Oh now I understand. I forgot about $f(0) = 0$ in my first case.

AkshajK wrote: we can divide to get $f(x) = \dfrac{f(0)}{(1 + f(-1))}$ for all $x$ You can divide by $1 + f(-1)$ only if it is not equal to zero.

semisimplicity wrote: AkshajK wrote: we can divide to get $f(x) = \dfrac{f(0)}{(1 + f(-1))}$ for all $x$ You can divide by $1 + f(-1)$ only if it is not equal to zero. Yes, thats why I made a separate case where there exists a root.

AkshajK wrote: Yes, thats why I made a separate case where there exists a root. $1 + f(-1) = 0$ does not imply that $f$ has a non-trivial root.

$f(0)=0$ and if f isnt the zerofunction, we get $f(-1)=-1$ then using this value for x and $y=-0.5$ we get $f(-0.5)=-0.5$ and then $ f(q)=q$ is with induction to naturals and so on an easy task. Next as $f(ab)=f(a)f(b)$ we get for positive values a positive dunctionvalue, so f is nondecreasing and cauchys equation gives there is only one way to complete. Hence there are two functions satisfying the question.

semisimplicity wrote: Find all functions $f$ from the set of real numbers to itself satisfying \[ f(x(1+y)) = f(x)(1 + f(y)) \] for all real numbers $x, y$. Let $P(x,y)$ be the assertion $f(x(1+y))=f(x)(1+f(y))$ $P(0,y) \Longrightarrow f(0)=f(0)(1+f(y))$ Now it's obvious $f(x)=0 \forall x \in \mathbb R$ is solution. So now we look for non-zeroes functions. $P(0,y) \Longrightarrow f(0)=0$ $P(x,-1) \Longrightarrow f(-1)=-1$ Set $P(\frac {x}{y} , \frac {1}{y})$ , then $\text {LHS}$ does not change so : \[ f(xy)+f(xy)f(\frac {1}{y})=f(x)+f(x)f(y) \] Now take $x=1$ in last equation we get : \[ f(x)(1+f(\frac {1}{x}))=(1+f(x))f(1) \] \[ \Longrightarrow f(1)f(x)-f(x)=f(x)f(\frac {1}{x})-f(1) \] Change $x$ with $\frac {1}{x}$ , since $\text {RHS}$ does not change we get : \[{ f(1)f(\frac {1}{x})-f(\frac {1}{x}})=f(1)f(x)-f(x) \] \[ \Longrightarrow f(1)(f(x)-f(\frac {1}{x}))= f(x)-f(\frac {1}{x}) \] Now two case : Case 1 : $f(1)=1$ $P(1,y) \Longrightarrow f(y+1)=f(y)+1$ So we have : \[ f(x(1+y))=f(x)(1+f(y))=f(x)f(y+1) \] $P(x,y) \Longrightarrow f(x+xy)=f(x)+f(x)f(y)=f(x)+f(xy)$ So we get $f(x)=x \forall x \in \mathbb R$ Case 2 : $f(x)=f(\frac {1}{x})$ Which is easy case. So two solutions : $1) f(x)=x$ $2) f(x)=0$

War-Hammer wrote: $2) f(x)=0$ for all $x \in \mathbb R$ except $-1$ and $f(-1)=-1$ This is false. Take $x=-2,y=-0.5$ to get $-1=0$

Sorry for my mistake ! Thanks ! Edited.

we have that $f(0)=0, f(-1)=-1$ and $f(y+1)=f(1)(1+f(y))$ and $f(y)+f(-y-1)=-1$. So $f (-4)=f(-2)(1+f(1))=-(1+f(1))^2 $ and $ f(-4)=-1-f(3)=-1-f(1)(1+f(2))=-1-f(1)-f(1)^2-f(1)^3 $. Hence $f(1)={-1,0,1}$. i) $f(1)=-1$; $P(x,1)$ $ \Rightarrow $ $f(2x)=0$. ii) $f(1)=0$; $f(y+1)=0$. iii) $f(1)=1$; $f(y+1)=f(y)+1$ and easily $f$ is multiplicative and additive. So, $f(x)=x$. Answer: 1) $f(x)=x$ for all $x\in R$. 2) $f(x)=0$ for all $x\in R$.

