Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Assume $f(1)=0$. Take $y=1$. We get $f^2(x)=0,\ \forall x$, so $f(x)=0,\ \forall x$. This is a solution, so we can take it out of the way: assume $f(1)\ne 0$. $y=1\Rightarrow f(x)[f(x)-f(1)]=(x-1)f(x)f(1)$. We either have $f(x)=0$ or $f(x)=f(1)x$, so for every $x$, $f(x)\in\{0,f(1)x\}$. In particular, $f(0)=0$. Assume $f(y)=0$. We get $f(x)f(xy)=0,\ \forall x$. This means that $f(a),f(b)\ne 0\Rightarrow f(\frac ab)\ne 0\ (*)$ ($\frac ab$ is defined because $f(b)\ne 0\Rightarrow b\ne 0$). Assume now that $x\ne y$ and $f(x),f(y)\ne 0$. We get $f(x)=f(1)x,\ f(y)=f(1)y$, and after replacing everything we get $f(xy)=f(1)xy\ne 0$, so $x\ne y,\ f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (**)$. Assume now $f(x)\ne 0$. From $(*)$ we get $f(\frac 1x)\ne 0$, and after applying $(*)$ again to $a=x,b=\frac 1x$ we get $f(x^2)\ne 0\ (***)$. We can now see that $(**),(***)$ combine to $f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (\#)$. Let $G=\{x\in\mathbb R|f(x)\ne 0\}$. $(*)$ and $(\#)$ simply say that $(G,\ \cdot)$ is a subgroup of $(\mathbb R^{*},\ \cdot)$. Conversely, let $G$ be a subgroup of the multiplicative group $\mathbb R^*$. Take $f(x)=\{\begin{array}{c}f(1)x,\ x\in G\\0,\ x\not \in G\end{array}$. It's easy to check the condition $f(xy)[f(x)-f(y)]=(x-y)f(x)f(y)$.

Please tell me if there is anything wrong with my solution: Put y=1. If $f(1)$ = 0, f is identically 0. Otherwise f(x) = f(1)x = kx for any real k.

Pascal96 wrote: Please tell me if there is anything wrong with my solution: Put y=1. If $f(1)$ = 0, f is identically 0. Otherwise f(x) = f(1)x = kx for any real k. Yes, just look at the previous post which gives a lot of other solutions. And since you just give the result and not the content of your own proof, we can not show you where is your error. The only thing we can say is that your proof is wrong.

I have the same solution as pascal, but i'll post it so you can point out what's wrong: let our assertion be $P(x,y)$. $P(x,1): f(x)(f(x)-f(1))=(x-1)f(x)f(1)$ Therefore, either $f(x)=0$ or $f(x)-f(1)=(x-1)f(1)$ Therefore, $f(x)=0$ is a solution Now assume $f(x) \not=0$ then $f(x)-f(1)=xf(1)-f(1) \implies f(x)=xf(1)$ Plugging in $x=1$, we see $f(1)=f(1)$, so we have no further restrictions on $f(1)$ Therefore, for all $c \in \mathbb{R}$, there is an $f(x)=cx$. Furthermore, $f(x)=0$ is merely $f(x)=cx, c=0$, so all solutions are in the form $f(x)=cx$ for all $c \in \mathbb{R} \blacksquare$ You said this is wrong; why?

It's wrong because you only have $f(x) = xf(1)$ or $f(x) = 0$ for each $x \in \mathbb{R}$. You have to do a little more work to restrict which $x$ you have fall under the first case and which fall under the second.

Oh, tired me might realize; Is it because this function is not continuous?

