(pq+1)=f(p+q) then at least one of p and q equals to 1. WLOG q=1. But then pq+1=p+q. Now, if f(a)=f(b), then f(a+n)=f(b+n). Note the quadratic x2−(a+n)x+(b+n−1). The discriminant is (a+n)2−4(b+n−1) which for large enough n would be positive. So its roots would be real. If the roots are p and q, then a+n=p+q and b+n=pq+1. Since f(a+n)=f(b+n) we get that a=b. So f is injective. Now, from f(f(x))=f(x−1) we get that f(x)=x−1. Plugging it into the FE, it works. So 0x−1 and 1−x are the only solutions.

This is an outline of a solution I got from a patriot of mine who took the test, can someone please check it?

A little bit lengthy, since I use the additivity to prove the injectivity D:

solution: the solutions are $f(x)=0,x-1,1-x$,let's show that they are the only ones. clearly the zero function is a solution,assume that $f$ is nonzero, put $x=y=0$,then $f(f(0)^2)=0$ set $y=f(0)^2$ so $f(0)+f(x+f(0)^2)=f(xf(0)^2)$ making the equality $x+f(0)^2=xf(0)^2$ with $f(0)^2\neq 1$ so $f(0)=0$ absurd so $f(0)=1$ or $f(0)=-1$ 1st case:$f(0)=1$ so $1+f(x+1)=f(x)$ so $f(x+1)-(x+1-1)=f(x)-(x-1)=cte$ thus $f(x)=x-1+c$ so $f(x)=x-1$ which is indeed a solution 2nd case;$f(0)=-1$ and since if $f$ is a solution then $-f$ is a solution thus $f(x)=1-x$ is a solution too.(or just do as case 1)

Muradjl wrote: 1st case:$f(0)=1$ so $1+f(x+1)=f(x)$ so $f(x+1)-(x+1-1)=f(x)-(x-1)$ thus $f(x)=x-1$ which is indeed a solution $f(x+1)-f(x)=1$ does not imply $f(x)=x-1$, since we are working in reals not integers

adamov1 wrote: Muradjl wrote: 1st case:$f(0)=1$ so $1+f(x+1)=f(x)$ so $f(x+1)-(x+1-1)=f(x)-(x-1)$ thus $f(x)=x-1$ which is indeed a solution $f(x+1)-f(x)=1$ does not imply $f(x)=x-1$, since we are working in reals not integers it's the slope of the line since there exist $y=x+1$ for all reals $x$

Muradjl wrote: adamov1 wrote: Muradjl wrote: 1st case:$f(0)=1$ so $1+f(x+1)=f(x)$ so $f(x+1)-(x+1-1)=f(x)-(x-1)$ thus $f(x)=x-1$ which is indeed a solution $f(x+1)-f(x)=1$ does not imply $f(x)=x-1$, since we are working in reals not integers it's the slope of the line since there exist $y=x+1$ for all reals $x$ No.

Muradjl wrote: it's the slope of the line since there exist $y=x+1$ for all reals $x$ what about $f(x)=\lfloor x \rfloor$

ignore that,what am saying!

Muradjl wrote: Hydrogen-Helium wrote: Muradjl wrote: it's the slope of the line since there exist $y=x+1$ for all reals $x$ what about $f(x)=\lfloor x \rfloor$ still true the slope is $1$(maybe i am mistaken) what does the slope of $\lfloor x \rfloor$ even mean? in particular $f(x)=\lfloor x \rfloor-1$ (which is not a solution to the functional equation) is a direct contradiction to your argument.

I heard that $4$ American and $5$ British didn't solve the problem fully, that's very surprising.

Nice one Case $1 : f(0) = 0$. $P(0,x) : f(x) = 0$. Case $2 : f(0) = c \neq 0$. $P(0,0) : f(c^2) = 0$. if $x+y = xy$ then $y = \frac{x}{x-1}$ so $P(x,\frac{x}{x-1} : f(f(x)f(\frac{x}{x-1})) = 0$ so if we put $x = c^2$ we will get contradiction unless $\frac{c^2}{c^2-1}$ doesn't exists so it means $c = \pm 1$. Case $2.1 : f(0) = 1$. $P(x,1) : 1 + f(x+1) = f(x) \implies f(x+n) = f(x) - n$. Now Assume $a,b$ such that $f(a) = f(b)$. putting $x = a,b$ in last equation and we will have $f(a + n) = f(b + n)$. $f(a+n+1) = f(a+n) + 1 = f(b+n) + 1$ so now put $n$ such that there exist $x',y'$ such that $x'+y' = a+n+1$ and $x'y' = b+n$ so now we would have $f(f(x')f(y')) + 1 = 0$ Note that $f(f(x')f(y')) + 1 = f(f(x')f(y') + 1) = 0$ so $f(x')f(y') + 1 = 1$(we had $f(1) = 0$ and $f(x+n) = f(x) - n$ so if $f(z) = 0$ then $z = 1$) so $f(x')f(y') = 0$ which means at least one of $x',y'$ are $1$ which means $a+n+1 = x + 1 , b+n = x \implies b = a$ so $f$ is injective. $P(x,-x) , P(x,1-x) : f(x)f(-x) = 1-x^2 , f(x)f(1-x) = x - x^2 \implies f(x)^2f(-x)f(1-x) = (1-x^2)(x-x^2)$ and Note that $f(1-x) = f(-x) - 1$ so $f(x) = 1-x$. Case $2.2 : f(0) = -1$. we use the same approach we used last part so $f(x+n) = f(x) + n$ and using that $f$ is injective we wolud get $f(x) = x-1$.