Let $P(x,y)$ be the solution. $P(0,y) \implies f(0)=f(0)(1+f(y)) \implies f(0)=0$ or $f(y)=0 \forall y$. The second case is a solution. So, let us continue with first case. Let there exist $a$ such that $f(a) \neq 0$. $P(a,-1) \implies f(a)(1+f(-1))=0 \implies f(-1)=-1$. $P(1,y) \implies f(1+y)=f(1)(1+f(y))$. Therefore, $P(1,1) \implies f(2)=f(1)^2+f(1)$. $P(1,2) \implies f(3)=f(1)^3+f(1)^2+f(1)$ $P(1,3) \implies f(4)=f(1)(1+f(1)+f(1)^2+f(1)^3)$ $P(2,1) \implies f(4)=f(1)(1+f(1))^2$ Comparing the last two, $(f(1)^3-f(1))(f(1))=0 \implies f(1)=-1,0,1$. $f(1)=-1 \implies P(x,1) \implies f(2x)=f(x)(1+(-1))=0 \forall x$, contradiction. $f(1)=0 \implies P(1,y) \implies f(1+y)=0 \forall y$, contradiction. $f(1)=1 \implies P(1,y) \implies f(1+y)=1+f(y) \implies f(x(1+y))=f(x)f(1+y)$ i.e $f$ is a multiplicative function. Also, $P(x,\frac{y}{x}) \implies f(x+y)=f(x)+f(x)f(y/x)=f(x)+f(y)$ i.e $f$ is additive. It is very well known that these $2$ conditions imply that $f(x)=x$.

Yet another approach. Putting $x=-1$, we get $f[-(y+1)]=-[f(y)+1]$................(1) Putting$x=1$,we get $f(y+1)=f(1)(-f(-[y+1]))$(putting value from (1)) Thus $-\frac{f(x)}{f(-x)}=f(1)$ Now putting $y=-2$ in the original equation $f(-x)=f(x)[f(-2)+1]$ $-\frac{1}{f(1)}=f(-2)+1$...............(2) Putting $y=1$ in (1) we get $f(-2)=-f(1)-1$ Putting this value in (2) we get $\frac{1}{f(1)}=f(1)$ Thus $f(1)=1$ or $f(1)=-1$ For $f(1)=-1$ we get $f(x)=0$ for all x.Thus contradiction. and once we get $f(1)=1$ it is easy to prove that the functions are additive and multiplicative.Rest is trivial.Please site any mistake. Sorry I Missed a part. Putting $x=0$ we get $f(0)=f(0)[f(y)+1]$.Thus $f(0)=0$ or $f(y)=0$ for $y$.Now the whole solutions are ready. Can you people site a simple way of proving that both additive and multiplicative solutions are nothing but $f(x)=x$.Actually to prove it I have to use the Sabdwich Theorem.But I think there is a elementary way to prove it and even with the Sandwich the logic is not too much clear.

I've proved additive and multiplicative of function $f$ on my solution (#3)

