WakeUp wrote: Let $R$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that \[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\] for all $x,y\in\mathbb{R}$. Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ $P(0,x)$ $\implies$ $f(0)(f(x)+x-2)$ If $f(0)\ne 0$, this implies $f(x)=2-x$ which indeed is a solution. Let us from know consider that $f(0)=0$ $P(x,0)$ $\implies$ $f(x^2)=xf(x)$ Then : $P(x,y)$ $\implies$ $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ Same : $P(y,x)$ $\implies$ $yf(y)+f(xy)=f(x)f((y)+xf(y)+yf(x+y)$ Subtracting implies $(x-y)f(x+y)-f(x)-f(y))=0$ and so $f(x+y)=f(x)+f(y)$ $\forall x\ne y$ Plugging this in $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$, we get $f(xy)=f(x)f(y)+yf(x)+xf(y)$ $\forall x\ne y$ $\iff$ $f(xy)+xy=(f(x)+x)(f(y)+y)$ Let then $g(x)=f(x)+x$. We got : $g(0)=0$ $g(x+y)=g(x)+g(y)$ $\forall x\ne y$ $g(xy)=g(x)g(y)$ $\forall x\ne y$ From the first, we get $g(-x)=-g(x)$ and so $g(2x+(-x))=g(2x)+g(-x)$ and so $g(2x)=2g(x)$ and so $g(x+y)=g(x)+g(y)$ $\forall x,y$ From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$ But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$ And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$ So : $g(x+y)=g(x)+g(y)$ $\forall x,y$ $g(xy)=g(x)g(y)$ $\forall x,y$ And so, very classical, $g(x)=x$ and $f(x)=0$ Hence the two solutions : $f(x)=2-x$ $f(x)=0$

pco wrote: From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$ But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$ And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$ Hi pco, could you please explain this part of the solution? Note also $f(x)=-x$ is a solution.

SO im right? x=y 2f(x^2)=f(x)^2+x(f(x)+f(2x)) 2f(0)=f(0)^2 f(0)=0 or f(0)=1/2 let x=x, y=0 f(x^2)+f(0)=f(x)f(0)+xf(x) f(0)=1/2 f(x^2)+1/2=f(x)(1/2+x) f(x)=ax+b ax^2+b+1/2=(ax/2+ax^2+b/2+bx) b/2=b+1/2 b=-1 a=2 f(x)=2x-1 is the solution for f(0)=0 f(x)=x

WakeUp wrote: pco wrote: From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$ But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$ And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$ Hi pco, could you please explain this part of the solution? Yes, , I wrote too quickly ! First we can see that $g(x)=0$ is a solution (and so $f(x)=-x$ is indeed ! If $g(x)$ is not the all zero function, let then $u$ such that $g(u)\ne 0$. If $u=1$, choose instead $u=-1$. Then the second equation gives us $g(u)(g(1)-1)=0$ and so $g(1)=1$ Then $g(x(x+1))=g(x)g(x+1)$ (using second equation since $x\ne x+1$) $=g(x)(g(x)+g(1))=g(x)^2+g(x)$ But $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$ And so, $g(x^2)=g(x)^2$ and so the second equation $g(xy)=g(x)g(y)$ is also true if $x=y$ ... WakeUp wrote: Note also $f(x)=-x$ is a solution. Yes, $g(x)=0$ is also a solution (I forgot it) And so : $f(x)=0$ $f(x)=-x$ $f(x)=2-x$

This is just about as interesting as a FE can get while still dying to the standard strategies of plugging stuff in, taking cases, and testing. Nevertheless, it was a fun problem.

put $ (x,y) = (0,0) $ then we take that 1) $ f(0) =0 $ 2)$ f(0) =2 $ 1) put $ (x,y) = (x,0) $ then we take that $ f(x^2) = x \cdot f(x) $ and this is odd function . put $ (x,y) = (x,-x) $ then we take that f(-x) = x , f(x) =0 2)put $ (x,y) = (0,x) $ then we take that f(x) = 2-x so answer is f(x)=0 , f(x) = -x ,f(x) = 2-x

@Onlygodcanjudgeme :I think you got "$\forall x$ : either $f(x)=0$, either $f(x)=-x$" and not "either $f(x)=0$ $\forall x$ , either $f(x)=-x$ $\forall x$ "

