Find all functions if $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$ for all $x,y\in\mathbb{R}$.

Question

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 $x,y\in\mathbb{R}$.

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If we put $x=0$ and mark $a=f(0)$ we get $$2a=af(y)+ya$$ so for $a\ne 0$ we get $f(y)=2-y$ for all $y$. Now say $a=0$, so $f(0)=0$. Leting $y=0$ we get $\boxed{f(x^2) = xf(x)}$. If we put this in starting equation we get $$xf(x) +f(xy)=f(x)f(y)+yf(x)+xf(x+y)\;\;\;\;(*)$$

From boxed equation we see that $f$ is odd: Edit $$-xf(-x) = f((-x)^2)= f(x^2)=xf(x) \implies f(-x)=-f(x)$$

Leting $y=-x$ in $(*)$ we get: $$xf(x)-f(x^2) =-f(x)^2-xf(x)\implies \underline{xf(x)= -f(x)^2}$$

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Edit 28. 06. 2022

Let $G$ be a set of all real $x$ such that $f(x)=-x$. Clearly $0\in G$.

Suppose exists $x,y\in G$ so that $x \neq 0$. Here we can assume both $x,y$ are nonnegative, since $f(-x)=-f(x) = -(-x)$. Then we have $$-x^2+f(xy) =xf(x+y)$$ Then we have 4 possibilities:

• $f(xy)=0$ and $f(x+y)=0$ so $-x^2=0$ which is not true.
• $f(xy)=0$ and $f(x+y)=-x-y$ so $-x^2=-x^2-xy$ which is true only if $y=0$.
• $f(xy)=-xy$ and $f(x+y)=0$ so $-x(x+y)=0$ which is not true since $x,y$ are nonegative and $x\ne0$.
• $f(xy)=-xy$ and $f(x+y)=-x-y$ so $-x^2-xy=x(-x-y)$ which is true.

So if $x,y\in G$ then also $x+y\in G$ wich means $G$ is aditive subgroup of $(\mathbb{R},+)$ if$|G|\geq 2$.

What to do now? Is this $G$ usefull at all?

Answer 1

Let $P(x,y)$ be the assertion that $$f(x^2)+f(xy)=f(x)\,f(y)+y\,f(x)+x\,f(x+y)\,.$$ If $f(0)=0$, then we already know that $$f(x^2)=x\,f(x)=\big(f(x)\big)^2\text{ and }f(-x)=-f(x)$$ for all $x\in\mathbb{R}$.

The condition $P(x,x)$ implies that $$2\,x\,f(x)=f(x^2)+f(x^2)=\big(f(x)\big)^2+x\,f(x)+x\,f(2x)=2\,x\,f(x)+x\,f(2x)\,.$$ Thus, $x\,f(2x)=0$ for all $x\in\mathbb{R}$. That is, $$f(2x)=0\text{ for all }x\neq 0\,.$$ Since $f(0)=0$, we conclude that $$f(x)=0\text{ for all }x\in\mathbb{R}\,.$$

Thus, there are two possible solutions. First, as the OP has discovered, we have $f(x)=2-x$ for all $x\in\mathbb{R}$. Second, we have the trivial solution $f\equiv 0$.

Answer 2

$x f(x) = f(x)^2$ says either $f(x) = 0$ or $f(x) = x$. Now if $f(x)=x$ and $f(y)=y$, $f(x^2) = x f(x) = x^2$, and your equation says $$ x^2 + f(xy) = 2 x y + x f(x+y) $$ But this does not work with any combination of $f(x+y)=0$ or $f(x+y)=x+y$ and $f(xy)=0$ or $f(x+y)=xy$ unless $x=0$ or $y=0$ or $x=y$ or $x=2y$. Ultimately the conclusion should be that $f(x)=0$.

Answer 3

As other answers have proven, $f(x) = 0$, and $2-x$ are solutions. But there is another.

If we assume that $f(x) \ne 2-x$ then $f(0) = 0$.

Rearranging the given equation, we get:

$f(xy) - f(x)f(y) = yf(x) + x f(x+y) - f(x^2) = xf(y) + yf(x+y)-f(y^2)$

Where the final equation was obtained by swapping $x\leftrightarrow y$ (The L.H.S. is symmetric in x, y).

Taking $y=-x$ we see from the right part of the equation that $f(-x) = -f(x)$ i.e. $f$ is odd.

Looking at the left side of the equation, we then get $f(x)^2 = - x f(x)$ (!!! this sign seems to be reversed in other "solutions"!!!)

Assuming that $f$ is not identically zero, then $f(x) = -x$, which is indeed a solution.

So the three solutions are $f(x) = \{0, 2-x, -x\}$.