First, suppose $f(0)\neq 0$. Then $f(0)\cdot [f(x)-f(0)] = x\cdot f(x)\cdot f(0)$. Taking $x=1$ we get $f(1)-f(0)=f(1)$, contradiction. Thus $f(0)=0$. Now, suppose there was a value of $x$ such that $f(x)\neq 0$. Then $f(x)\cdot [f(x)-f(1)] = (x-1)\cdot f(x)\cdot f(1)$, or $f(x)=x\cdot f(1)$. Clearly $f(1)\neq 0$, or this gives a contradiction. Moreover, if $f$ is a solution, then $c\cdot f$ is a solution for any constant $c$. Then we either have $f(1)=0$, in which case $f\equiv 0$, or $f(1)=1$, in which case $f(x)$ is either $0$ or $x$. Let $S$ be the set of values for which $f(x)=x$. If $S$ is nonempty, then $1\in S$. Moreover note that $0\not\in S$. Now, suppose $|S|>1$. We claim that $S$ is a subgroup of $\mathbb{R}^*$. To do this, we need to show that $S$ is closed under multiplication, and that $x\in S\rightarrow 1/x\in S$. Suppose $x\in S$ but $1/x\not\in S$. Clearly $x\neq -1$. Then $f(1/x)=0$. It follows that $f(1)\cdot [f(x)-f(1/x)] = 0$. But $f(x)=x$ and $f(1/x)=0$, so $f(1)\cdot x = 0$, implying $x=0$, contradiction as $0\not\in S$. Thus for every $x\in S$, we have $1/x\in S$. Now we need to check for closure under multiplication. First, if $x,y$ are distinct and in $S$, then $f(xy)\cdot (x-y) = (x-y)\cdot x\cdot y$, so $f(xy)=xy$. It remains to check that for $x\in S$, $f(x^2)=x^2$. This is clear if $x=1$. But note that $f(x)\cdot [f(x^2)-f(1/x)] = (x^2-1/x)\cdot f(x^2)\cdot f(1/x)$. If $f(x^2)=0$, then it follows that $f(x)\cdot (0-1/x)=0$, or $f(x)=0$, contradiction! Thus $f(x^2)=x^2$. It follows that $S$ is also closed under multiplication. Thus, the solutions are of the form $\boxed{f(x)=f(1)\cdot x \mid x\in S; f(x) = 0 \mid x\not\in S}$ when $S$ is a subgroup of $\mathbb{R}^*$. It is easy to check that these solutions work.

We claim the solutions are $f(x)=kx$ for $x\in G$ and $f(x)=0$ for $x\not\in G$ where $G$ is any multiplicative subgroup of $\mathbb{R}^*$. It is easy to check that this works. Let $P(x,y)$ be the FE and suppose $f$ satisfies it. Note that \[P(x,1)\implies f(x)(f(x)-f(1))=(x-1)f(x)f(1),\]or that $f(x)^2=xf(x)f(1)$. Thus, $f(x)=xf(1)$ if $f(x)\ne 0$. If $f(1)=0$, then we get $f\equiv 0$, so suppose $f(1)\ne 0$, and since we can scale $f$, WLOG suppose $f(1)=1$. Let $G=\{x:f(x)\ne 0\}$. Note that $1\in G$. Also, if $x\in G$, then \[P(x,1/x)\implies x-f(1/x)=(x-1/x)xf(1/x)\implies f(1/x)=1/x,\]so $1/x\in G$. Furthermore, if $x\ne y\in G$, then \[P(x,y)\implies f(xy)(x-y)=(x-y)xy\implies xy\in G.\]It simply remains to be shown that if $x\in G$, then $x^2\in G$. Note that \[P(x^2,1/x)\implies x(f(x^2)-1/x)=(x^2-1/x)f(x^2)/x\implies f(x^2)=x^2,\]so $x^2\in G$. This shows that $G\le\mathbb{R}^*$, so we're done.

Wait, how would you write such a solution in a real olympiad? After you get the relatively trivial $f(x)=xf(1)$ or $f(x)=0$ for any $x$ real, and some relations like if $f(x)$ is a solution, then so is $g(x)=cf(x)=>$Let $f(1)=1$, $P(t,\frac{1}t) => f(\frac{1}{t})=\frac{1}{t}$ if $f(t)=t$, if $f(x)=x$ and $f(y)=y$ $=>$ $f(xy)=xy$... now how do you construct a "multiplicative sub-group" from this, and more importantly how would you write down such a function anyways?