My proof: It is easy to derive that $f(q)=q$ for all rational numbers $q$. $f(x)f(y)=f(xy)$ Putting $x=y=z$ we get $f(z^2)=f(z)^2$ Putting $z^2=x$ we get $f(x)=f(\sqrt{x})^2\ge0$ Clearly for all $x\ge0$ we have $f(x)\ge0$ Now $f(y)=f(y-x+x)=f(y-x)+f(x)$ If $y\ge x$ then then $f(y-x)\ge0$ as $y-x\ge0$ So $f(y)=f(y-x)+f(x)\ge f(x)$. Thus $f$ is a strictly increasing function. Now take a real number $x$ and two infinite sequences of rationals {$q_i$} and {$r_i$} such that $q_1<q_2<.....<x$ and $r_1>r_2>......>x$ and $q_i<x<r_i$ Now $\lim_{i\to\infty}\ q_i=\lim_{i\to\infty}\ r_i=x$ Now $x=(\lim_{i\to\infty}\ q_i)=\lim_{i\to\infty}\ f(q_i)\le\lim_{i\to\infty}\ f(r_i)=(\lim_{i\to\infty}\ r_i)=x$Here it is clear where I have used the monotonicity of the function.So $\lim_{i\to\infty}\ f(q_i)=\lim_{i\to\infty}\ f(r_i)$.....(1) But as $q_i< x <r_i$ so using monotonicity $f(q_i)\le f(x) \le f(r_i)$Using this and (1) we apply the Sandwich theorem $\lim_{i\to\infty} f(x)=x$.But as $x$ is fixed so $\lim_{i\to\infty} f(x)=f(x)$.Thus $f(x)=x$

$f(x(1+y))=f(x)(1+f(y))$......(1) Let us first consider the case when $f(1)=1$. Taking $x=1$ in (1) we get $f(y+1)=f(y)+1$....(2) This along with $f(1)=1$ implies $f(n)=n$ for all $n%Error. "belongs" is a bad command. \mathbb{Z}$ Take an arbitrary rational $\frac{p}{q}$ where $p,q %Error. "belongs" is a bad command. \mathbb{Z}$ Then we have $p=f(p)=f(\frac{p}{q}q)=f(\frac{p}{q}(1+(q-1)))=qf(\frac{p}{q})$ $\Rightarrow f(r)=r$ for all rationals $r$. Also an easy induction using (2) yeilds $f(y+n)=f(y)+n$ for all natural numbers n. (1) can also be written as $f(xz)=f(x)f(z)$ for all $x,z %Error. "belongs" is a bad command. \mathbb{R}$. Taking $z=x$ we get $f(x^2)=(f(x))^2 \Rightarrow f(x)>0$ for all positive reals x.(It is easy to prove that $f(m)=0\Rightarrow m=0$) Using the multiplicativity and (1) we get $f(x+y)=f(x)+f(y)$ for all $x\neq0,y %Error. "belongs" is a bad command. \mathbb{R}$ Now suppose $x\neq0$ and $y$ are two arbitrary reals with $x>y$.Then $f(x)=f(x-y+y)=f(x-y)+f(y)>f(y)$ So $f$ is a strictly increasing function.(It is trivial when x is 0) Let $f(x)<x$ for some real x(which is not rational) Then there exists a rational r between $x$ and $f(x)$ such that $f(x) \le r <x$. But $r<x \Rightarrow r=f(r)<f(x)$ which contradicts our assumption. Similarly,one can see that $f(x)>x$ is also not possible. Therefore $f(x)=x \forall x %Error. "belongs" is a bad command. \mathbb{R}$ It is easy to see that $f(1)=0 \Rightarrow f(x)=0 \forall x%Error. "belongs" is a bad command. \mathbb{R}$.

Clearly, $ f(x)=0 $ is solution for all real $ x $.Now we find that non-zero functions.Let $ P(x,y) $ be the assertion of the function.By assumptation $ f(y)+1 \neq 1 $ . $ P(0,y) \rightarrow f(0)=0 $. $ P(x,-1) \rightarrow f(-1)=-1 $ .Let $ f(1)=a $. $ P(1,1) \rightarrow f(2)=a^2+a $. $ P(-1,-2) \rightarrow a= -(f(-2)+1 ) $ $ P(2,-2) \rightarrow f(-2)=f(2)(f(-2)+1) ... \rightarrow a=1 $.Thus we have $ f(y)+1=f(y+1) $ and then $ f $ is multiplicative and then $ f $ is addite. So $ f(x)=x $. Answers: 1) $ f(x)=0 $ . 2) $ f(x)=x $ .