Let $P(x,y)$ denote the assertion. Clearly if $f$ is a solution then $-f$ is also a solution. The trivial case is $f(0)=0$ which implies $f\equiv 0,$ which fits. We assume $f(0)<0.$ If $f(k)=0$ and $k\neq 1,$ then $P(k,\frac{k}{k-1})\implies f(0)=0,$ absurd. If $k=1,$ then $P(0,0) \implies f(f(0)^2)=0,$ so $f$ has a zero and $f(1)=0, f(0)=-1.$ Thus $f(k)=0$ iff $k=1.$ $P(x,1)\implies f(x+1)=f(x)+1,$ and by induction $f(x+n)=f(x)+n. \qquad (*)$ If $f$ is injective then compare $P(k,-k)$ and $P(k,1-x)$ to get $f(k)=k-1,$ which works. And we also get $f(k)=1-k,$ since $-f$ is also a solution. Both of these work.

Let $P(x,y)$ denote the assertion. $P(0,0)$ gives $f(f(0)^2)=0.$ Then $P(2,2)$ gives $f(f(2)^2)=0.$ $P(\frac{x}{x-1},x)$ for $x\neq 1$ gives $f(f(x)f(\frac{x}{x-1}))=0.$ Now, let $f(x)=0$ then $f(0)=0$ and so $P(0,y)$ gives $f(y)=0$ which is a valid function. Now assume $f(0)\neq 0$ then no $f(x)\neq 0$ which means $f(0)^2=f(2)^2=1$. We claim that $f$ is injective. First, $P(1,x)$ gives $1+f(1+x)=f(x)$ so $f(x+n)=f(x)-n.$ Suppose $f(a)=f(b)$ then $f(a+n)=f(b+n).$ Thus, we can make $a,b$ sufficient large so that $xy=a,x+y=b+1$ has solutions. Then \[f(f(x)f(y))=f(xy)-f(x+y-1)-1=-1.\]Now, $f(f(x)f(y)+1)=0$ so $f(x)f(y)=0.$ Thus, $x$ or $y$ is $1.$ WLOG $x=1$ which implies $y=a,1+y=b=1$ implying $a=b$ as desired. Suppose $f(0)=1$ then $P(0,y)$ gives \[f(f(y))+f(y)=f(0)\]so $f(f(y))=1-f(y)$ letting $y$ be $f(y)$ gives $f(f(f(y)))=f(y).$ Now, since $f(f(f(y)))=f(y)$ we have $f(f(y))=y$ but on the other hand, it equals $1-f(y)$ so $f(y)=1-y$ and we check that this is a valid function. Next case: suppose $f(0)=-1$ then $P(0,y)$ gives $f(-f(y))+f(y)=1$ so $f(-f(y))=1-f(y)$ so $f(-f(-f(y)))=1-f(-f(y))=f(y)$ so we have by injectivity $-f(-f(y))=y$ so we have $f(y)-1=y$ so $f(y)=y-1$ and this is a valid function.