I think all the above solutions explain this itself. Basically all you have to do is to show that there is a set of real numbers which satisfy the following relations : (1) Show that the sub-group contains the identity element i.e $1$ (2) Now if $x \in G \implies \frac{1}{x} \in G$ which implies the inverse element exists (3) For $x,y \in G \implies x \cdot y \in G$ which is the obvious definition of a group Showing all this gives $G$ is a multiplicative sub-group of $\mathbb{R}^*$ Now if we partition $\mathbb{R}^*$ into two mutually disjoint subsets $\{G\}$ and $\{\mathbb{R}^* - G\}$, then it is fairly clear to see that all such all such solutions work. Actually it is similar to the case of a point-wise trap!

I claim that $f(x)=cx$ when $x\in G$ and $f(x)=0$ otherwise is always a solution, where $G$ is a multiplicative group over $\mathbb{R}\setminus\{0\}$, and $c$ is an arbitrary constant. It is easy to verify that such solutions work. To show that all solutions are of this form, denote the assertion $P(x,y)$. We have that $P(1,0)$ yields $f(1)f(0)=(f(1)-f(0))f(0)$. So, either $f(0)=0$, or $f(1)-f(0)=f(1)\implies f(0)=0$. So, $f(0)=0$. Now, $P(1,x)$ yields $f(x)(f(x)-f(1))=(x-1)f(x)f(1)\implies f(x)(f(x)-xf(1))=0$. So, either $f(x)=0$ or $f(x)=xf(1)$. Assume $f(1)\neq 0$, or else $f(x)=0$. If we have nonzero $x\neq y$ such that $f(x)=xf(1),f(y)=yf(1)$, then $P(x,y)$ gives that $f(xy)\neq 0\implies f(xy)=xyf(1)$, for $x\neq y$. Similarly, we get that if $f(x)=xf(1),f(xy)=xyf(1)$, then $f(y)=yf(1)$. Denoting the set $G$ to be all $x$ such that $f(x)=xf(1)$, we have $x,y\in G\implies xy\in G$ for $x\neq y$, $x,y\in G\implies \frac{x}{y}\in G$, and $1,x\in G\implies x,\frac{1}{x}\in G\implies x^2\in G$. So, $G$ is a multiplicative group, as desired.

william122 wrote: I claim that $f(x)=cx$ when $x\in G$ and $f(x)=0$ otherwise is always a solution, where $G$ is a multiplicative group over $\mathbb{R}\setminus\{0\}$, and $c$ is an arbitrary constant. It is easy to verify that such solutions work. To show that all solutions are of this form, denote the assertion $P(x,y)$. We have that $P(1,0)$ yields $f(1)f(0)=(f(1)-f(0))f(0)$. So, either $f(0)=0$, or $f(1)-f(0)=f(1)\implies f(0)=0$. So, $f(0)=0$. Now, $P(1,x)$ yields $f(x)(f(x)-f(1))=(x-1)f(x)f(1)\implies f(x)(f(x)-xf(1))=0$. So, either $f(x)=0$ or $f(x)=xf(1)$. If we have nonzero $x\neq y$ such that $f(x)=xf(1),f(y)=yf(1)$, then $P(x,y)$ gives that $f(xy)\neq 0\implies f(xy)=xyf(1)$. So, if we define $G$ to be the set of all $x$ such that $f(x)=x$, we get that $G$ is a multiplicative group, as desired. technically, all you have shown is that $x,y\in G$ when $x\ne y$ implies $xy\in G$, so you still have to show $x^2\in G$, but this is not hard. I just think you might lose a point for that since it is a nontrivial required step.

Setting $y=1$, we have the following: $$f(x)^2=xf(x)f(1)$$If $f(1)=0$, then we have that $f(x)=0$ for all $x \in\mathbb{R}$, in the case that it isn't equal to $0$, we have that $f(x)=f(1)x$ Let's set $f(1)=c\neq 0$ and let's define the set $G=\{ y \mid f(y) \neq 0\}$ So we want to prove the following: $$ f(x) = \begin{cases} cx & x\in G \\0 & x \notin G \end{cases} $$ If this holds, then $G$ must be closed under multiplication and under division So now let's see some characteristics of $G$ 1) Obviously $1 \in G$ 2)If $x \in G$ and $y \notin G$, we see by the functional equation that $f(xy)f(x)=0$, since $f(x) \neq 0$, we have that $f(xy) = 0$, that means that $xy \notin G$ 3)If $x,y \in G$, we have that $\frac{x}{y} \in G$, because if this didn't hold, then by the second condition we have that $y \frac{x}{y} = x \notin G$ 4)From all of this we have that if $x,y \in G$, we have that $x^{-1} \in G$, and from this we have that $xy=\frac{y}{x^{-1}}\in G$ We have seen that $G$ is closed under multiplication and division, which contains $1$. So from this we have that the solution for this functional equation: $$ f(x) = \begin{cases} cx & x\in G \\ 0 & x \notin G \end{cases} $$Where $G \subseteq \mathbb{R}$ . . .

