Functional_equation wrote: $P(x,-x)\implies f(f(x))=f(0)-xf(x)$ No. $P(x,-x)$ $\implies$, $f(f(x))=f(0)-xf(-x)$

Tintarn wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all real numbers $x, y$. Let $P(x,y)$ be the assertion $f(f(x)+x+y)=f(x+y)+yf(y)$ Let $a=f(0)$ $P(0,-1)$ $\implies$ $f(a-1)=0$ $P(a-1,x)$ $\implies$ $xf(x)=0$ $\forall x$ and so $f(x)=0$ $\forall x\ne 0$ If $a\ne 0$, $P(0,-a)$ $\implies$ $a=0$, contradiction And so $\boxed{f(x)=0\quad\forall x}$, which indeed fits.

Tintarn wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all real numbers $x, y$. Let $P(x, y)$ be the assertion. We claim that all solutions to the given functional equation are $\boxed{f(x)=0}$ for all reals $x$. It can be easily verified that the aforementioned functions work, hence if suffices to show that they are the unique functions satisfying the given conditions. $P(0, -1) \implies f(f(0)-1) = 0$ and $P(f(0)-1, x) \implies f(x)=0$ for all $x \neq 0$. If $f(0) \neq 0$, then $P(0, 0) \implies f(f(0))=f(0)$, but $f(0) \neq 0$ so $f(f(0))=f(0)=0$, which is a contradiction implying that $f(0)=0$. Therefore, all solutions to the given functional equation are $\boxed{f(x)=0}$ for all reals $x$. @below thanks for notifying me, apparently I deleted the negative sign accidentally while proofwriting

DapperPeppermint wrote: $P(0, 1) \implies f(f(0)-1) = 0$ Typo : $P(0,1)$ $\implies$ $f(f(0)+1)=2f(1)$ You meant $P(0,-1)$

Pco wrote: If $a\ne 0$, $P(0,-a)$ $\implies$ $a=0$, contradiction Can you please explain this part?

Tintarn wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all real numbers $x, y$. Nice problem... My Solution Case 1.$f(0)=0$ $P(0,x)\implies xf(x)=0\implies f(x)=0$ Case 2.$f(0)\neq 0$ $P(0,0)\implies f(f(0))=f(0)$ $$P(f(0),-f(0))\implies f(f(f(0))+f(0)-f(0))=f(0)-f(0)f(-f(0))\implies f(0)-f(0)f(-f(0))=f(f(f(0))+f(0)-f(0))=f(f(0)+f(0)-f(0))=f(f(0))=f(0)\implies f(0)-f(0)f(-f(0))=f(0)\implies f(-f(0))=0$$$P(-f(0),f(0))\implies f(0)=f(0)+f(0)f(f(0))\implies 0=f(0)f(f(0))=f(0)^2\implies f(0)=0$ $0\neq f(0)=0\implies$impossible $\boxed{f(x)=0\quad\forall x}$

Hard Version:Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all positive numbers $x, y$.

Arabian_Math wrote: Pco wrote: If $a\ne 0$, $P(0,-a)$ $\implies$ $a=0$, contradiction Can you please explain this part? $P(0,-a)$ is $f(a+0-a)=f(-a)-af(-a)$ and so $a=0$ (remember $f(-a)=0$ since $a\ne 0$)

Functional_equation wrote: Tintarn wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all real numbers $x, y$. Nice problem... My Solution Case 1.$f(0)=0$ $P(0,x)\implies xf(x)=0\implies f(x)=x$ Case 2.$f(0)\neq 0$ $P(0,0)\implies f(f(0))=f(0)$ $$P(f(0),-f(0))\implies f(f(f(0))+f(0)-f(0))=f(0)-f(0)f(-f(0))\implies f(0)-f(0)f(-f(0))=f(f(f(0))+f(0)-f(0))=f(f(0)+f(0)-f(0))=f(f(0))=f(0)\implies f(0)-f(0)f(-f(0))=f(0)\implies f(-f(0))=0$$$P(-f(0),f(0))\implies f(0)=f(0)+f(0)f(f(0))\implies 0=f(0)f(f(0))=f(0)^2\implies f(0)=0$ $0\neq f(0)=0\implies$impossible $\boxed{f(x)=0\quad\forall x}$ Typo in your first assertion, please correct

@above sorry.I made a mistake while writing.