Great problem finally solved this The only solutions are $\boxed{f(x) = 0, f(x)=x-1, f(x)=1-x}$, which upon checking works. Let $P(x, y)$ denote the given assertion. $P(0,0)$ yields $f(f(0)f(0)) + f(0) = f(0)$. Letting $z=(f(0))^2$ yields $f(z)=0$. $P(z, y)$ then yields $$f(0) + f(z + y) = f(zy).$$ Note that whenever $z \neq 1$, there exists a value of $y$ such that $f(z+y) = f(zy)$, resulting in $f(0)=0$. In that case, $P(0,x)$ yields $\boxed{f(x)=0}$ for all $x$, one of our solutions. From now on, assume that $z=1$ is the only value whose output regarding function $f$ is $0$, as otherwise the entire function is $0$ as aforementioned. Thus $f(0) = \pm 1$. Since we can observe that if $f$ is a solution to the given, $-f$ must also be a solution, WLOG assume $f(0)=1$. Claim: $f^3(x) = f(x)$. Proof: $P(0, x)$ and $P(0, f(x))$ respectively yields $$f^2(x) +f(x) = 1$$$$f^3(x) + f^2(x) = 1$$from which the result is obvious. $\fbox{}$ Claim: $f$ is injective. Proof: First, $P(1, x)$ yields $$1 + f(1+x) = f(x) \implies k + f(k+x) = f(x) \forall k \in \mathbb{N} \qquad (\clubsuit)$$Now, assume FTSOC that there exists $f(a)=f(b)$ for $a \neq b$. We claim that there exists reals $x, y$ such that $x+y = a+k$ and $xy+1 = b+k$ for sufficiently large $k \in \mathbb{N}$. This is equivalent to saying that the discriminant $(a+k)^2-4(b+k-1)$ is nonnegative for sufficiently large $k$, which works if we take $k$ large enough such that $a+k, b+k-1 \ge 10^{100}$ and $a+k \ge 0.999 (b+k-1)$, for example. Furthermore, we also claim that the exact values of $x,y$ are both not $1$, as if either one is then $x+y$ equals $xy+1$, violating the assumption $a \neq b$. Using this and $(\clubsuit)$, this means that there exists $x_1, y_1 \neq 1$ such that with $f(x_1+y_1) = f(a+k)=f(b+k)= f(x_1y_1+1)$. $P(x_1, y_1)$ yields $$f(f(x_1)f(y_1)) +f(x_1+y_1) = f(x_1y_1) = f(x_1y_1+1) + 1 \implies f(f(x_1)f(y_1)) = 1.$$ As we established the only $z$ such that $f(z)=0$ is $z=1$, $(\clubsuit)$ allows us to easily deduce that the only $z_0$ such that $f(z_0)=1$ is $z_0 = 0$. Combining these two regarding the previous condition yields either $x_1$ or $y_1$ is $1$, which is a contradiction. Thus, $f$ is injective. $\fbox{}$ Using this on $f^3(x) = f(x)$ yields $f^2(x) = x$. Then $f^2(x) + f(x) = 1$ turns into $\boxed{f(x) = 1-x}$. Since we WLOG assumed that $f(0) = + 1$ because every function in the solution set's additive inverse is also in the solution set, $\boxed{f(x) = x-1}$ is also a solution. We are done.

) If anyone has more examples of tricks along those lines in olympiad FEs, please PM me as these are fun

We claim that the only solutions are $f\equiv 0, x-1, 1-x$. These clearly work. Let $P(x,y)$ be the statement. Now, note that if we let $\alpha=f(0)^2$, then $P(0,0)$ gives us that $f(\alpha)=0$. Let $Z$ be the set of $z$ such that $f(z)=0$. Clearly, $Z$ is nonempty. Note that if $0\in Z$, then $P(x,0) \Longrightarrow f(x) = 0$ for all $x$, giving $f\equiv 0$. From now on, assume $0\notin Z$. Now, assume that there exists some $z\in Z$ where $z\neq 1$. Then, we may take $w= \frac{z}{z-1}$, which gives $P(z,w)\Longrightarrow f(0) = 0$ since $z+w=wz$. This has already been ruled out, so the assumption that $0\notin Z$ implies that $Z = \{1\}$ (since we know that Z is nonempty). Circling back, this means that $f(0)^2 = \alpha =1$, so $f(0)=\pm 1$. Note that if $f$ works, so does $-f$. Thus, we assume $f(0)=1$. (this will eventually give $f\equiv -x+1$, and reversing everything gives $f\equiv x-1$). Next, $P(1,y)\Longrightarrow f(y+1) = f(1\cdot y) - f(0) = f(y)- 1$. Thus, for all $x\in \mathbb{Z}$, we have $f(x) = -x+1$. Additionally, $f(x+N) = f(x)-N$ for all $x\in \mathbb{R}$. Next, $P(0,y) \Longrightarrow f(f(y)) = f(0) - f(0+y) = 1- f(y)$. Using the triple involution trick, this becomes $f(1-f(y)) = 1- f(f(y)) = f(y)$. Thus, if we prove injectivity than we finish. $\textbf{Claim: }$ $f$ is injective $\textbf{Proof: }$ Consider $f(a)= f(b)$. AFTSOC that $a-b \neq 0$. Now, we can find $x,y \neq 1$ such that $(x-1)(y-1) = b-a$. This means that $xy - (x+y) = (x-1)(y-1) -1 = b-a-1 \Longrightarrow b-xy = 1 + a - (x+y)$. Since we still have a degree of freedom, we can clearly select $x,y$ such that $x+y \equiv a \pmod{1}$. Thus, the quantity $b-xy = 1+a - (x+y)$ is an integer. THen, \[f(xy) - f(b) = b - xy = 1+a - (x+y) = 1- f(a) + f(x+y)\]Thus, combining this with $P(x,y)$, we get \[f(f(x)f(y)) = f(xy) - f(x+y) = f(b) - f(a) + 1 = 1\]using the $f(a)=f(b)$ assumption. Then, note that $f(f(x)f(y)+1) = f(f(x)f(y)) - 1 = 0$, so $f(x)f(y)+1=1$, so $f(x)f(y)=0$, and this means that $1\in \{x,y\}$. But we created $x,y$ not satisfying this, so contradiction and $a$ must equal $b$. $\square$. Using this on $f(1-f(y)) = f(y)$ gives $1-f(y) = y$, so $f(y) = -y+1$ and after doing the symmetrical thing for $f(0)=-1$ which gives $f(y) = y-1$, we are done. $\blacksquare$.