Let $P(x,y)$ denote the assertion, and let $f(1)=k.$ $\textbf{Claim: }$ $f(x)\in\{0,kx\}$ for all $x\in\mathbb{R}.$ $\textbf{Proof: }$ Assume $f(x)\ne 0.$ Then, $P(x,1)$ gives $$f(x)(f(x)-f(1))=(x-1)f(x)f(1)\implies f(x)-f(1)=(x-1)f(1).$$This implies that $f(x)=kx,$ as desired. $\blacksquare$ Let $S=\{x| f(x)=kx\}.$ $\textbf{Claim: }$ If $a\in S,$ then $\frac{1}{a}\in S.$ $\textbf{Proof: }$Let $x\in S.$ The assertion $P(x,1/x)$ simplifies to $$k(kx-f(1/x))=(x-1/x)kxf(1/x)$$$$\implies f(1/x)=k/x,$$which implies the desired conclusion. $\textbf{Claim: }$ If $a,b\in S$ such that $a\ne b,$ then $ab\in S.$ $\textbf{Proof: }$ Suppose $f(x)=kx$ and $f(y)=ky$ for some $x,y\in\mathbb{R}.$ From $P(x,y),$ we have $$f(xy)(kx-ky)=(x-y)(kx)(ky)\implies f(xy)=kxy,$$as desired. $\blacksquare$ It is easy to check that any function associated with a set $S$ satisfying the two conditions described above satisfies the equation.

Any working function $f$ is a particular $f_T$ where $T$ is a multiplicative group contained in $\mathbb{R}$ and \[f_T(x)=\begin{cases} cx&x\in T\\ 0&x\not\in T \end{cases}\]for some nonzero constant $c$. Work with $f_T(x)/c = f_T'$ instead for convenience. We check that such $f_T$ work in three steps; if $x\in T,y\in T$, then the condition rewrites as $xy(x-y)=(x-y)xy$. If $x\not\in T,y\not\in T$, then the condition rewrites as $f_T'(xy)\cdot0=(x-y)\cdot0$. If $x\in T,y\not\in T$, then the condition rewrites as $f_T'(xy)f_T'(x)=(x-y)\cdot f_T'(x)\cdot0$, so it is sufficient to show that $xy\not\in T$. But this follows from the observation that otherwise we would have $y=x^{-1}xy\in T$. Take $y=0,x=1$ to see that $f(0)f(1)-f(0)^2=f(1)f(0)\implies f(0)=0.$ Observe that if $f(x)=f(y)$, then $0=(x-y)f(x)f(y)$ implies at least one of $f(x)$ and $f(y)$ is zero, so both are. That is, the only $x$ for which $f^{-1}(x)$ could contain more than one element is $x=0$. Now, take $y\in f^{-1}(0)$ and observe that $f(xy)f(x)=0$, so either $f(x)=0$ or $f(xy)=0$ for each $x$. We now analyze the structure of $f^{-1}(x)=S$. We have observed that for each $y\in S$ and $x\in\mathbb{R}$, either $x$ or $xy$ is in $S$. Now, note that if there existed $x,y$ such that $xy\in S$ and $x,y\not\in S$, we would get a contradiction by substituting $x,y$ in. Now, observe that if $1\in S$, then any $x$ would be in $S$, so $f\equiv0$ or $1\not\in S$. Now, note that for any $y\not\in S$, $y^{-1}\not\in S$ because otherwise one of $1$ and $y$ would be in $S$. Hence, $T=\mathbb{R}\setminus S$ is a multiplicative subgroup of $\mathbb{R}$. Now, let $f(1)=c$ and define $g(x) = f(x)/c$ which satisfies the same properties as $f$ except $g(1)=1$. Take $x\in T$ and observe that the given information applied to $x,1$ yields $g(x)-f(1)=x-1$, so $x=g(x)$, giving us the claimed $f$.