Let $P(x,y)$ denote the statement $f(x(1+y))=f(x)(1+f(y))$. I am leaving out the calculations, as they are easy to notice. $P(0,0)\Rightarrow f(0)=0$. Note that the only constant solution is $f(x)\equiv 0\forall x\in \mathbb{R}$. So wlog assume that $f$ is non-constant. Then $\exists w\in \mathbb{R}$ such that $f(w)\neq 0$. Now, $P(w,-1)\Rightarrow 1+f(-1)=0\Rightarrow f(-1)=-1$. $P(-1,y)\Rightarrow f(-1-y)=-1-f(y)..........(i)$ $P(-1,1)\Rightarrow f(-2)=-1-f(1)\Rightarrow 1+f(-2)=-f(1)$ $P(x,-2)\Rightarrow f(-x)=-f(x)(1+f(-2))=-f(x)f(1)..........(ii)$ So, $-1=f(-1)=-f(1)^2\Rightarrow f(1)=1$ or $f(1)=-1$. If $f(1)=-1$, then, $P(x,1)\Rightarrow f(2x)=0\Rightarrow f(x)\equiv 0\forall x\in \mathbb{R}$, contradicting our assumption. So, $f(1)=1$. Then, from $(ii)$, we have $f(-x)=-f(x)\forall x\in \mathbb{R}$. Then from $(i)$, we can write that $f(1+y)=1+f(y)$. Now, $P(x,y-1)\Rightarrow f(xy)=f(x)(1+f(y-1))=f(x)f(y)\forall x,y\in \mathbb{R}$, i.e. $f$ is multiplicative. Again, $P(x,y)$ can be re-written as $P(x,y)\Rightarrow f(x+xy)=f(x)+f(xy)\forall x,y\in \mathbb{R}$. Now, $P(x,\dfrac{z}{x})\Rightarrow f(x+z)=f(x)+f(z)$ whenever $x\neq 0$. Also, since $f(0)=0$, we have $f(x+y)=f(x)+f(y)\forall x,y\in \mathbb{R}$. This means that $f$ is additive too. Now by Cauchy, $f(x)=cx$ for some real $c$. Substituting it in $(i)$ yields that $c=1$. So the solutions are: $f(x)\equiv 0 \forall x\in \mathbb{R}$ and $f(x)=x \forall x\in \mathbb{R}$.

Note that $f(x)\equiv 0$ is the only constant solution, so let us look for non-constant solutions from now on. Let $P(x,y)$ denote the assertion that $f(x(1+y)) = f(x)(1 + f(y))$. Claim 1. $f(0)=0$. Proof: $P(0,y)\implies f(y)f(0)=0$, and since non constant, $f(0)=0$. Claim 2. $f(x)=0\implies 0$ or that $x\ne 0\implies f(x)\ne 0$. Proof. Suppose there exists nonzero $a$ such that $f(a)=0$. Then $P\left(a,\frac{x}{a}-1\right)\implies f(x)=0\forall x$, which is already ruled out. Claim 3.$f(1)=1$. Proof.Suppose $f(1)=c$. Note that $P(1,x)\implies f(x+1)=c(f(x)+1)$. Then putting $x=1,2,3$ respectively in this equation we have : \begin{align*} f(2)=& c^2+c\\ f(3)=& c^3+c^2+c\\ f(4)=& c^4+c^3+c^2+c=c(c^2+1)(c+1)\end{align*}But $P(2,1)\implies f(4)=f(2)(1+f(1))\implies c(c+1)(c^2+1)=(c^2+c)(c+1)\implies c(c^2+c)(c-1)=0$. But $c=f(1)\ne 0$ and $c^2+c=f(2)\ne 0$ by claim 2. Thus $f(1)=c=1$. Claim 4. $f(x+1)=f(x)+1$. Proof. This follows from $P(1,x)$ and Claim 3. Claim 5. $f$ is multiplicative. Proof. From $P(x,y)$ and Claim 4, we have $f(x(1+y))=f(x)f(1+y)$. Now setting $y\mapsto y-1 $ here, the claim follows. Claim 6. $f$ is additive. Proof. From the original equation and Claim 5, we infer that $f(x+xy)=f(x)+f(x)f(y)=f(x)+f(xy)$. Now putting $y\mapsto \frac yx$ here , the claim follows. Conclusion. It is well known that the only non constant function which is both addiive and multiplicative is the identity function. So all solutions are $\boxed{S1:f(x)=0\forall x\in\mathbb{R}}$ and $\boxed{S2:f(x)=x\forall x\in\mathbb{R}}\quad\blacksquare$.