Let $P(x,y)$ be the given FE. One can directly check that $f\equiv 0$, $f(x)=x-1$, $f(x)=1-x$ are solutions. Case 1. $f(0)=0$. Then $P(x,0)$ gives us $f \equiv 0$ which is indeed a solution. Case 2. $f(0)\neq 0$. $P(0,0)$ gives us that $f(f(0)^{2})=0$, so there is at least one $c$: $f(c)=0$. Assume that $c\neq 1$. Then $P(c, \frac{c}{c-1})$ gives us that $f(0)=0$, contradiction $\implies f(x)=0 \iff x=0 \implies f(0) \in \{-1;1\}$. It is easy to see that if $f$ is a solution, then $-f$ is also a solution $\implies$ WLOG, let $f(0)=-1$. $P(x,-1): f(x+1)=f(x)+1$. The last one means that $f(n)=n-1, \forall n \in \mathbb{Z}$. Let's take $g=f+1$ and rewrite $P(x+1,y+1)$ in terms of $g$. Then we end up with $$Q(x,y): g(g(x)g(y))+g(x+y)=g(xy+x+y)$$$Q(x,-1): g(-g(x))=-g(x)$ (1) $Q(x,1): g(g(x))=g(2x)-g(x)$ (2) In (2) we take $x=-g(x)$ to conclude that $-g(x)=g(-2g(x))+g(x)\iff g(-2g(x))=-2g(x)$ (3) $Q(x,-2): g(-2g(x))=g(-x)-g(x)$. Combining the last one with with (3), we get that $g(x)=-g(-x)$, so $g$ is odd. Now (1) gives us $g(g(x))=g(x)$/(4)/ and putting it in (2) we find that $g(2x)=2g(x)$. $Q(-x,-y): g(g(x)g(y))-g(x+y)=g(xy-x-y)$. Subtracting that from $Q(x,y)$ we get $$2g(x+y)+g(xy-x-y)=g(xy+x+y) \implies g(x+y-xy)+g(x+y+xy)=g(2x+2y).$$Now we'll prove that $g$ is additive. Notice that if $v\leq 0$ and $u$ are reals, then there are $x,y \in \mathbb{R}: x+y=u, xy=v$. It means that $g(u-v)+g(u+v)=g(2u)$ whenever $v\leq 0$. Now let $a$ and $b$ be two arbitrary reals, WLOG $a\leq b$. We take $u=\frac{a+b}{2}, v=\frac{a-b}{2}\leq 0$ to get that $g(a)+g(b)=g(a+b)$, as desired. Now in (4) we get that $0=g(g(x))-g(x)=g(g(x)-x)$, but the only zero of $g$ is $0\implies g(x)=x$, so $f(x)=x-1$. Also, $f(x)=1-x$ is a solution, so we are done!

this is a terrible problem. The only solutions are f(x)={0,x-1,1-x}, which clearly work. If f(0)=0 then y=0 gives f(x)=0. Otherwise, $f(f(0)^2)=0$, let $f(c)=0$ (and c is hence unique); if $c\ne1$ then $(c,\frac c{c-1})$ is evident contradiction since f(0)=0. Follows that f(1)=0, f(0)=$\pm1$, which we assume 1 since if f is sol, then -f is. We claim that f is injective. Proof. Assume FTSOC f(a)=f(b),$a\ne b$. The idea is that solutions to $x+y=a,xy=b$ ALMOST works, but since $1+f(x+1)=f(x)$ ideally we'd want $x+y=a+1,xy=b$ (later equation shows why). Since $f(a+n)=f(b+n)$ for all n, we can find such a pair $(a+n,b+n)$ such that there are real solutions for x and y with $x+y=a+n+1,xy=b+n$ (which obviously exists when they're sufficiently large since the resulting quadratic is $x^2-(a+n)x+b+n=0$ and $(a+n)^2>4(b+n)$); henceforth let the a+n and b+n be the new a,b. Then $$f(f(x)f(y)+1)=f(f(x)f(y))-1=f(xy)-f(x+y)-1=f(b)-f(a+1)-1=f(b)-f(a)+1-1=0\implies f(x)f(y)=0\Rightarrow1\in\{x,y\}\implies a=b,$$contradiction. $\square$ To finish, simply note that $f(f(x))+f(x)=1$ so $f(1-f(x))=f(f(f(x)))=1-f(f(x))=f(x)\Rightarrow f(x)=1-x$. Now -f is x-1.