Let $P(x, y)$ denote the assertion. Note that $P(x, 1)$ yields\[f(x)(f(x) - f(1)) = (x-1)f(x)f(1)\]and if we let $f(1) = c$ for some real constant $c$, then we get\[f(x)(f(x) - c) = c(x-1)f(x)\]which in fact, nicely rearranges to $f(x)(f(x) - cx) = 0 \implies f(x) \in \{0, cx\}$. First, note that $f(x) = 0$ is a solution; for the rest of the proof, we will assume $c \neq 0$. Clearly $f(0) = 0$. Let $S$ denote the set of all nonzero reals $r$ for which $f(r) = cr$. By definition, $1 \in S$. Furthermore, from $P(x, \tfrac{1}{x})$, we can see that\[c\left(f(x) - f\left(\frac 1x\right)\right) = \left(x - \frac 1x\right)f(x)f\left(\frac 1x\right)\]so if $x \in S \iff f(x) = cx$, then upon manipulation we get\[cx - f\left(\frac 1x\right) = x\left(x - \frac 1x\right)f\left(\frac 1x\right) \implies cx = x^2f\left(\frac 1x\right) \implies f\left(\frac 1x\right) = \frac{c}{x}\]so $\tfrac1x \in S$ as well. Furthermore, if we have $x \neq y$ both $\in S$, then from $P(x, y)$, we get that\[f(xy) = \frac{(x - y)c^2xy}{c(x - y)} = cxy\]hence $xy \in S$. Furthermore, if $x \in S$, we can use $P(x^2, \tfrac{1}{x})$ to deduce that\[cx\left(f(x^2) - \frac cx\right) = \left(x^2 - \frac 1x\right)f(x^2)\left(\frac cx\right)\]which upon solving for $f(x^2)$, yields $f(x^2) = cx^2$, so $x^2 \in S$ as well. Thus, the set $S$ is completely closed under multiplication, and division as well. It turns out that this is actually enough: all functions\[\boxed{f(x) = \begin{cases} 0 & \text{ if } x \not\in S\\ cx & \text{ if } x \in S \end{cases}}\]where $S$ is a set containing $1$ closed under both multiplication and division, work. This can be case-checked to work for all $(x, y) \in \mathbb{R}^2$, so we are done. $\blacksquare$

ISL 01A4. Find all functions $f: \mathbb{R}\mapsto\mathbb{R}$ such that \[\forall (x, y)\in\mathbb{R}^2: f(xy)(f(x)-f(y))=(x-y)f(x)f(y)\stackrel{\text{def}}{:} P(x, y).\] Solution. $P(x, 0): f(0)(f(x)-f(0))=xf(x)f(0)$. This motivates us to divide into two cases as follows: Case 1. $f(0)\ne 0$ Then $(1-x)f(x)=f(0)$. Plugging in this to the original equation we get $x+y=2 \forall (x, y)\in\mathbb{R}^2$, which is clearly a contradiction. Case 2. $f(0)=0$ If $f=0\forall x\in\mathbb{R}$ then it obviously satisfy $P$. It suffices to consider the case when $\exists c\ne 0: f(c)=0$. $P(x, 1): (x-1)f(x)f(1)=f(x)(f(x)-f(1))$. For $x\ne 0$, $f(x)=xf(1)$, which obviously satisfy $P$ again. In conclusion, $f\in\{f(x)=cx\forall x\in\mathbb{R}: c\in\mathbb{R}\}$.