Note that $f(x) \equiv 0$ is a solution. Suppose that $f$ is not identically zero. Taking $x=y=0$, we have $f(0) = f(0) + f(0)^2$, so $f(0)=0$. Taking $y=-1$, we have $0 = f(x)(1+f(-1))$. Since $f$ is not identically zero, we have $f(-1) = -1$. Taking $x=y$, we have $f(x+x^2) = f(x)+f(x)^2$ for all $x$. Then for all $t \ge -1/4$, we have $f(t) \ge -1/4$. Taking $y=-2$, we have $f(-x) = f(x)(1+f(-2))$ for all $x$. Replacing $x$ with $-x$, we have $f(x) = f(-x)(1+f(-2))$ for all $x$. Since $f$ is not identically zero, we must have $(1+f(-2))^2 = 1$. Thus $f(-2) \in \{0, -2\}$. If $f(-2) = 0$, then $f(-x) = f(x)$ for all $x$ (i.e. $f$ is even), so $f(1) = f(-1) = -1$. But we need $f(1) \ge -1/4$, a contradiction. Thus $f(-2) = -2$, and $f(-x) = -f(x)$ for all $x$ (i.e. $f$ is odd). Taking $x=-1$, we have $f(-1-y) = -(1+f(y))$ for all $y$, or $f(y+1) = 1+f(y)$ for all $y$. Then the original equation gives \[f(x(1+y)) = f(x)f(1+y)\]for all $x, y$. That is, $f(ab)=f(a)f(b)$ for all $a, b$. Then \[f(x+xy) = f(x)+f(x)f(y) = f(x) + f(xy)\]for all $x, y$, which is equivalent to Cauchy's functional equation. Since $f$ is bounded below by $-1/4$ on the interval $[-1/4, \infty)$, this forces $f(x) = cx$ for some constant $c \neq 0$. Substituting into the original equation, we need $cx(1+y) = cx(1+cy)$, forcing $c=1$. Thus, the solutions to the equation are $f(x) = 0$ for all $x$ and $f(x) = x$ for all $x$ (these obviously work).

Nice FE. Let $P(x,y)$ be the given assertion. Note $P(0,0)\implies f(0)=0$. $P(x,-1)$ gives $0=f(x)(1+f(-1))$, meaning $\boxed{f=0}$ is a solution, or $f(-1)=-1$. From here on assume $f$ is non-trivial. Taking, $P(1,-2)$ and $P(-1,-2)$, we see $$-1=f(1)(1+f(-2))\text{ and } f(1)=-1-f(-2)$$Thus, $f(1)^2=1\implies f(1)=\pm 1$. If $f(1)=-1$, then taking $y=1$, we get $f(2x)=0$ for all $x$, impossible since $f(1)\neq 0$. Therefore, $f(1)=1$. Now take $P(x,1)$ to get $f(1+y)=1+f(y)$. Then, $$f(x(1+y))=f(x)(1+f(y))=f(x)f(1+y),$$which means $f(ab)=f(a)f(b)$ for all $a,b$. Note further that $$f(a+b)=f(a(1+b/a))=f(a)(1+f(b/a))=f(a)+f(a)f(b/a)=f(a)+f(b).$$This is Cauchy's equation, and $f(x^2)=f(x)^2\ge 0$ by multiplicativity, which means $f$ is bounded below on the interval $[0,\infty)$. Thus, $f(x)$ is linear, and the non-trivial solution is $\boxed{f(x)=x}$