I'll talk about the following generalization. Quote: Let $F$ be a field. Determine all functions $f : F \to F$ such that, for any real numbers $x$ and $y$, \[ f(f(x) f(y)) + f(x + y) = f(xy). \] I'll remark that the solution in #75 (https://artofproblemsolving.com/community/c6h1480146p8693244) works when $\mathbb{R}$ is replaced by any field $F$ of characteristic not equal to $2$, and it almost works when $\mathbb{R}$ is generalized to division rings of characteristic not equal to $2$ (just dropping commutativity.) The main thing I will be talking about here is the solution in the case $char(F) = 2$. We need a few properties that I believe is used in every single solution above (and is again provable in the same way): 1. If $f(c) = 0$ for some $c \neq 1$, then $f \equiv 0$. Otherwise, we have $f(x) = 0$ iff $x = 1$, and $f(0)^2 = 1$. 2. If $f$ is a solution, then $-f$ is a solution; thus WLOG assume that $f(0) = 1$. (This actually doesn't matter for my solution since $char(F) = 2$.) 3. For any $x \in F$ non-zero, we have $f(x + 1) f(x^{-1} + 1) = 1$. 4. For any $x \in F$, we have $f(x + 1) = f(x) - 1$. 5. If $f$ is injective, then $f(x) = 1 - x$ for any $x \in F$. Thus, the goal is now to show that $f$ is injective. I'll stop using subtraction and just use addition instead; they are the same thing over fields of characteristic $2$. Consider an arbitrary $a, b \in F$ such that $f(a) = f(b)$. Note that \[ f(x) = 1 \iff f(x + 1) = f(x) + 1 = 0 \iff x + 1 = 1 \iff x = 0, \]so if one of $a$ or $b$ is zero, the other one is zero as well, and they are equal. From now, we assume that $a, b \neq 0$, and work towards proving that $a = b$. From property 3 (and 4), we get $f(a + 1) f(b^{-1} + 1) = f(b + 1) f(b^{-1} + 1) = 1$, so the original equality yields \[ f((a + 1) + (b^{-1} + 1)) = f((a + 1)(b^{-1} + 1)) \implies f(a + b^{-1}) = f(ab^{-1} + a + b^{-1} + 1) \implies f(1 + a + b^{-1}) = f(ab^{-1} + a + b^{-1}). \]Similarly, we have \[ f(1 + b + a^{-1}) = f(ba^{-1} + b + a^{-1}). \]We now make use of the identity \[ (1 + a + b^{-1})(1 + b + a^{-1}) = (ab^{-1} + a + b^{-1})(ba^{-1} + b + a^{-1}), \]which gives us \[ f(f(1 + a + b^{-1}) f(1 + b + a^{-1})) + f(1 + a + b^{-1} + 1 + b + a^{-1}) = f(f(ab^{-1} + a + b^{-1}) f(ba^{-1} + b + a^{-1})) + f(ab^{-1} + a + b^{-1} + ba^{-1} + b + a^{-1}). \]By the previous two equalities, the first addition term are equal, so \[ f(1 + a + b^{-1} + 1 + b + a^{-1}) = f(ab^{-1} + a + b^{-1} + ba^{-1} + b + a^{-1}), \]\[ f((a + b + 1) + (a^{-1} + b^{-1} + 1)) = f((a + b + 1)(a^{-1} + b^{-1} + 1) + 1) = f((a + b + 1)(a^{-1} + b^{-1} + 1)) + 1. \]Using the original equality and property 1 (and 4), we get \begin{align*} f(f(a + b + 1) f(a^{-1} + b^{-1} + 1)) = 1 \\ f(a + b + 1) f(a^{-1} + b^{-1} + 1) = 0 \\ f(a + b) = 1 \vee f(a^{-1} + b^{-1}) = 1 \\ a + b = 0 \vee a^{-1} + b^{-1} = 0. \end{align*}Both yields $a = b$, so $f$ is injective, as desired.

