Let $P(x,y)$ be the assertion of $x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$. $P(0,0)\implies f(0)^2=0\implies f(0)=0$ $P(0,x)\implies f(x)f(x^2)=xf(x^2) \implies f(x^2)(f(x)-x)=0$ Thus, $f(x^2)=0$ or $f(x)=x$ for all $x\in\mathbb{R}$. Suppose there exist at the same time some $a,b\neq 0$, where $f(a^2)=0$ or $f(b)=b$: $P(b,a)\implies f(a+b)f(a^2-ab+b^2)=b^3\implies\text{Contradiction!}$ Contradiction, because none of them is $0$ and $(a+b)(a^2-ab+b^2)=a^3+b^3$. Thus, $f(x^2)=0$ for all $x\in\mathbb{R}$ or $f(x)=x$ for all $x\in\mathbb{R}$. If $f(x)=x$ for all $x\in\mathbb{R}$, obviously it is a solution. If $f(x^2)=0$ for all $x\in\mathbb{R}$, then $f(z)=0$ for all $z\in\mathbb{R^+}$ Thus, we have that $x^2f(x)=f(x+y)f(x^2-xy+y^2)$. Let $x<0$ and $y>0$, so that $x+y>0$, hence $f(x+y)=0$. $P(x,y)\implies x^2f(x)=0\implies f(z)=0$ for all $z\in\mathbb{R^-}$. We conclude that $f(x)=0$ for all $x\in\mathbb{R}$, this also obviously work. Answer. $\boxed{f(x)=0\forall x\in\mathbb{R} \text{ or } f(x)=x\forall x\in\mathbb{R}}$

Korea National Olympiad 2020 Problem 1 wrote: Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$$for all $x,y\in\mathbb{R}$. Can someone please verify my solution? Thanks!

MNJ2357 wrote: Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$$for all $x,y\in\mathbb{R}$. Nice problem... My Solution $f(0)=0$ $x\neq 0$ $P(x,0)\implies f(x^2)=xf(x)$ $P(0,x)\implies xf(x^2)=f(x)f(x^2)$ Then $f(x)(f(x)-x)=0$ Case 1.$f(0)=0,f(x)=0,x\in R_{\neq 0}$ Case 2.$f(0)=0,f(x)=x,x\in R_{\neq 0}$ Case 3.$f(a)=a,f(b)=0,(a,b)\in R_{\neq 0}$ $P(b,a)\implies a^3=(a+b)(a^2+b^2-ab)=a^3+b^3\implies b=0\implies impossible$

Hard Version: Determine all functions $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ such that $$x^2f(x)+y^2f(y)=f(x+y)f(x^2-xy+y^2)$$for all $x,y\in\mathbb{R^+}$.