Let $P(x,y)$ be the given assertion, $P(0,0)\implies 2f(0)=f(0)^2\implies f(0)=0,2$. If $f(0)=2$, then $P(0,x)\implies 4=2f(x)+2x\implies f(x)=2-x \ \ \forall x\in\mathbb{R}$. So, if $f(0)=0,$ $P(x,0)\implies f(x^2)=xf(x) \ \ \ (1)$ $P(x,-x)\implies f(x^2)+f(-x^2)=f(x)f(-x)-xf(x) \ \ \ (2)$ Using $P(x,y)$ and $P(-x,-y)$, we can arrive at \[f(x)f(y)+yf(x)+xf(x+y)=f(-x)f(-y)-yf(-x)-xf(-x-y),\]letting $y=0$, we get $f(x)=-f(-x)$ where $x\ne0$, so $f$ is odd. From $(2)$, using the fact that $f$ is odd, \[f(x)(f(x)+x)=0\implies f(x)=0,-x.\]Now, assume that there exists $a,b\in \mathbb{R}, a,b\ne 0$ such that $f(a)=0$ and $f(b)=-b$, using $(1)$, $f(a^2)=0, f(b^2)=-b^2$, using $P(a,b)$ we get \[f(ab)=af(a+b).\]If $f(ab)=-ab$, \[-b=f(a+b).\]If $f(a+b)=0$, then $b=0$, a contradiction. If $f(a+b)=-(a+b)$, then $a=0$, a contradiction. Thus, when $f(ab)=0$, $f(a+b)=0$ as $a=0$ is a contradiction. Take $P(a,b)$ and $P(b,a)$, subtracting one from another gives \[f(a^2)-f(b^2)=bf(a)+af(a+b)-af(b)-bf(a+b)\]which simplifies to \[b^2=ab\implies b(a-b)=0\]and so $a=b$ since $b\ne 0$ but this means $f(b)=f(a)=0=-b$, a contradiction. Hence, we have \[f(x)\equiv 0, 2-x, -x\]as solutions and plugging them into the equation, we see that they indeed satisfy. $\blacksquare$

Solved together with @blastoor Let $P(x,y)$ denote assertion of given functional equation. Note that $P(0,0)$ gives us $2f(0) = f(0)^2$, which means that $f(0)=0$ or $f(0)=2$. If $f(0) =2$, then $P(0,x)$ gives us: \begin{align*} f(0) + f(0) = f(0)f(x) + xf(0) \\ 4 = 2f(x) + 2x \\ f(x) = 2 -x \end{align*}It is easy to check that function $f(x) =2-x$ works. From now we assume that $f(0) =0$. Note that $P(x,0)$ gives us: $$ f(x^2) = xf(x) \qquad (1) $$In relation $(1)$ replacing $x$ by $-x$ yields: $$ f(x^2) = xf(x) = -xf(-x) \implies -f(x) = f(-x) $$Also note that $P(x,-x)$ gives us: \begin{align*} f(x^2) + f(-x^2) = f(x)f(-x) -xf(x) \\ f(x^2) -f(x^2) = -f(x)^2 - xf(x) \\ f(x)^2 = -xf(x) \end{align*}We conclude that $f(x) = 0$ or $f(x) =-x$. Now we are left to escape pointwise trap. Assume that there exist nonzero real numbers $a, b$ such that $f(a) =-a$ and $f(b) = 0 $. Note that $f(b^2) = bf(b) = 0$ and that $P(b,a-b)$ gives us: \begin{align*} f(b^2) +f(b(a-b))= f(b)f(a-b) + (a-b)f(b) + bf(a) \\ f(b(a-b)) = -ab \end{align*}If $f(b(a-b)) = b^2 -ab$, then $b^2 =0 $, which is contradiction since $b \ne 0$. On another hand id $f(b(a-b)=0=ab$, then one of the numbers $a,b$ is zero, which is again contradiction. We conclude that $f(x)=0$, $f(x) = 2 -x$, $f(x) =-x$ are only solutions.

WakeUp wrote: Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that \[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$. Let $P(x, y)$ be the assertion. We claim that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$, $\boxed{f(x) = -x}$ for all reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$. It is not hard to see they work. Now we show that these are the only such functions. We see that $P(x, 0) \implies f(0)(f(x) + x - 2) = 0$, so if $f(0) \neq 0$, then $\boxed{f(x) = 2-x}$ which is indeed a solution. Otherwise, let $f(0) = 0 \dots (1)$. Then by $P(x, 0)$, we yield that $f(x^2) = xf(x) = -xf(-x)$ (by replacing $x$ by $-x$) and so $f$ is odd function. Now $P(x, y) - P(y, x)$ along with $f(x^2) = xf(x)$ gives that $(x-y)(f(x+y)-f(x)-f(y)) = 0$ and so if $x \neq y$, definitely $f(x+y) = f(x) + f(y)$ and so $f$ is additive. Now, we re-arrange few terms in $P(x, y)$. $P(x, y) \implies f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y) \implies xf(x) + yf(x) = f(x)f(y) + yf(x) + xf(x) + xf(y) \implies 0 = f(x)f(y) + xf(y) = f(y)(f(x) + x)$ which means that $f(x) = -x$ or $f(x) = 0$ for all reals $x$. Let us say that $A = \{ x \lvert f(x) = -x, x \neq 0 \}$ and $A = \{ x \lvert f(x) = 0, x \neq 0 \}$. Let $a \in A, b \in B$. We see that $f(ab) = bf(a) + af(b)$, and so here in this case, $f(ab) = b \times -a = a \times 0 = 0$, so either of $a$ or $b$ is $0$, a contradiction to definition of elements belonging to sets $A$ and $B$. Hence, $\lvert A \rvert = 0$ or $\lvert B \rvert = 0$. We see that $f(0) = 0 = -0$. Therefore, we see that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$, $\boxed{f(x) = -x}$ for all positive reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$

Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ $P(0,0)\Rightarrow 2f(0)=f(0)^2$ If $f(0)=2$ then: $P(0,x)\Rightarrow\boxed{f(x)=2-x}$, which works. Now assume $f(0)=0$. $P(x,0)\Rightarrow f(x^2)=xf(x)$ $P(1,x)\Rightarrow f(x+1)=f(x)+f(1)-f(x)f(1)-xf(1)$ We use this recurrence to find $f(1)$. $P(1,1)\Rightarrow f(2)=f(1)-f(1)^2$ $P(1,2)\Rightarrow f(3)=f(2)-f(1)-f(2)f(1)=f(1)^3-2f(1)^2$ $P(1,3)\Rightarrow f(4)=f(3)-2f(1)-f(3)f(1)=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$ But $f(4)=2f(2)=2f(1)-2f(1)^2$, so we find that $2f(1)-2f(1)^2=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$. Solving, we have $f(1)\in\{-1,0,2\}$. $\textbf{Case 1: }f(1)=0$ $P(1,x)\Rightarrow f(x)=f(x+1)$ $P\left(x,\frac yx+1\right)-P\left(x,\frac yx\right)\Rightarrow f(x+y)=f(x)+f(y)$ if $x\ne0$, but since it holds for $x=0$, $f$ is additive. By USAMO 2002/4, since $f(x^2)=xf(x)$ and $f$ is additive, we must have $f(x)=xf(1)$, hence $\boxed{f(x)=0}$ which works. $\textbf{Case 2: }f(1)=-1$ $P(1,x)\Rightarrow f(x+1)=2f(x)+x-1$ $P(x,1)\Rightarrow f(x^2)+f(x)=xf(x+1)\Rightarrow xf(x)+f(x)=2xf(x)+x^2-x\Rightarrow\boxed{f(x)=-x}$ since $f(1)=-1$, which works. $\textbf{Case 3: }f(1)=2$ $P(1,x)\Rightarrow f(x+1)=-f(x)-2x+2$ $P(x,1)\Rightarrow(x-1)f(x)=-x^2+x\Rightarrow f(x)=\begin{cases}-x&\text{if }x\ne1\\2&\text{if }x=1\end{cases}$ which doesn't work.

Here is my way to solve the first case where $f(0)=0$ $P(x,0)$ gives us $$f(x^2) = xf(x), \forall x \quad (1)$$$P(y,x)$ gives us $$f(y^2) + f(xy)= f(x)f(y)+xf(y)+yf(x+y), \forall x,y \quad (2)$$By pluging (1) into (2) then subtracting (2) and the original FE, we got $$(x-y) (f(x)+f(y)) = (x-y) f(x+y), \forall x,y$$, which implies $$ f(x) + f(y) = f(x+y), \forall x \neq y \quad (3)$$From $(1)$, we also have $f(x^2)= -x f(x)$ which leads to $$f(x)= f(-x), \forall x \quad (4)$$By $P(x,-x)$ and using $(4)$, we got $$f(x)^2 = -xf(x), \forall x (5)$$Using $(1)$ and $(3)$, from the origina FE, we have $$f(xy) = f(x)f(y)+xf(y) + yf(x), \forall x \neq y$$, which equivalent to $$ f(x)(f(y)+y) = f(xy) - xf(y), \forall x \neq y \quad (6)$$Case 1: There is a number $k \neq 0$ such that $f(k) = -k$. In $(6)$, let $y=k$, we have $f(kx)=-kx, \forall x \neq k$. Thus $f(x) = -x, \forall x \neq k^2$. In the other hand, $f(k^2) = kf(k) = -k^2$. So, $f(x)=-x, \forall x$. Case 2: There is no number $k$ other than $0$ such that $f(k) = -k$, which means $f(x) \neq -x, \forall x \neq 0$. With $(5)$, we implies $f(x) = 0, \forall x \neq 0$. In the otherhand, $f(0)=0$, thus $f(x) = 0$.