The only solutions are $f(x)\equiv 0$, $f(x)\equiv x-1$, and $f(x)\equiv 1-x$, which clearly work. From now on, assume $f$ is not of this form. Let $P(x,y)$ denote the assertion in the problem statement. $P(x,0)\implies f(f(0)f(x)) = f(0)-f(x)$. In particular, $f(0)\neq 0$. Now $P(0,0)\implies f(f(0)^2) = 0$, so $f$ has a root. Conversely, suppose $f(c)=0$. Then $P(x-c,c)\implies f(cx-c^2) = f(x) + f(0)$. If $c\neq 1$, then setting $x=\frac{c^2}{c-1}$ gives $f(0)=0$, a contradiction, so $f(1)=0$ and $f$ is injective at $0$. This gives a lot of information: $f(0)=\pm 1$, $f(x-1)=f(x)+f(0)$, and $f$ is injective at all integer outputs. $\textbf{Case 1:}$ $f(0)=1$. Then $f(f(x))=1-f(x)$ and $f(x-1)=f(x)+1$. In particular, $f(n)=1-n$ for all integers $n$. $P(f(x),f(y)) - P(x+1,y+1)\implies f(f(x)+f(y)) = f(xy) - f(xy+x+y)-1$. Setting $y=-x$ gives $f(f(x)+f(-x))=-1$, so $f(-x)=2-f(x)$. Then $P(x,2)\implies f(2x) = f(-f(x)) + f(x+2) \implies f(2x)=2f(x)-1$. $P(2-x,2-y) - P(x,y)\implies f(xy) - 2f(x+y) = f(xy-2(x+y))-2\implies f(xy)+f(-2(x+y)) = f(xy-2(x+y))+1$. In particular, $f(a+b)=f(a)+f(b)-1$ for all real numbers $a,b$ such that $a>0$ and $b \le -4\sqrt{a}$, and from $f(x-1)=f(x)-1$ it follows that $f(a+b)=f(a)+f(b)-1$ for all real numbers $a$ and $b$. Now if $f(x_1) = f(x_2)$, then $f(x_1-x_2)=1$. Since $f$ is injective at integer outputs, it follows that $f$ is injective. From $f(f(x)) = f(1-x)$ and injectivity we get $f(x)=1-x$ for all $x$, as desired. $\textbf{Case 2:} f(0)=-1$. Then $f(-f(x)) = -1-f(x)$ and $f(x-1) = f(x)-1$. In particular, $f(n) = n-1$. The argument for this case will be parallel to the previous, but we include it for completeness. $P(-f(x),-f(y)) - P(x+1,y+1)\implies f(-f(x)-f(y)) = f(xy) - f(xy+x+y)+1$. Setting $y=-x$ gives $f(-f(x)-f(-x))=1$, so $f(-x)=-2-f(x)$. Then $P(x,2)\implies f(2x) = f(f(x))+f(x+2)\implies f(2x) = 2f(x)+1$. $P(2-x,2-y)-P(x,y)\implies f(xy) - 2f(x+y)=f(xy-2(x+y))+2\implies f(xy) + f(-2(x+y)) = f(xy-2(x+y))-1$. In particular, $f(a+b)=f(a)+f(b)+1$ for all real numbers $a,b$ such that $a>0$ and $b \le -4\sqrt{a}$, and it again follows that $f(a+b)=f(a)+f(b)+1$ for all real numbers $a$ and $b$. Now if $f(x_1) = f(x_2)$, then $f(x_1-x_2)=-1$. Since $f$ is injective at integer outputs, it follows that $f$ is injective. From $f(-f(x)) = f(-x+1)$ and injectivity we get $f(x)=x-1$ for all $x$, as desired.