Yet another approach worked out by me and user Omega18. Let, $P(x,y)$ be the proposition that, $f(x(1+y)) = f(x)(1 + f(y))$ $\qquad \longrightarrow (1)$ Now, it is trivial to show that $f(0)=0$ and that the only constant function satisfying $(1)$, is the function $f \equiv 0$. Hence, from now on I assume that $f$ is non-constant. Thus, there should exist a $c \in \mathbb{R} $ such that $f(c) \neq 0 $. $P(c,-1)$ $\implies f(0)=f(c)(1+f(-1))$ $\implies f(-1)=-1 $ Now, $P(-1,y-1)$ $\implies -f(-y)=f(y-1)+1$ $ \qquad \longrightarrow (2) $ Then, we have, $P(x,y-1)$ $\implies f(xy)=-f(x)f(-y) $ $\implies f(-xy)=-f(x)f(y) $ $\qquad \longrightarrow (3) $ Also, $P(-x,y-1)$ $\implies f(-xy)=-f(-x)f(-y)$ $\qquad \longrightarrow (4) $ Now, if we put $x=y=-1$ in $(4)$, we have, $f(-1)=-f(1)^2$. Therefore, $f(1)=1 \; or \; -1$. Case $I$:$f(1)=-1$ Therefore, $P(x,1)$ $\implies f(2x)=0 $. Thus, we have $f \left( 2 \times \frac{1}{2} \right) = f(1)=0 $ [An outright contradiction] Therefore, there are no functions in this case. Case $II$: $f(1)=1$ Now, we compare $(3)$ and $(4)$. Therefore, $f(x).f(y)=f(-x).f(-y)$ $\qquad \longrightarrow (5) $ Putting $y=1$ in $(5)$, $\implies f(x) = -f(-x)$ $ \qquad \longrightarrow (6) $ Thus, $(6)$ and $(3)$ together conclude that, $f(x).f(y)=f(xy)$ that is $f$ is multiplicative. Rearranging $P(x,y)$ and using the multiplicative property, we have $f(x+xy)=f(x)+f(xy) $. Therefore, we have $f(x+y)=f(x)+f(y)$, $\forall x,y \in \mathbb{R} $ [$x \neq 0 $] Therefore, along with the fact that $f(0)=0$ and the above result we also conclude that $f$ is additive. But then, it is a known fact that if a function $f$ is both additive and multiplicative on $\mathbb{R}$, then $f \equiv 0 $, or $f(x)=x \; \forall x \in \mathbb{R} $.

It is obvious that $f(x)=0$ for all real $x$ is a solution. So, $\boxed{\textrm{S1:} f(x)=0 \ \forall x\in \mathbb{R}}$. Now assume that $f(x)$ isn't the zero function, then denote $P(x,y)$ by the given assertion, we have \[P(0,0): f(0)^2=0 \implies f(0)=0.\]Moreover, as \[P(x,-1): f(x)(1+f(-1))=0 \ \forall x\in \mathbb{R} \implies f(-1)=-1,\]we have \[P(-1,1):f(1)=-1-f(-2)\]and \[P(1,-2): -1=f(1)(1+f(-2))\]so $f(1)^2=1\implies f(1)=\pm 1$. However, notice that if $f(1)=-1,$ then $f(-2)=0$ and \[P\left(-2,-\frac{x}{2}-1\right): f(x)=0 \ \forall x\in \mathbb{R}\]which is a contradiction. Therefore, $f(1)=1$ and \[P(1,x): f(x+1)=1+f(x)\]\[P(x,x-1): f(x^2)=f(x)(1+f(x-1))=f(x)^2.\]As $f$ satisfies $f(x+1)=1+f(x)$ and $f(x^2)=f(x)^2$, $\boxed{\textrm{S2:} f(x)=x \ \forall x\in \mathbb{R}}$. $\quad \blacksquare$