WakeUp wrote: Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that \[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$. Case 1: $f$ is constant. We will set $f(x)=c$ where $c$ is some real constant. Plugging this on the F.E. we get: $$2c=c^2+cy+cx \implies c=0 \implies f(x)=0$$Cade 2: $f$ is non-constant. Let $P(x,y)$ the assertion of the given F.E. $P(0,0)$ $$f(0)^2=2f(0) \implies f(0)=0 \; \text{or} \; f(0)=2$$Case 2.1: $f(0)=2$ $P(0,x)$ $$4=2f(x)+2x \implies f(x)=2-x$$Case 2.2: $f(0)=0$ $P(x,0)$ $$f(x^2)=xf(x) \implies f \; \text{odd}$$$P(x,-x)$ where $x$ is any non-cero real $$f(x)^2+xf(x)=0 \implies f(x)=-x$$Since $f(0)=0$ we have that $f(x)=-x \; \forall x \in \mathbb R$ Thus the solutions are: $\boxed{f(x)=0 \; \forall x \in \mathbb R}$ $\boxed{f(x)=2-x \; \forall x \in \mathbb R}$ $\boxed{f(x)=-x \; \forall x \in \mathbb R}$ Thus we are done

WakeUp wrote: Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that \[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$. A good problem for oddness of a function and how to tackle pointwise traps. Other steps are natural.

A nice exercise for pointwise trap! The answers are $f\equiv 0, 2-x, -x$. These work. Denote the assertion by $P(x,y)$. $P(0,0)$ gives that $f(0)=0,2$. Case 1: $f(0)=2$. $P(0,y)$ gives that $f(y)=2-y.$ Case 2: $f(0)=0$ $P(x,0)$ gives that $f(x^2)=xf(x),$ so we have that $f$ is odd. Then $P(x,-x)$ gives that $0=f(x)(f(x)+x),$ so $f(x)=0,-x$. To avoid pointwise trap, suppose that $f(x)=0$ and $f(y)=-y$ for some $x,y\neq 0$. Then from $P(x,y)$ and $P(y,x)$ and subtracting, we get that $xf(x)-yf(y)=yf(x)-xf(y)+(x-y)f(x+y)$. Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$. If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.

JustKeepRunning wrote: Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$. If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well. How did you simplify the equation into this?

They used the properties $f(x)=0$ and $f(y)=-y$.

Let $P(x,y)$ denote the assertion. Quickly $P(0,x)$ gives $f(x)\equiv 2-x$ or $f(0)=0.$ The former works, so we explore the latter. $P(x,0)$ gives $f(x^2)=xf(x).$ And so comparing $P(x,y)$ with $P(y,x)$ shows that $f(x+y)=f(x)+f(y)$ for all $x\neq y.$ To conclude $P(x,-x)$ implies $f(x)\in \{0,-x\}$ but since additive $f\equiv 0$ or $f\equiv -x$ and both satisfy.

This dies to basically anything. Setting $x=y=0$ yields $2f(0) = f(0)^2$. If $f(0) = 2$, setting $x=0$ yields $f(x) = 2-x$ immediately. If $f(0) = 0$, setting $y=0$ yields $f(x^2) = xf(x)$, implying $f$ is odd. Setting $y=-x$ in the original, \[0=f\left(x^2\right) + f\left(-x^2\right) = f(x)f(-x)-xf(x) = -f(x)^2 - xf(x).\]So for each $x$, either $f(x) = 0$ or $f(x) = -x$. There are many ways to resolve the pointwise trap, but here is a really stupid way. By setting $y=x$ we get $f(x)^2 + xf(2x) = xf(x)$, i.e. $xf(2x) = 2xf(x)$ or $f(2x) = f(x)$. Furthermore, by swapping $x$ and $y$, we get \[(x-y)f(x) - xf(x+y)=(y-x)f(y) - yf(x+y)\]so $f(x+y) = f(x)+f(y)$ for all $x \neq y$. Combining this with the previous equation yields that $f$ is Cauchy and bounded below on $x \geq 0$, thus $f$ is linear. We can check that only $f \equiv 0$ works here.

case 1. f is a constant function : c+c=c^2+cy+xc x=y=1 plug in we get c=0 so f(x)=0 case 2. f is not a constant function : P(0,y) :2f(0)=f(0)f(y)+yf(0) 2-1. if f(0) is not 0 then y=1 plug in we get f(1)=1 P(1,y): 1+f(y)=f(y)+y+f(y+1) and we know that f(y)=2-y 2-2.f(0)=0 : P(0,0): f(x^2)=xf(x) this imply f is a odd function P(x,-x) : 0=-(f(x)^2)-xf(x) we get f(x)=-x so the answer is f(x)=0 or 2-x or -x