The only solution is $f(x)=0~\forall x\in\mathbb{R}$ which is easy to check. Let $P(x,y)$ denote the assertion. We claim $f(0)=0$. By $P(0,0)$ we have $f(f(0))=f(0)$. Combining this with $P(f(0),0)$ we have $f(2f(0))=f(0)$. Subtracting $P(2f(0),f(0))$ from $P(f(0),2f(0))$ we have $f(0)^2=0$. Hence $f(0)=0$ as desired. To finish, by $P(0,x)$ we have $xf(x)=0$ so $f(x)=0~\forall x\in\mathbb{R}-\{0\}$ but then we already have $f(0)=0$. Hence $f(x)=0~\forall x$ as desired. $\square$

So here is solution, which my teammate found during the contest. Let $P(x,y)$ as usual denotes assertion of given functional equation. Observe that $P(0,-1)$ gives us that: $$ f(f(0) -1) = f(-1) -f(-1) =0 $$This implies that there exists real number $a$ such that $f(a) =0$. Note that $P(a,y)$ gives us: $$ f(fa) + a +y) = f(a+y) + yf (y) \implies yf(y) = 0 $$Thus we must have $f(y) =0$ for all real numbers $y \neq 0$. But from $P(x,0)$ we know that $f(f(x)+x) = f( x)$. Putting $x=0$ yields $f(f(0)) = f(0)$. If $f(0) \neq 0$, then we get that $f(f(0)) = 0 = f(0)$. So in both cases $f(0) = 0$. We conclude that $f(y)=0 $ for all real numbers.

The only solution is $f(x)=0$ for any real $x$. Let $c=f(0)$. If we plug in $x=0$ into our equation we have that $f(c+y)=(y+1)f(y)$. Now we plug in $y=-1$ into the upper result to get that $f(c-1)=0$. Into our original equation we plug in $y=-x$ to get that $f(f(x))=f(0)-xf(-x)$. Into this we plug in $x=c-1$ to get that: $$(1-c)f(1-c)=0$$ So we have two different cases: The first case is when $1-c=0$, which implies that $c=1$, this implies that $f(1)=0$ along side with $f(0)=0$. Now we just plug in $x=0$ into the original equation to get that $yf(y)=0$, which easily implies that $f(y)=0$ for any real $y$. The second case is when $f(1-c)=0$. Into the original equation we just plug in $x=1-c$ to get that $yf(y)=0$, which again implies that $f(y)=0$ for any real $y$.

Here's another solution: $P(x ,y): f(f(x) + x + y) =f(x+y) + yf(y)$ if $f(0) =c$ ,$P(0,0)$ gives us $f(c)=c$ $P(c,0) :f(2c) =c$ $P(c,c) :f(3c) =c^2 +c$ $P(2c ,c) :f(4c) =2c^2 + c$ $P(c ,2c) :f(4c) =3c^2 + c$ so $c =0$. now putting $y =0$ gives us $f(y) = f(y) + yf(y)$. so $f(y) =0$ and we're done....

How does everyone have a different solution? redacted

EpicNumberTheory wrote: Let $f(x) = f(y) $, for some $x,y \in \mathbb{R}$ Combining $P(x,y)$ and $P(y,x)$ $\implies (x-y)f(x)=0$ Case 1: $\boxed{f(x) = 0 \quad \forall x \in \mathbb{R}}$ Case 2: $x =y$ and so $f$ is injective. You can not conclude from "$(x-y)f(x)=0$ $\forall x,y$ such that $f(x)=f(y)$" that "either $f(x)$ is allzero, either $f(x)$ is injective". Example : $f(x)=x+|x|$ is neither allzero, neither injective, although we have $(x-y)f(x)=0$ $\forall x,y$ such that $f(x)=f(y)$

We claim the only solution is $\boxed{f(x)=0\quad\forall x}$ which works. Suppose there exists $a$ such that $f(a) = b \ne 0.$ $P(a,0) \implies f(f(a) + a) = f(a) = b,$ then $P(f(a) + a,a) \implies f(2b+2a) = f(b+2a) + ab$ and $P(a,f(a) + a) \implies f(2b+2a) = f(b+2a) + (f(a) + a)b.$ Subtracting these two equations yields $f(a)b = 0$ which is contradiction. $\square$