semisimplicity wrote: Find all functions $f$ from the set of real numbers to itself satisfying \[ f(x(1+y)) = f(x)(1 + f(y)) \]for all real numbers $x, y$. Nice problem. Now let $P(x,y)$ be the assertion Then $P(0,0)\implies f(0)=0$ Now $P(1, y)\implies f(1+y)=f(1)(1+f(y))$ there are $2$ possible cases we will examine both. Case 1-: $f(1)=0$ then we get $f(x)=0 \forall x\in R$ Case 2-: $f(1)\neq 0$ So $f(0)=f(1)(1+f(-1))\implies f(-1)=-1$ So $P(-1, y)\implies f(-(1+y))+f(y) =-1 --*$ Now Substituting $y=\frac{-1}{2}$ in Eq * gives $f(\frac{-1}{2})=\frac{-1}{2}$ Now $P(\frac{-1}{2},y)\implies 2f(\frac{-(1+y)}{2})+f(y)=-1 --**$ Now Comparing Eq $*, **$ we get that $2f(\frac{x}{2})=f(x)\implies 2f(x)=f(2x)$ Now $P(x, 1)\implies f(2x)=f(x)+f(1)f(x)\implies f(1)=1$ So $P(1, y)\implies f(1+y)=f(1)+f(y)$ and so $f(x(1+y))=f(x)(f(1+y))$ Hence we get that $f$ is additive and multiplicative so $f(x)=x$ in this case $\blacksquare$

semisimplicity wrote: Find all functions $f$ from the set of real numbers to itself satisfying \[ f(x(1+y)) = f(x)(1 + f(y)) \]for all real numbers $x, y$. Set x=0. f(0) = f(0) * (1 + f(y)) --> f(0) =0 or f(y) = 0. f(x) = 0 is therefore a possible function. The equation is always true for y=0.

$\textbf{Case 1: }\exists j:f(j)\ne0$ $P(0,0)\Rightarrow f(0)=0$ $P(j,-1)\Rightarrow f(-1)=-1$ $P\left(-1,-\frac12\right)\Rightarrow f\left(-\frac12\right)=-\frac12$ $P\left(2x,-\frac12\right)\Rightarrow f(2x)=2f(x)$ $P(j,1)\Rightarrow f(1)=1$ $P(1,x)\Rightarrow f(x+1)=f(x)+1$ $P\left(x,\frac yx\right)-P\left(x,\frac yx-1\right)\Rightarrow f(x+y)=f(x)+f(y)\forall x\ne0\Rightarrow f(x+y)=f(x)+f(y)$ $P(x,y-1)\Rightarrow f(xy)=f(x)f(y)\Rightarrow\boxed{f(x)=x}$, works. $\textbf{Case 2: }\boxed{f(x)=0}$, works.

$$f(xy)=f(x)f(y),f(x+1)=f(x)+1$$$x \implies x+1 : f(xy + y) =(f(x)+1)f(y)=f(xy)+ f(y) \implies f(x+y)=f(x)+f(y)$ $$\boxed{f(x)=x}$$

lazizbek42 wrote: $$f(xy)=f(x)f(y),f(x+1)=f(x)+1$$$x \implies x+1 : f(xy + y) =(f(x)+1)f(y)=f(xy)+ f(y) \implies f(x+y)=f(x)+f(y)$ $$\boxed{f(x)=x}$$ I think by RMM 2013 P1, if u have f(x+1)=f(x)+1 and f(x)^2=f(x^2) you are done

Certainly $f\equiv 0$ works, so we disregard this. Denote the assertion by $P(x,y)$ we claim that $f$ is the identity, which works. $P(0,x)$ implies $f(0)=0.$ Now if $f(x)=0$ for some $x\neq 0$ then $P(x,yx^{-1}-1)$ forces $f(y)=0,$ absurd. i.e. \ injective at zero. Let $f(1)=a$ then: \begin{align*} P(1,1)\implies &f(2)=a^2+a \\ P(1,2)\implies & f(3)=a^3+a^2+a \\ P(1,3)\implies & f(4)=a^4+a^3+a^2+a \\ P(2,1)\implies &a\in \{0,1,-1\}.\end{align*}Note that if $a=0$ is not possible and $a=-1$ means $f(2)=0$ also not possible, so $a=1.$ Now $P(1,x)$ yields $f(x+1)=f(x)+1.$ Then $P(x,y)$ gives $f(x(y+1))=f(x)f(y+1)$ i.e. \ $f$ is multiplicative. Again $P(x,y)$ gives $f(x+xy)=f(x)+f(xy)$ i.e. \ $f$ is additive. As desired.