Korea National MO 2020 P1 wrote: Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$$for all $x,y\in\mathbb{R}$. Slightly convoluted solution Let $P(x,y)$ be the given functional equation. To start, $P(0,0) \Rightarrow f(0)=0$, therefore $$P(x,0) \Rightarrow x^2f(x)=f(x)f(x^2),$$and $$P(0,x) \Rightarrow xf(x^2)=f(x)f(x^2).$$Hence, $$x^2f(x)=f(x)f(x^2)=xf(x^2),$$implying $f(x^2)=xf(x)$ if $x \neq 0$, and since $f(0)=0$ we obtain $f(x^2)=xf(x)$ for all $x$. In addition, $$x^2f(x)=f(x)f(x^2)=xf(x)^2,$$so $f(x) \in \{0,x \}$ for all $x$. Also, note that $-xf(-x)=(-x)^2=x^2=xf(x)$ hence $f$ is odd. We distinguish two cases: Case 1: There exists $A>1$ such that $f(A)=0$. We prove the following Claims: Claim 1: $f(x)=0$ for all $0 \leq x \leq A$. Proof: Let $0 < k \leq A$, and take $\ell >0$ such that $k+\ell=A$. Then, $P(k,\ell)$ implies $k^2f(k)+\ell^2f(\ell)=0$ - using $f(x^2)=xf(x)$ for $x=\ell$. But, $f(k) \in \{0, k \}$ and $f(\ell) \in \{0, \ell \}$ so $f(k) \geq 0$ and $f(\ell) \geq 0$. Therefore, $$0=k^2f(k)+\ell^2f(\ell) \geq 0,$$hence equality must hold. So, $f(k)=f(\ell)=0$ which proves the Claim $\blacksquare$. Claim 2: $f$ is actually zero everywhere. Proof: Note that, since $f(x^2)=xf(x)$ and $f(A)=0$ we may obtain $f(A^{2^n})=0$ for all $n \in \mathbb{N}$. But, since $A>1$, the sequence $\{A^{2^n}, \, n \in \mathbb{N} \}$ is clearly strictly increasing. Combining this with the result of Claim 1 (i.e. if $f(t)=0 \Rightarrow f(k)=0$ for all $0 \leq k \leq t$), we obtain that $f(x)=0$ on the positives. Since $f$ is odd, $f$ is zero everywhere $\blacksquare$. Case 2: $f(x) \neq 0$ for all $x>1$. Then, since $f(x) \in \{0,x \}$, we obtain $f(x)=x$ for all $x>1$. To conclude, just take a pretty large $y$ - say, $y>10^{2020}$ - and take a $x \leq 1$. Then, $f(y^2)=y^2$, $f(x+y)=+y$ and $f(x^2-xy+y^2)=x^2-xy+y^2$, since all of the three terms are $>1$ - for the third one specifically, note that $$x^2-xy+y^2=(x-\frac{y}{2})^2+\frac{3y^2}{4} >1$$since $y$ is pretty large. Hence $P(x,y) \Rightarrow x^2f(x)+y^3=(x+y)(x^2-xy+y^2) \Rightarrow x^2f(x)=x^3$ hence $f(x)=x$ if $x \neq 0$, implying that $f(x)=x$ for all $x$. To end, it is trivial to check that both of the functions we found, namely $f \equiv 0$ and $f \equiv x$, satisfy the given functional equation.

$\clubsuit \color{green}{\textit{\textbf{ANS:}}}$ $f(x)=0 \quad \textrm{and} \quad f(x)=x \quad \forall x\in \mathbb{R}.$ $\blacklozenge \color{red}{\textit{\textbf{Proof:}}}$ It's easy to see that these are indeed solutions to the above FE. Let $P(x,y)$ be the given assertion, we have \[P(0,0): f(0)=0\]and \[P(x,0) \quad \textrm{and} \quad P(0,x): xf(x)=f(x^2) \quad \forall x\in \mathbb{R} \setminus \{0\}.\]Since $f(0)=0$, $f(x^2)=xf(x) \quad \forall x\in \mathbb{R}$. Then, \[P(0,x): f(x^2)(f(x)-x)=0 \implies f(x)(f(x)-x)=0 \quad \forall x\in \mathbb{R} \setminus \{0\}.\]Let $a,b \ne 0$ such that $f(a)=0, f(b)=b$, we have $f(b^2)=b^2$, so \[P(a,b): b^3=f(a+b)f(a^2-ab+b^2)\]notice that $f(a+b), f(a^2-ab+b^2)\ne 0$ otherwise, this would imply $b=0$ which is a contradiction. Therefore, $f(a+b)=a+b, f(a^2-ab+b^2)=a^2-ab+b^2$ but \[b^3=(a+b)(a^2-ab+b^2)=a^3+b^3 \implies a=0\]which is also a contradiction. Thus, \[\boxed{\color{blue}{f(x)=0 \quad \textrm{and} \quad f(x)=x \quad \forall x\in \mathbb{R}.}}\]