Tintarn wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all real numbers $x, y$. Here's a funny approach, mainly motivated by symmetry abuse. The only solution is $\boxed{f(x)= 0}$, which could be easily verified. Now, let $P(x,y)$ be the assertion of $x$ and $y$ to the given functional equation. $P(0,x)$ gives us $f(x+f(0)) = (x + 1)f(x)$, which gives us $f(f(0)) = f(0)$ and $f(f(0) - 1) = 0$. Now, $P(x,y+f(y))$ gives us \[ f(x+f(x)+y+f(y)) = f(x+y+f(y)) + (y + f(y))f(y+f(y)) = f(x+y) + xf(x) + (y+f(y))f(y+f(y)) \]Switch $x$ and $y$, we obtain \[ (x+f(x))f(x+f(x)) = xf(x) + c \]for a constant $c$, for which $x = 0$ gives us $c = f(0)^2$. Now, take $x = f(0) - 1$, then we have \[ f(0)^2 = 0 \Rightarrow f(0) = 0 \]Therefore, $f(y) = (y + 1)f(y) \Rightarrow yf(y) = 0 $ for all $y \in \mathbb{R}$. Combining this with $f(0) = 0$, we get $f(x) = 0$ for all $x \in \mathbb{R}$.

Tintarn wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all real numbers $x, y$. $f(f(x) + x + y) = f(x + y) + yf(y), \forall x, y \in \mathbb{R}$ We have $P(0, 0): f(f(0)) = f(0)$ $P(f(0), 0): f(2f(0)) = f(0)$ $P(0, f(0)): f(2f(0)) = f(0) + f(0)^2$ Then $f(0) = 0$ $P(0, y): yf(y) = 0$. So $f(y) = 0, \forall y \in \mathbb{R}$

\[P(0,0) \implies f(f(0))=f(0) \quad (1)\]now if $f(x_1)=f(y_1)$ \[P(x_1,y_1),P(y_1,x_1) \implies x_1=y_1 \quad \text{So the function is injective}\]now in the equation $(1)$ we get $f(0)=0$ now inserting $y=0$ we get $\boxed{f(x) \equiv 0}$ which works.

Let $P(x,y)$ be the assertion $f(f(x)+x+y)=f(x+y)+yf(y)$. $P(0,-1)\Rightarrow f(f(0)-1)=0$ $P(f(0)-1,x)\Rightarrow xf(x)=0\Rightarrow f(x)=0\forall x\ne0$ $P(0,-f(0))\Rightarrow f(0)=f(-f(0))-f(0)f(-f(0))$ If $f(0)\ne0$ then $f(-f(0))=0$, so $f(0)=0$, contradiction. Thus $\boxed{f(x)=0}$, which fits.

Functional_equation wrote: Hard Version:Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all positive numbers $x, y$. $P(x ,y):f(f(x)+x+y) = f(x+y) + y f(y)$ . now by $P(f(x) + x + 1 , y)$ and original relation we have : $f(f(f(x) + x + 1) + f(x) + x + 1 + y) = f(f(x) + x + 1 + y) + yf(y) = f(x+y+1) + (y+1)(f(y)+1) + yf(y)$ $*$ on the other hand , $f(f(x) + x +1) = f(x+1) + f(1)$ so $f(f(f(x) + x + 1) + f(x) + x + 1 + y) = f(f(x+1) + x +1 + f(1) + y + f(x)) = f(f(x) + x + 1 + y + f(1)) + (f(1) + y + f(x))f(f(1) + f(x) + y) = f(f(1) + 1 + x+y) + (y+1+f(1))f(f(1) + 1 + y) + (f(1) + y + f(x))f(f(1) + f(x) + y)$ but $f(f(1) + 1 + t) = f(t+1) + tf(t)$ so: $f(f(1) + 1 + x+y) + (y+1+f(1))f(f(1) + 1 + y) + (f(1) + y + f(x))f(f(1) + f(x) + y) = f(x+y+1) + (x+y)f(x+y) + (f(y+1) + yf(y))((y+1) + f(1)) f(f(1) + x + y)(f(1) + f(x) +y)$ $**$ now , we have $*$ $=$ $**$ . but $**$ have $f(x+y+1)$ , $(y+1)(f(y)+1)$ and $yf(y)$ and more stuff too..... contradiction .

Let $P(x,y)$ denote the assertion in the problem. We try to something to 0, as that getting a zero off the polynomial can help us eliminate the inner $f(x)$ on the LHS. Thus, we want $y+1=0\implies y=-1,$ so $x-1=-1\implies x=0$. Therefore, $P(0,-1)\implies f(a-1)=0,$ where $a=f(0)$. Now, as promised, plug in $a-1$ for $x$, so the inner $f$ vanishes and we just have $f(a-1+y)=f(a-1+y)+yf(y),$ and thus $yf(y)=0.$ For nonzero $y,$ we have that $f(y)=0.$ All we have left to show is that $f(0)=0$ as well. It is natural to plug in $P(0,-a)$ to get the LHS to equal $a$. $P(0,-a)\implies a=f(-a)(-a+1)\implies f(-a)=\frac{a}{-a+1}$. Assume for contradiction $f(0)\neq 0$. Then this means $-a$ is also nonzero, so $f(-a)=0$ and we have that $a=0.$ Contradiction. We are done.