The only solutions are $f\equiv 0$ and $f(x) = x$, which work. Let $P(x,y)$ denote the given assertion. Since $0$ is the only constant solution, assume $f$ isn't constant. Claim: $f$ is injective at $0$. Proof: If $f(k) = 0$ for some $k \ne 0$, then $x = k$ gives $f(k(1 + y)) = 0$, so $f \equiv 0$, absurd. $\square$ $P(0,0): f(0) = f(0)(f(0) + 1)$, so $f(0) = 0$. $P(x,-2): f(-x) = f(x) (1 + f(-2))$. $P(-x,-2): f(x) = f(-x) (1 + f(-2)) = f(x) (1 + f(-2))^2$. Now choosing $x$ with $f(x) \ne 0$ gives $(1 + f(-2))^2 = 1$. Now, $f(-2) \ne 0$ by our claim so $1 + f(-2) = -1$, meaning $f(-2) = -2$ and therefore $P(x,-2)$ now implies $f$ is odd. $P(-2, -1): 0 = -(1 + f(-1))$, so $f(-1) = -1$ and thus $f(1) = 1$. $P(1, x): f(x+1) = f(x) + 1$. $P(x, y + 1) - P(x,y): f(xy + 2x) - f(xy + x) = f(x)$. Hence \[f(x(y+1)) + f(x) = f(x(y+1) + x) \ \ \ \ \ \ \ (1)\] Claim: $f(a) + f(b) = f(a+b)$ for all reals $a,b$ Proof: If $a,b$ are nonzero, fix $x = a$ and $x(y+1) = b$ in $(1)$. If $ab = 0$, the result is obvious. $\square$ Hence the LHS becomes $f(xy) + f(x)$, so $f(xy) = f(x) f(y)$. Since $f$ is additive, multiplicative, and not constant, $f(x) = x$ for all reals $x$, as desired.

\[ f(x(1+y)) = f(x)(1 + f(y)) \]This one is similar. Answers are $f\equiv 0$ and $f(x)=x$ which clearly work. $P(0,0)$ gives us $f(0)=0$. Also $P(x,-1)$ implies $0=f(x)(1+f(-1))$ hence if $f(-1)\neq-1$, then $f\equiv 0$. Now we suppose that $f(-1)=-1$. Plug $x=-1$ to verify $f(-y-1)=-1-f(y)\iff f(y-1)+1=-f(-y)$. Plugging $x,-y$ yields $f(x-xy)=f(x)+f(x)f(-y)=f(x)-f(x)-f(x)f(y-1)=-f(x)f(y-1)$ thus, $-f(-xy)=f(x)f(y)$. Note that $f(1)^2=-f(-1)=1$. If $f(1)=-1,$ then $-f(x)=-f(-x)\iff f(x)=f(-x)$. Since $f(x-1)+1=-f(-x)=f(x),$ we observe that $f(2k)=0$ for any positive integer $k$. However, plugging $2k,\frac{y}{2k}-1$ gives $f(y)=0$ so $f$ is equavilent to $0$ again. Assume that $f(1)=1$. We can see that $f(x)=-f(-x)$ so $f$ is odd. $f(x-1)+1=-f(-x)=f(x)$ implies $f(x+1)-f(x)=1$. This gives us $f(n)=n$ for all positive integers. Also since $f$ is multiplicative, $f(q)=q$ for all rationals. \[f(x)f(y)+mf(x)=f(x)f(y+m)=f(xy+xm)\]Plug $\frac{x}{m},\frac{ym}{x}$ to get $f(y)+f(x)=f(x+y)$ thus, $f$ is additive. When $f$ is both multiplicative and additive, we get that $f(x)=x$ for all reals as desired.$\blacksquare$