Call the assertion $P(x,y)$. We first find all constant/linear solutions. Let $f(x)=ax+b$. Then, we have \[a(ax+b)(ay+b)+b+a(x+y)+b=axy+b\]Comparing coefficients, we see that the only possible solutions are $(a,b)=\{(1,-1),(-1,1),(0,0)\}$. Hence, the constant/linear solutions are $f(x)\in\{0,1-x,x-1\}$. We now show that there are no other solutions. Suppose there are non-linear solutions. $P(2,2)$ gives $f(f(2)f(2))=0$. So, there is some $z$ for which $f(z)=0$. If $z=0$, then $f(0)=0$. However, $P(x,0)$ gives $f(x)=0$ for every $x\in \mathbb R$ which has already been listed. So, assume $f(0)\ne0$. Note that if $f$ is a solution then so is $-f$. So, take $f(0)>0$. There exists $z\ne0$ such that $f(z)=0$. If $z\ne1$, then $P\left(z,\frac z{z-1}\right)$ gives $f(0)+f\left(z+\frac z{z-1}\right)=f\left(\frac {z^2}{z-1}\right)=f\left(z+\frac z{z-1}\right)$, which means $f(0)=0$, a contradiction. Hence, if $f(z)=0$ then $z=1$ must be true. So, $f(1)=0$. Now, $P(0,0)$ gives $f\left(f(0)^2\right)=0$ which forces $f(0)^2=\pm1$. Since $f(0)>0$ we have $f(0)=1$. Now, plugging $P(x,1)$ gives $f(0)+f(x+1)=f(x)$, which gives $f(x+1)=f(x)-1$. A simple forward-backward induction gives $f(x)=1-x$ for all integers $x$. Then, plugging $P(x,0)$ gives $f(f(x))=1-f(x)$. Also, \[f(f(f(x)))=f(1-f(x))=f(f(f(x)))=1-f(f(x))=1-(1-f(x))=f(x)\]So, $f(1-f(x))=f(x)$. We now show that $f$ is injective after which we'd be done because $1-f(x)=x\Longrightarrow f(x)=1-x$. Suppose there are $a,b$ such that $f(a)=f(b)$. Note that for sufficiently large integer $N$, there exists $x,y$ such that $x+y+1=a+N$ and $xy=b+N$. This because we just need $(a+N-1)^2\geq 4(b+N)$, which is a quadratic in $N$ and is true for sufficiently large $N$. Plugging $P(x,y)$ we get that $f(f(x)f(y))+f(x+y)=f(xy)=f(f(x)f(y))+f(a)-N-1=f(b)-N$, which means $f(x)f(y)=0$ as $f(f(x)f(y))=1$. It means that either $x=1$ or $y=1$ or both. So, we have $2+y=a+N$ and $y=b+N$, which means $f(a+N)-f(b+N)=-2$, but $f(a+N)=f(b+N)$ which means $-2=0$, a contradiction. Hence, the function is injective and hence the only solutions are \[f(x)\in\{0, 1-x, x-1\}\]

The answers are $\boxed{f(x) = 0}$, $\boxed{f(x) = x - 1}$, and $\boxed{f(x) = 1 - x}$. It is easy to verify that these work. It remains to show that they are the only solutions, for which the key claim is that $f$ is injective. Claim 1: If $f(x) f(\frac{x}{x - 1})$ is not always $1$, then $f(x) = 0$ for all $x$. Proof. $\underline{\text{P}}(x, \frac{x}{x - 1}) \implies \boxed{f(f(x) f(\frac{x}{x - 1})) = 0 \text{ } \forall x \neq 1}$ If $f(x) f(\frac{x}{x - 1})$ is not always $1$, then the substitution $x \rightarrow f(x) f(\frac{x}{x - 1}) \implies \boxed{f(0) = 0}$. $\underline{\text{P}}(x, 0)$ then gives $\cancel{f(0)} + f(x) = \cancel{f(0)} \implies \boxed{f(x) = 0}$ for all $x$ as desired. It remains to deal with the case where $f(x) f(\frac{x}{x - 1}) = 1 \text{ }$ for all $x \neq 1$. Claim 2: $x=0 \Longleftrightarrow f(x)=-1$, $x=1 \Longleftrightarrow f(x)=0$, and $f(x + n) = f(x) + n$ for all integers $n$. Proof. Substituting $x \rightarrow 0$ in this equation yields $f(0)^2 = 1 \implies f(0) = \pm 1$. However, these two cases are almost exactly identical, so for the sake of compactness I will assume $f(0) = -1$. $\underline{\text{P}}(0, 0)$ then gives $f(1) + \cancel{f(0)} = \cancel{f(0)} \implies f(1) = 0$. $\underline{\text{P}}(x, 1)$ gives $f(0) + f(x + 1) = f(x) \implies f(x + 1) = f(x) + 1$. In fact, the fact that $f(x + n) = f(x) + n$ for any integer $n$ immediately follows from this. A formal proof is possible by induction. Now assume, for the sake of contradiction, that there exists some $\alpha$ such that $f(\alpha) = 0$ but $\alpha \neq 1$. Then $0 \cdot f(\frac{\alpha}{\alpha - 1}) = 1$ which is a contradiction. Then assume there exists some $\beta$ such that $f(\beta) = -1$ but $\beta \neq 0$. Then $f(\beta) + 1 = 0 \implies f(\beta + 1) = 0 \implies \beta + 1 = 1 \implies \beta = 0$, which is another contradiction. So $\boxed{x = 0 \iff f(x) = -1}$, $\boxed{x = 1 \iff f(x) = 0}$, and $\boxed{f(x + n) = f(x) + n \text{ } \forall n \in \mathbb{Z}}$ as desired. We now claim the following: Claim 3: $f$ is injective. Notice that the claim immediately kills the problem: $\underline{\text{P}}(x, 1 - x) \implies f(f(x) f(1 - x)) = f(x (1 - x)) \overset{\text{Claim 3}}{\implies} \boxed{f(x) f(1 - x) = x (1 - x)}$ $\underline{\text{P}}(x, -x) \implies f(f(x) f(-x)) = f(-x^2) + 1 \overset{\text{Claim 2}}{\implies} f(f(x) f(-x)) = f(1 - x^2) \overset{\text{Claim 3}}{\implies} \boxed{f(x) f(- x) = 1 - x^2}$ This is a system of 2 linear equations with two unknowns ($f(x)$ and $f(-x)$). Solving it yields $f(x) = x - 1$. Notice that if we had chosen $f(0) = 1$ instead of $f(0) = -1$ in Claim 2, then by a similar reasoning, $f(x)$ would equal $1 - x$. This proves that the only answers are $\boxed{f(x) = 0}$, $\boxed{f(x) = x - 1}$, and $\boxed{f(x) = 1 - x}$, or at least reduces it to Claim 3. Proof of Claim 3. Assume, for the sake of contradiction, that there are some $a, b$ such that $f(a) = f(b)$ and $a \neq b$. Substituting $x_0 + y_0 = a + 1$ and $x_0 y_0 = b$, we see that $\underline{\text{P}}(x_0, y_0)$ finishes the claim.