Redacted

@above, once you get $f(xz)f(x)=0$, how do you get $f(x)=0\forall x$? Let $P(x,y)$ be the given assertion. $P(x,1)\Rightarrow f(x)^2=xf(x)f(1)\Rightarrow f(x)=\begin{cases}xf(1)&\text{if }x\in S\\0&\text{if }x\notin S\end{cases}$ where $S\subseteq\mathbb R\setminus\{0\}$. WLOG $f(1)=1$ (for ease). Case 1: $x,y\in S$ $P(x,y)\Leftrightarrow f(xy)=xy$, so $xy\in S$ Case 2: $x\in S,y\notin S$ $P(x,y)\Leftrightarrow xf(xy)=0$ and since $x\ne0$, we must have $f(xy)=0$. Case 3: $x\notin S,y\in S$ Same as Case 2 with $x,y$ switched Case 4: $x\notin S,y\notin S$ $P(x,y)\Leftrightarrow0=0$ So the entire solution set is: $$\boxed{f(x)=\begin{cases}cx&\text{if }x\in S\\0&\text{if }x\notin S\end{cases}}$$where $S\subseteq\mathbb R\setminus\{0\}$ and $xy\in S\Leftrightarrow x,y\in S$ and $c$ is any real constant. This works by the casework described above.

Let $P(x,y)$ be the assertion. $P(x,1) \implies f(x)^2=xf(x)f(1).$ Assume $f(1)\neq 0,$ and let $S$ be the set of all real $x$, s.t. $f(x)=xf(1).$ We then have the following (possible) solutions: \[\boxed{f(x) = \begin{cases} xf(1) & \text{ if } x \in S\\ 0 & \text{otherwise}, \end{cases}}\]where $f(1)$ is any real constant not $0,$ it's easy to check both of these fit. Case 1: $1\in S$ since $f(1)\neq 0.$ Case 2: $x\in S, y\notin S,$ then $P(x,y) \implies f(x)f(xy)=0,$ so $xy \notin S.$ Case 3: $x,y \in S$ then $\frac{x}{y} \in S.$ And by Case 2, $\frac{1}{x}, \frac{y}{x} \in S.$ To conclude, $S$ is a set containing $1$ but not $0$ and is closed under multiplication and division.

Let $P(x,y)$ denote the assertion. $P(x,1)$ gives $f(x)(f(x)-f(1))=(x-1)f(x)f(1).$ For each $x,$ we have $f(x)=0$ or $f(x)=xf(1).$ Clearly, if $f(1)=0$ then $f(x)^2=0$ which implies $f(x)=0$ for all $x$ which is a valid function. Let $\mathbb{S}$ be the set of $x$ such that $f(x)\neq 0.$ Suppose $\mathbb{S}$ is nonempty. We have $1\in \mathbb{S}$ but $0\notin \mathbb{S}.$ However, $\mathbb{S}$ is closed under multiplication because if $f(x)=xf(1)$ and $f(y)=yf(1)$ then $f(xy)f(1)(x-y)=(x-y)xy f(1)^2$ so if $x\neq y$ then $f(xy)=f(1)xy.$ Also, $\mathbb{S}$ is closed under division because if $x\in \mathbb{S}$ then $P(x,\frac{1}{x})$ gives \[f(1)(xf(1)-f(y))=(x-y)xf(1)f(y)\]but $f(1)\neq 0$ so $(x-y)xf(y)=xf(1)-f(y).$ If $f(y)=0$ then $xf(1)=0$ which is a contradiction. Thus, $\frac{1}{x}\in \mathbb{S}.$ Therefore, $\frac{y}{x}\in \mathbb{S}$ if $y\in \mathbb{S}$ which implies that $\mathbb{S}$ is closed under division. Note that all functions satisfying this also works: $f(x)=0$ unless $x\in \mathbb{S}$ in which case $f(x)=kx$ where $\mathbb{S}$ has the property that two of the following properties implies the third: $a\in \mathbb{S}, b\in \mathbb{S},ab\in \mathbb{S}$ and $1\in \mathbb{S}$ and $0\notin \mathbb{S}.$