Tintarn wrote: Find all functions $f:\mathbb{R} \to \mathbb{R}$ so that \[f(f(x)+x+y) = f(x+y) + y f(y)\]for all real numbers $x, y$. Let$P(x,y)$Be a claim$f(f(x)+x+y)=f(x+y)+yf(y)$ $P(0,-1)\Rightarrow f(f(0)-1)=f(-1)-f(-1)=0$ $P(f(0)-1,x)\Rightarrow f(f(0)-1+x)=f(f(0)-1+x)+xf(x)\Rightarrow xf(x)=0$ $\Rightarrow $so we have$f(x)=0 (\forall x\in {R})$

Assume that f(x)\neq 0 for all x. We first prove that f is injective. Let f(x)=f(y)\neq 0, then f(x)+x+y=f(y)+x+y take f on both sides, we get that f(x+y)+yf(y)=f(x+y)+xf(x) or xf(x)=yf(y) implying that x=y. Now, let P(x,y) be the assertion above. then, P(x,0) gives f(f(x)+x)=f(x) or f(x)+x=x implying that f(x)=0, a contradiction so the only solution is f(x)=0 for all x. Oh and also if for some x, f(x)=0 and for some y\neq 0 f(y) is not 0, then do P(x,y) and we will get f(y)=0. For a contradiction! I hope I did not make any mistakes, if I did please correct me.

Oops, for some reason I did this problem again without knowing it, but this time I have a different solution, so I'll post anyways. The only solution is $f\equiv 0$. It works. Let $P(x,y)$ denote the assertion. $P(0,0)$ gives $f(f(0))=f(0)$, and $P(0,y)$ gives that $f(y+f(0))=(y+1)f(y),$ or setting $y=-1,$ that $f(f(0)-1)=0$. Therefore, $P(f(0)-1, y)$ gives that $yf(y)=0\forall y\in \mathbb{R},$ so we know that $f(y)=0\forall y\neq 0$. Now suppose $f(0)\neq 0$. Then we have that $f(f(0))=0$(because $f(0)$ is not equal to zero), but $f(f(0))=f(0),$ contradiction.

medman1086 wrote: Assume that f(x)\neq 0 for all x. We first prove that f is injective. Let f(x)=f(y)\neq 0, then f(x)+x+y=f(y)+x+y take f on both sides, we get that f(x+y)+yf(y)=f(x+y)+xf(x) or xf(x)=yf(y) implying that x=y. Now, let P(x,y) be the assertion above. then, P(x,0) gives f(f(x)+x)=f(x) or f(x)+x=x implying that f(x)=0, a contradiction so the only solution is f(x)=0 for all x. Oh and also if for some x, f(x)=0 and for some y\neq 0 f(y) is not 0, then do P(x,y) and we will get f(y)=0. For a contradiction! I hope I did not make any mistakes, if I did please correct me.

If $f$ is constant, then only $f\equiv 0$ works, now assume $f$ is not constant. Let $P(x, y)$ be assertion $P(1, y)$: $f$ is surjective Pick $u\in \mathbb{R}$ s.t. $f(u)=0$ $P(u, y)$: $yf(y)=0$ for all real $y$, but this doesn't have non-constant solution, so only solution is $\boxed{f(x)=0 \forall x\in \mathbb {R}}$

Let $P(x,y)$ denote the given assertion. $P(0,-1): f(f(0)-1)=0$. $P(f(0)-1,x): xf(x)=0$ so either $x=0$ or $f(x)=0$. Case 1: $f(0)\ne0$ $P(0,0): f(f(0))=f(0)$ so if $f(0)=k$, where $k\ne0$, then $f(k)=0$, so $f(f(0))=f(0)=0$, which is a contradiction. Case 2: $f(0)=0$. Here the function is $\boxed{f(x)=0}$ for all $x\in \mathbb{R}$, which works.

Let $a,b$ be such that $f(a) = f(b)$ then $P(a,b) : f(f(a) + a + b) = f(a+b) + bf(b)$. $P(b,a) : f(f(b) + a + b) = f(a+b) + af(a)$. so $a = b$ which implies $f$ is injective. $P(x,0) : f(f(x)+x) = f(x) \implies f(x) = 0$ which fits the main equation. Answers $: f(x) = 0$.