MNJ2357 wrote: Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$$for all $x,y\in\mathbb{R}$. Let $P(x,y)$ the assertion of the given FE. $P(0,0)$ $$f(0)^2=0 \implies f(0)=0$$$P(x,0)$ $$x^2f(x)=f(x)f(x^2) \implies f(x) \equiv 0 \; \text{or} \; f(x^2)=x^2$$$P(0,x)$ $$xf(x^2)=f(x)f(x^2) \implies f(x^2) \equiv 0 \; \text{or} \; f(x)=x$$. Then the given equations gives to us that $f(x) \equiv 0$ or $f(x)=x$. Assume that exists $a,b \ne 0$ such that $f(a)=0$ and $f(b^2)=b^2$ $P(a,b)$ $$b^3=f(a+b)f(a^2-ab+b^2) \; \text{contradiction!!}$$. Assume that exists $m,n \ne 0$ such that $f(n^2)=0$ and $f(m)=m$. $P(m,n)$ $$m^3=f(m+n)f(m^2-mn+n^2) \; \text{contradiction!!}$$. Then the solutions are: $f(x) \equiv 0$ and $f(x)=x$ thus we are done

Let $P(x,y)$ be the assertion. $P(0,0)\implies f(0)=0$. $P(-x,x)\implies xf(-x) = f(x^2) \forall x\neq 0$. $P(x,-x)\implies xf(x) = f(x^2) \forall x\neq 0 \implies f(x)=-f(-x)$ so $f$ is odd. $P(x,0)\implies xf(x)(x-f(x)) = 0 \forall x\in \mathbb{R}$. So from there we get $\boxed{f(x)=x \forall x\in \mathbb{R}}$ and $\boxed{f(x)=0 \forall x\in \mathbb{R}}$ as solutions which work. Now suppose for contradiction that there exist $a,b\neq 0$ such that $f(a)=a$ and $f(b)=0$. $P(a,b)\implies a^3 = f(a+b)f(a^2 - ab + b^2)$. But this is a contradiction since $a,b\neq 0$. So there is no solution in this case.

MathLuis wrote: MNJ2357 wrote: Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$$for all $x,y\in\mathbb{R}$. Let $P(x,y)$ the assertion of the given FE. $P(0,0)$ $$f(0)^2=0 \implies f(0)=0$$$P(x,0)$ $$x^2f(x)=f(x)f(x^2) \implies f(x) \equiv 0 \; \text{or} \; f(x^2)=x^2$$$P(0,x)$ $$xf(x^2)=f(x)f(x^2) \implies f(x^2) \equiv 0 \; \text{or} \; f(x)=x$$. Then the given equations gives to us that $f(x) \equiv 0$ or $f(x)=x$. Assume that exists $a,b \ne 0$ such that $f(a)=0$ and $f(b^2)=b^2$ $P(a,b)$ $$b^3=f(a+b)f(a^2-ab+b^2) \; \text{contradiction!!}$$. Assume that exists $m,n \ne 0$ such that $f(n^2)=0$ and $f(m)=m$. $P(m,n)$ $$m^3=f(m+n)f(m^2-mn+n^2) \; \text{contradiction!!}$$. Then the solutions are: $f(x) \equiv 0$ and $f(x)=x$ thus we are done I don't really get why you assumed that exists $a,b \ne 0$ such that $f(a)=0$ and $f(b^2)=b^2$

$f(x)=0$ is the only constant that works, assume $f$ to be non constant from here on. $P(0,0) \implies f(0)=0. P(x,0) \implies x^2f(x)=f(x)f(x^2)$. Work in positive reals for now. So we get $f(x)=0$ or $f(x^2)=x^2 \forall x \geq 0$. Suppose we have $f(a)=a, f(b)=0$ for some $a,b >0$. $P(a,b) \implies a^3=f(a+b)f(a^2-ab+b^2)$. Since $a >0$, $f(a+b)$ or $f(a^2-ab+b^2)$ can’t be $0$. We get $a^3=a^3-b^3 \implies b=0$, contradiction. Now choose $x>0$. $P(-x,x) \implies f(-x)x^2=-xf(x^2) \implies f(-x)=-x$, so we get $f(x)=x \forall x \in \mathbb{R}$