The answer is $$f(x)=\begin{cases}cx& x \in S\\0 & x \not \in S,\end{cases}$$where $S$ is a subset of $\mathbb{R}$ closed under multiplication and division. A case check shows that these all work. First, $P(x,1)$ yields $$f(x)(f(x)-f(1))=f(x)f(1)(x-1) \implies f(x)(f(x)-xf(1))=0.$$Letting $x=0$ yields $f(0)=0$, henceforth all variables are nonzero unless otherwise specified. Further, if $f(1)=0$, then $f \equiv 0$, which is described by taking $S=\emptyset$. Henceforth suppose $f(1) \neq 0$, and note that by scaling we may assume that $f(1)=1$ (so $1 \in S$), so $f(x) \in \{0,x\}$ for all $x$. Suppose $x,y \in S$, where $x \neq y$. Then $P(x,y)$ implies $f(xy)=f(x)f(y)=xy$, so $xy \in S$ and $S$ is closed under multiplication of distinct elements. Likewise, if $x,xy \in S$, $P(x,y)$ implies (as variables are nonzero) $$xy(x-f(y))=(x-y)xf(y) \iff xy-yf(y)=xf(y)-yf(y) \iff f(y)=y,$$hence $y=\tfrac{xy}{y} \in S$ as well, so $S$ is (unconditionally) closed under division. We now show that $S$ is also closed under multiplication of non-distinct elements, which is a fancy way of saying that $x \in S \implies x^2 \in S$. Indeed, note that due to the properties of $S$ we discovered, including the fact that $1 \in S$, we have $x \in S \implies \tfrac{1}{x} \in S \implies \tfrac{1/x}{x}=\tfrac{1}{x^2} \in S \implies x^2 \in S$, which implies that $S$ is closed under multiplication and division, as desired. $\blacksquare$

Please tell me if my solution is wrong.

Let the set of all sets of real numbers that are closed over division containing $1$ be $S$. Let $S' = S \cup \emptyset$. Then the set of all valid functions $f$ can be characterized as the functions such that for some set $A \in S'$, we have $f(x) = cx$ for $x \in A$ and $x = 0$ otherwise. It is trivial to see that all such functions work. $P(x, 0)$ gives either $f(0) = 0$, or $f(x) = \frac{f(0)}{1-x}$, which doesn't work, so we can conclude $f(0) = 0$. Now consider the values which upon $f$ is nonzero. Clearly if $f(x)$ is nonzero, we have $P(x,1)$ gives $f(x)(f(x) - f(1)) = (x-1)f(x)f(1)$. If $f(1) = 0$, this forces $f(x) = 0$, which is a contradiction, otherwise we have $f(x) = f(1)x$. Now, we show that the set of values which $f$ is nonzero is closed under divison. Firstly, note that $f$ is injective when it is nonzero. For division, notice that for distinct $a,b$ (the case where $a = b$ just wants us to show $1$ is in the set, which has already been shown), we have $f(a)(f(b) - f(\frac ab)) = (b- \frac ab)f(a)f(b)$, where the left side is clearly nonzero. We are done.

Solved with cxyerl. Let $G$ be any multiplicative subgroup of $R^{*}$. The answer is $f(x)=Cx$ when $x\in G$ and $f(x)=0$ otherwise, where $C$ is any real. This "clearly" works. $P(x,1)$ gives $f(x)=0$ or $f(x)=Cx$ where $C=f(1)$ for each individual $x$, so $f(0)=0$. Now let $S$ be the set of nonzero reals $x$ such that $f(x)=Cx$. Clearly $S$ is closed under multiplication and $x\in S$ iff $1/x\in S$. This means that $S$ is a multiplicative subgroup of $R^{*}$. $\blacksquare$

Claim: $f$ is injective on nonzero outputs. Proof. Suppose that $f(a) = f(b) \ne 0$. Then $P(a, b)$ implies that $a = b$. $\blacksquare$ Let $S$ be the kernel of $f$. First suppose that $a \in S$ such $a \ne 0$. If $1 \in S$ then by $P(x, 1)$ it follows that $f(x)$ is uniformly $0$, so suppose not. Then, by $P(x, 1)$ \[ f(x)(f(x) - f(1)) = (x - 1)f(x)f(1) \]so either $f(x) = 0$ or \[ f(x) = xf(1) \] Claim: ${\mathbb R} \setminus S$ forms a multiplicative subgroup. Proof. For $x, y \not\in S$ then \[ f(x)\left(f(y) - f\left(\frac{x}{y}\right)\right) = \left(y - \frac{x}{y}\right)f(y)f\left(\frac{x}{y}\right) \]so either $y^2 = x$ or $f\left(\frac{x}{y}\right)$ is nonzero, but the first implies the second. Then, since $1 \not\in S$ it follows that \[ \frac{x}{\frac{1}{y}} = xy \not\in S \]$\blacksquare$ As such, the solution set is then \[ f(x) = \begin{cases} cf(x) & x \in H \\ 0 & x \not\in H \end{cases} \]where $H < {\mathbb R}$ is a multiplicative subgroup.