Let $P(x,y)$ denote the assertion $f(f(x)+x+y) = f(x+y) + y f(y)$. $P(0,-1)\implies f(f(0)-1)=f(-1)-f(-1)=0$. Let $a=f(0)-1$. $P(a,x)\implies f(a+x)=f(a+x)+xf(x)\implies f(x)=0$ where $x\neq 0$. $P(0,0)\implies f(f(0))=f(0)$. If $f(0)\neq 0$ then $f(f(0))=0$ implying that $f(0)=0$, a contradiction. Hence, $f(0)=0$. We conclude that $f(x)=0$ for all reals $x$ which indeed fits.

We claim the only solution is $f(x)=0$. Let $P(x, y)$ be the given assertion. $P(0, -1) \rightarrow f(f(0)-1)=0 \rightarrow \exists n \in \mathbb{R}$ such that $f(n)=0$. $P(x, 0) \rightarrow f(f(x)+x)=f(x)$. Plugging in $n$ for $x$, we get $f(x)=0$, as needed. $\blacksquare$

@above When I plug $n=x$ I get $0=0.$

ZETA_in_olympiad wrote: My Solution: Note that $f$ is injective. Setting $y=0$ and using injectivity we get $f(x)=0$ which fits. Hemmm, how, proving first injectivity (how?, you did not explain) and then using it, you get .... a non-injective solution ??????

pco wrote: ZETA_in_olympiad wrote: My Solution: Note that $f$ is injective. Setting $y=0$ and using injectivity we get $f(x)=0$ which fits. Hemmm, how, proving first injectivity (how?, you did not explain) and then using it, you get .... a non-injective solution ?????? I was seeing stars... Let $P(x,y)$ be the assertion and $x_0=f(0).$ First $P(0,0)$ implies $f(x_0)=x_0.$ Then calculating $f(2x_0)$ in two ways (e.g. $P(0,x_0)$ and $P(x_0,0)$) imply $x_0=0.$ And $P(0,x)$ implies $f\equiv 0$ which works.

Let $P(x, y)$ be the given assertion. \begin{align*} P(x,x) &\implies f(f(x) + 2x) = f(2x) + xf(x), \\ P(0,0) &\implies f(f(0)) = f(0), \\ P(x,1) &\implies f(f(x) + x+1) = f(x+1) + f(1), \\ P(x,0) &\implies f(f(x) + x) = f(x), \\ P(x,-x) &\implies f(f(x)) = f(0) - xf(-x), \\ P(0,x) &\implies f(f(0) + x) = f(x) + xf(x),\\ &\implies f(f(0) + x) =(1+x)f(x),\\ P(0,-1) &\implies f(f(0) - 1) = f(-1) - f(-1) = 0, \\ P(f(0)-1,x) &\implies f(f(f(0)-1) + f(0)-1+x) = f(f(0)-1+x) + xf(x), \\ &\implies 0 = xf(x), \end{align*}which implies $f(x)$ is a constant function. Let $f(x) = c$, plugging back we get: \begin{align*} f(f(x) + x + y) &= f(x+y) + yf(y),\\ f(c + x + y) &= c + yc,\\ c &= c + yc,\\ 0 &= yc, \end{align*}since $y$ can be any number on the reals, this implies $c = 0$ and thus $f(x) = 0 \forall x\in \mathbb{R}$ is the only solution.

If $f(x) = 0$ for any $x$, $P(x,y)$ gives $f(y) = 0$ or $y = 0$, but then $P(0,0)$ gives $f(f(0)) = f(0)$, so $f(0)$ cannot be nonzero, so we have $f(x) = 0$. Now assume $f(x) = 0$ for no real $x$. We then have that $f(0)$ is a fixed point of the function, so we plug $P(f(0), -f(0))$ to get $-f(0)f(-f(0)) = 0$, implying that $f(x) = 0$ for some value of $x$, so we are done by the above statement.

$P(0; 0)$ $\Rightarrow$ $f(f(0))=f(0)$ $P(f(0); -f(0))$ $\Rightarrow$ $f(f(0))=f(0)-f(0)f(-f(0))$ $\rightarrow$ $f(0)f(-f(0))=0$ $\rightarrow$ $\exists$$c$, $f(c)=0$ $P(c; x)$ $\Rightarrow$ $xf(x)=0$ $\rightarrow$ $\boxed{f(x)=0}$ $\forall$$x$ $\in$ $\mathbb{R}$