$P(0,0)\Rightarrow f(0)=0$ $P(x,-x)\Rightarrow xf(x^2)=x^2f(x)$ $P(-x,x)\Rightarrow xf(x^2)=-x^2f(-x)$ By these two, $f$ is odd. $P(x,0)\Rightarrow x^2f(x)=f(x)f(x^2)\Rightarrow x^3f(x)=f(x)\cdot xf(x^2)\Rightarrow x^3f(x)=x^2f(x)^2\Rightarrow f(x)^2=xf(x)$ Now $\forall x$, $f(x)\in\{0,x\}$. Assume that $\exists a,b\ne0:f(a)=0,f(b)=b$. \begin{align*} P(b,a)&\Rightarrow b^3+af(a^2)=f(a+b)f(a^2-ab+b^2)\\ &\Rightarrow b^3+a^2f(a)=f(a+b)f(a^2-ab+b^2)\\ &\Rightarrow b^6=f(a+b)^2f(a^2-ab+b^2)^2\\ &\Rightarrow b^6=(a+b)(a^2-ab+b^2)f(a+b)f(a^2-ab+b^2)\\ &\Rightarrow b^6=(a^3+b^3)b^3\\ &\Rightarrow b^3=a^3+b^3\\ &\Rightarrow a=0\end{align*}and we finally have a contradiction. Then the only solutions are $\boxed{f(x)=0}$ and $\boxed{f(x)=x}$, which both work.

BVKRB- wrote:

This seems correct to me.

MNJ2357 wrote: Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$$for all $x,y\in\mathbb{R}$.

Let $P(x,y)$ denote the given assertion. $P(0,0): f(0)^2=0\implies f(0)=0$. $P(x,-x): x^2f(x)-xf(x^2)=0\implies x^2f(x)=xf(x^2)\implies f(x^2)=xf(x)$. $P(0,x): x^2f(x)=xf(x)^2\implies xf(x)(f(x)-x)=0$, so either $f(x)=0$ or $f(x)=x$. Suppose that $f(a)=a$ and $f(b)=0$ with $a,b\ne0$. Then $P(a,b): a^3=(a+b)(a^2-ab+b^2)=a^3+b^3$, a contradiction. So the answer is $\boxed{f(x)=x}$ or $\boxed{f\equiv0}$, which both work.

Let $P(x,y):=x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$ $P(0,0)$ yields $f(0)^2=0\Longrightarrow f(0)=0$ $P(x,0)$ yields $x^2f(x)=f(x)f(x^2)$ $P(0,x)$ yields $xf(x^2)=f(x)f(x^2)\Longleftrightarrow f(x^2)(x-f(x))=0$ Furthermore notice that combining $P(x,0)\text{ and }P(0,x)$ yields $x^2f(x)=xf(x^2)$ thus $f(x^2)=xf(x), \forall x\neq0$ however notice that this also holds for $x=0$ since $f(0)=0$ Therefore $f(x^2)=xf(x), \forall x\in\mathbb{R}$, plugging this result into $P(0,x)$ yields $xf(x)(x-f(x))=0\Longrightarrow f(x)(x-f(x))=0, \forall x\neq0$ however this is also true when $x=0$ Thus either $f(x)=x\text{ or }f\equiv0, \forall x\in\mathbb{R}$ both of which are solutions to the functional equation. Now, to avoid a pointwise trap assume that there exist $a,b\neq0$ such that $f(a)=0\text{ and }f(b)=0$ at the same time $P(b,a)$ yields $bf(b)+af(a^2)=f(a+b)f(a^2-ab+b^2)\Longrightarrow a^3=f(a+b)f(a^2-ab+b^2)$ however notice that $\text{RHS}=(a+b)(a^2-ab+b^2)\text{ or }0$ If $\text{RHS}=0$, this forces $a^3=0\Longrightarrow a=0$ which clearly contradicts our assumption. If $\text{RHS}=(a+b)(a^2-ab+b^2)=a^3+b^3$ this forces $a^3=a^3+b^3\Longrightarrow b^3=0\Longleftrightarrow b=0$ which also contradicts our original assumption. In conclusion $\boxed{f\equiv0\text{ or }f(x)=x, \forall x\in\mathbb{R}}$ $\blacksquare$.

easy $f(x)=0$ $f(x)=x$ the case which is $f(x)=0$ for some xs $f(y)=y$ for some ys is very obvious contradicition!!