This is sweet. Let $S$ be a multiplicative subgroup of $\mathbb R$. Then $f(x) = cx$ for a real $c$ and $x \in S$ and $f(x)=0$ otherwise works; this is evident through casework. Firstly, if $f(a) = f(b) \neq 0$, we get \[ 0 = f(ab) (f(a)-f(b)) = f(a)f(b)(a-b) \]which means $a=b$. Secondly,\[ f(0)(f(0) - f(a)) = -af(0)f(a) \implies f(0)^2 = 0. \](by setting $a=1$) Let $S$ be the set of reals such that $\forall s \in S$ we have $f(s) \neq 0$ and suppose it is nonempty. We verify that it is a multiplicative subgroup. $\textbf{Identity.}$ Note that if $f(a) \neq 0$, \[ f(a)(f(a) - f(1)) = f(a) f(1) (a - 1) \]which implies $f(a) = af(1)$. As such, $f(1) \neq 0$ which means $1 \in S$. $\textbf{Closure.}$ We have that if $a,b$ are in $S$ and are distinct, $f(a) - f(b) \neq 0$ and $(a-b) \neq 0.$ It follows that $f(ab) \neq 0$. $\textbf{Inverse.}$ If $a \neq 1/a$ is in $S$, \[ f(1)(f(a) - f(1/a)) = (a - 1/a) f(a) f(1/a) \]and after supposing $f(1/a) =0 $ we derive a contradiction. If $a = 1/a$ then $a$ is its own inverse and is in $S$. Associativity is trivial. This readily solves the problem.

Let $K \in \mathbb{R}$ s.t. it is closed under multiplication and division then $f(x)=cx$ works for $c \in \mathbb{R}$, verification is easy. By $P(x,0)$, we get $f(x)=0$ or $f(x)=cx$ for every value of $x$, then by some value putting we get the claimed set as solution.

Let $P(x,y)$ denote the assertion. $P(1,0)$ implies that $f(0) = 0$. First, assume that there are no $a \neq 0$ such that $f(a) = 0$. For $x \neq 0$, $P(x,1)$ implies (since $f(x) \neq 0$): \[f(x) - f(1) = (x-1) \cdot f(1) \implies f(x) = cx\]which satisfies $P(x,y)$. Now, assume that there exists at least one value of $a \neq 0$ such that $f(a) = 0$. Taking $P(x,a)$ implies \[f(ax) \cdot f(x) = 0 \text{ for all $x$}\]Thus if $f(b) \neq 0$, then $f(ab) = 0$. Repeating this with $a = ab$, we get $f(ab^2) = 0$, and so on. Note this holds for any $b$ such that $f(b) \neq 0$. Now let $S_0$ be the set of reals such that $f(x) = 0$, and let $S$ be the set of all other reals. Due to what we previously said, we must have: 1) For any $x \in S$, $P(x,1)$ implies that $f(x) = cx$ for some fixed constant $c$. 2) For any $x \in S_0$ and any $a \in S_0$, we have $ax \in S_0$. Thus the answer is \[f(x) = \begin{cases} 0 \text{ for any $x \in S_0$} \\ cx \text{ for any $x \not \in S_0$} \\ \end{cases}\]where for any $x \not \in S_0$, we have $ax \in S_0$ for any $a \in S_0$ where $S_0$ can be any set satisfying these conditions.

Note that $P(x, 1) \implies f(x) \in \{0, cx\}$. Let $A = \{x : f(x) = 0\}, B = \{x: f(x) \ne 0\}$. Some cases give us that $A$ and $B$ satisfy the condition that $b \in B \implies \chi(ab) = \chi(a) \forall a$, where we define $\chi(x)$ to be zero if $x \in A$ and one if $x \in B$. We can check that such functions work, qed (??)