plug in $x=y=0$ to get that $f(f(0)^2)=0$ now, either $f(0)=0$ or it doesn't if $f(0)=0$, plug in $y=0$ to get that all $f(x)=0$, which is a possible solution if not, let $f(z)=0$ and let $x=z, y=\frac{z}{z-1}$, which should be defined as long as $z \neq 1$ however, that leads to $f(f(z)f(\frac{z}{z-1}))=0$ and $f(0)=0$, which is a contradiction thus, $z=1$, and $f(0)^2=1$ so $f(0)=1$ or $f(0)=-1$ now, plug in $y=1$ to get that $f(x+1)=f(x)-f(0)$, and $f(x+n)=f(x)-nf(0)$ for all $n$ plug in $x=0$ to get $f(n)=f(0)(1-n)$ for all $n$, so we can have $f(x)=1-x$ or $f(x)=x-1$ for integers $x$ so all the possible $f(x)$ are $f(x)=0$, $f(x)=1-x$, and $f(x)=x-1$

let \(P(x,y)\) denote the notation do, \(P(0,0)\) get \(f(f(0)^2) = 0\) now suppose \(f(0)^2\) is not equal to 1. plug \(x = f(0)^2, y = \frac{f(0)^2}{f(0)^2 - 1} \) to get \(f(0) = 0\) then \(P(x,0)\) gives \(f(x) = f(0) = 0 \) so thats one answer. if \(f(0)^2 = 1\) then just \(P(x,1)\) gives \(f(x+1) = f(x) - f(0)\) meaning the function is linear then plugging it in the original equation we get \(f(x) = 1 - x\) and \(f(x) = x-1\)so its done!

We claim that the only solutions are $f \equiv 0, 1-x, x-1$. One can check that they indeed works. We will prove that they are the only solutions. If $f(0)=0$, then $P(x, 0)$ gives: $f \equiv 0$. Thus, assume that $f(0) \neq 0$. If $f$ is a solution, then $-f$ also works. Therefore, assume that $f(0) > 0$. We will prove that: $f(x)=1-x$ for all $x \in \mathbb R$. Plugging $P(x, \tfrac{x}{x-1})$ (for some $x \neq 1$) gives us: $f(z)=0$ for some $z \in \mathbb R$. If $z \neq 1$, then $P(z, \tfrac{z}{z-1})$ gives us $f(0)=0$ which leads to contradiction. Notice that: $P(0, 0)$ gives us: $f(f(0)^2)=0$. Thus: $f(0)=1$ (we assumed $f(0)>0$) and have: Fact: $f(z)=0$ if and only if $z=1$ and also: $f(z)=1$ if and only if $z=0$ Notice that: $P(x, 1)$ gives: $$f(x+1)=f(x)-1.$$Therefore, by Induction, we have: $f(n)=1-n$ and $f(x+n)=f(x)-n$ for all $n \in \mathbb Z$ and $x \in \mathbb R$. Notice that: $P(x, 0)$ gives us: $$f(f(x))=1-f(x).$$ Note that: $$f(f(f(x))) = f(1-f(x)) = 1-f(f(x)) $$$$= 1-(1-f(x)) = f(x).$$Therefore: $$f(1-f(x)) = f(x).$$ The following claim would finish that problem with $f(x)=1-x$: Claim: $f$ is injective. Proof: If $f(a)=f(b)$, then, note that: $f(a+n)=f(b+n)$ for all $n \in \mathbb N$. For sufficiently large enough value of $n$, there exists $x, y$ such that: $$x+y=a+n+1, xy=b+n.$$Note that: $P(x, y)$ gives right away: $f(f(x)f(y))=1$. Thus, either $x=1$ or $y=1$. Either of it, gives us $a=b$ and we are done. Thus, we conclude that $f \equiv 0, 1-x, x-1$ are the only possible solutions.