Really easy but you have to be careful. The answers are $f(x) = x$ for all $x \in \mathbb{R}$ and $f(x)=$ for all $x \in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We denote by $P(x,y)$ the assertion that $x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)$. Then, we immediately have by $P(0,0)$ that $f(0)^2 =0$ , so $f(0)=0$. Then, comparing $P(x,0)$ and $P(0,x)$ gives, \[x^2f(x)=f(x)f(x^2)\]Thus, $f(x)=0$ or $f(x^2)=x^2$ and \[xf(x^2)=f(x)f(x^2)\]which implies $f(x^2)=0$ or $f(x)=x$. Now, for a given real number $x \neq 0$ , either $f(x)=0$ or $f(x)\neq 0$. If this is the case, from the first equation we have that $f(x^2)=x^2 \neq 0$. Thus, from the second equation it follows that $f(x)=x$. Thus, all for $x\in \mathbb{R}$, $f(x)=0$ or $f(x)=x$. Now, we sort out the pointwise trap. Say there exists a real number $x\neq 0$ such that $f(x)=0$. We now show that $f$ must be constant 0. Consider some positive real $t=y^2$ such that $f(t)=t$. Then, $P(x,y)$ implies \[y^3=x^2f(x)+yf(y^2)=f(x+y)f(x^2-xy+y^2)\]Then, it is clear that neither of $f(x+y)$ and $f(x^2-xy+y^2)$ can be zero, so \[y^3 = (x+y)(x^2-xy+y^2)=x^3+y^3\]which is a clear contradiction. Thus, for all $t >0$, $f(t) = 0$. Now, looking back at $P(x,0)$ yields, \[x^2f(x)=f(x)f(x^2)=0\]which implies that $f(x)=0$ for all $x\in \mathbb{R}$ as claimed, and we are done.

Sketch: Denote the above assertion by $P(x,y)$. $P(0,0)$ gives $f(0)=0$ and $P(0,x)$ gives $f(x)f(x^2)=xf(x^2)$. Therefore, possible cases are, $f(x)=x$ pr $f(x^2)=0$ or $f(x)=x$ for some $x\in A, f(x)=0$ for $x=\mathbb{R} -{A}$. From the first case, we get a solution $f(x)=x$. From the second case, we get that $f(x)=0$ for all positive reals $x$. Let $a<0, b>0$. Then $P(a,b)$ gives: $a^2f(a)+bf(b^2)=f(a+b)f(a^2-ab+b^2) \implies a^2f(a)=0$, as $b^2>0$ and $a^2-ab+b^2>0$. Therefore, $f(x)=0$ for all negative reals $x$ also. Therefore, $f(x)=0$ $\forall x\in \mathbb{R}$, which is also a solution. Now we deal with the third case. If $f(x) = 0$ or $x$, $f(a)=a, f(b)=0$ and $a,b\neq 0$. Then $P(a,b)$ gives $a^3=f(a+b)f(a^2-ab+b^2)$ which means that $RHS$ is not zero, as $a\neq 0$. Therefore, $f(a+b)$ must be $(a+b)$ and $f(a^2-ab+b^2)$ must be $(a^2-ab+b^2)$. Then $a^3=a^3+b^3$, but $b$ is non-zero, which is a contradiction! Therefore, the third case is not possible and our only solutions are $\boxed{f(x)=x}$ $\forall$ $x\in \mathbb{R}$ or $\boxed{f(x)=0}$ $\forall$ $x\in\mathbb{R}$

$P(0,0)$ $P(x,-x)$ $P(0,x)$ this substitution kills the problem

good question, you know, I love you MNJ2357

Me to.......

From P(0;y) we get f(y)=y or f(y²)=0 Main case is f(y²)=0 ,other case is obviously solution. We get x²f(x)=f(x+y)f(x²-xy+y²)=y²f(y) plug y=0 we get f(x)=0 Answers:f(x)=0 and f(x)=x