I found that $f(x)=-x$ is the solution, but I don't have time to write it now, it may also be wrong. Anyways, I analysed $P(x,0),P(x,-x)$ and $P(-x,x)$.

Wrong solution. Edit: Yes. Thanks pco.

Diamondhead wrote: ... $P\left(\frac{x}{f(k)},k-x\right)$: $f\left(\left(\frac{x}{f(k)}\right)^2+\frac{x}{f(k)}f(k-x)\right)=x$ for all $x\in\mathbb{R}$, so $f$ is a bijective function.... No, you can only conclude from this that $f(x)$ is surjective. This is not enough to prove injectivity (and you need injectivity for your final step)

Put $x,y=0$ to get $f(0)=0$ also put $y=0$ to obtain $f(x^{2})=xf(x)$ Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$. Putting $y=k$ we obtain $f(x^{2})=xf(x)=xf(x+k)$ which means for $x\not= 0$ we have $f(x)=f(x+k)$. Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution. Suppose no such $k$ exists. Then putting $y=-x$ we get $f(x^2+xf(-x))=0$ $\implies x^2 +xf(-x)=0$ which gives $f(-x)=-x$ replacing $-x$ with $x$ we get $f(x)=x$ Hence $f(x)=x$ and $f(x)=0$ are the only solutions. Edit- Wrong solution. Thanks @Ashutoshmaths for pointing it out.

darthsid wrote: Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$. we have $f(x)=f(x+k)$. Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution. This is wrong. I don't think you can solve this problem without using surjectivity.

Ashutoshmaths wrote: darthsid wrote: Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$. we have $f(x)=f(x+k)$. Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution. This is wrong. I don't think you can solve this problem without using surjectivity. I don't get it. Why is it wrong? Edit: I get it now.

Note: This problem had a lot of fake solves at the Camp and was by far the most troll problem there. And btw only 4-5 students solved it correctly in the time limit (excluding fake solves which almost all others did).

darthsid wrote: Ashutoshmaths wrote: darthsid wrote: Now let us suppose there exists a constant $k$ such that $k\not=0$ and $f(k)=0$. we have $f(x)=f(x+k)$. Since $x$ can take any value in $\mathbb R \implies f(x)$ is a constant function $\implies f(x)=0$ which is indeed a solution. This is wrong. I don't think you can solve this problem without using surjectivity. I don't get it. Why is it wrong? $f(x+k)=f(x)$ does not mean $f(x)$ is constant, $f$ can be periodic with period $k$

Let $E(x,y)$ be the statement of the functional equation. From $E(0,0)$ we get $f(0)=0$. Clearly $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ is a solution of the equation, henceforth suppose that $f\not\equiv 0$. Let $a$ be such that $f(a)\neq 0$. Then $E(x, a-x)$ implies $f(z)=xf(a)$, where $z=x^2+xf(a-x)$. Since $f(a)\neq 0$, it follows that $f$ is surjective. Suppose that $f(n)=f(m)$ for real numbers $n,m\in\mathbb{R}$. From $E(x,n)$ and $E(x,m)$ we get $xf(x+n)=f(x^2+xf(n))=f(x^2+xf(m))=xf(x+m)$, therefore $f(x+n)=f(x+m), \forall x\in\mathbb{R}$. Set $k=m-n$; then with $z=x+n$ the previous equation can be written as $f(z)=f(z+k),\forall z\in\mathbb{R}\,\,(*)$. Choose $y$ such that $f(y)=-k$, which exists because $f$ is surjective. Now using $E(k,y)$ we have \begin{align*} 0=f(k^2-k^2)=f(k^2+kf(y))=kf(k+y)\stackrel{*}{=}kf(y)=-k^2 \end{align*}hence $k=0$, so $n=m$. Therefore $f$ is injective. $E(1,x)$ gives $f(1+f(x))=f(1+x),\forall x\in\mathbb{R}$, which is just $\boxed{f(x)=x,\forall x\in\mathbb{R}}$ because of injectivity.

Let $P(x,y)$ be the assertion. If f is constant. we get $f(x)=0$ $P(0,0) \implies f(0)=0$ Suppose there 's $f(k)=0$ and$ k\not=0$ $P(x,0) \implies f(x^2)=xf(x) $ set $x=k \implies f(k^2)=0 $ $P(x,k^2)$and$P(x,0) \implies f(x^2)=xf(x+k^2)=xf(x) \implies f(x)=f(x+k^2)$ $P(x,k)$and$P(x,0) \implies f(x^2)=xf(x+k)=xf(x) \implies f(x)=f(x+k)$ $P(k,x) \implies f(k^2+kf(x))=kf(x+k) \implies f(kf(x))=kf(x)$ _(1) And easy to see $f$ is bijections. from $(1)$ we get $f(kx)=kx \implies f(x)=x $contradictions if $x=k$ So $f(z)=0$ if only if $z=0$ $P(x,-x) \implies f(x^2+xf(-x))=0 so f(x)=x$

guys it is very easy.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$. $p(0,0)\Rightarrow\ f(0)=0$.If there is f($\omega$)=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$(constant) Let there is no real such $f(x)=0$ then $P(x,0)\Rightarrow\ f(x^2)=xf(x)$ and $P(x,-x)\Rightarrow\ f(x^2+xf(-x))=0\Rightarrow\ x^2=-xf(-x)$ wich is$\boxed {f(x)=x}$

TRYTOSOLVE wrote: guys it is very easy.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$. $p(0,0)\Rightarrow\ f(0)=0$.If there is f($\omega$)=0 then $p(x,\omega)$ implies that $f(\omega)=f(\omega+y)$ For me, $p(x,\omega)$ implies $f(x^2)=xf(x+\omega)$ and not $f(\omega)=f(\omega+y)$

ups oh yes I will сhange it now

it is not easy like I thought.

Is my solution true guys?

.Let the assertion will be $p(x,y)=f(x^2+xf(y))=xf(x+y)$. $p(0,0)\Rightarrow\ f(0)=0$.If there is f($\omega$)=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$(constant) Let there is no real such $f(x)=0$ then $P(x,0)\Rightarrow\ f(x^2)=xf(x)$ and $P(x,-x)\Rightarrow\ f(x^2+xf(-x))=0\Rightarrow\ x^2=-xf(-x)$ wich is$\boxed {f(x)=x}$

guys I can not find mistake in my solution.What about you?

TRYTOSOLVE wrote: If there is f($\omega$)=0 then $p(\frac{\omega}{2}\ ,\frac{\omega}{2})$ implies that $f(\omega)=f(\frac{\omega^2}{4}+\frac{\omega}{2}f(\frac{\omega}{2}))$ and continue to infinity , which means $f(x)=0$(constant) And how might this show that $f(x)=0$ for all $x\in\mathbb{R}$?

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Here's a solution in the vain of INMO 2015 P3. (proving injectivity at $0$)

@2above and @3above, your solution seems to be true but needlessly complicated in places. Great solution nevertheless.

Can I say this: $f(x)=0$ is solution. Suppose that $\boxed{f\equiv 0}$ and $f(a)\neq 0$. $P(x;a-x)$ then $f(x^2+xf(a-x))=xf(a)$ and $f$ is surjective

abbosjon2002 wrote: Can I say this: $f(x)=0$ is solution. Suppose that $\boxed{f\equiv 0}$ and $f(a)\neq 0$. $P(x;a-x)$ then $f(x^2+xf(a-x))=xf(a)$ and $f$ is surjective I think you mean $f \not\equiv 0$. Then, yes you can.

Probably the silliest way to solve this problem: As usual, let $P(x,y)$ denote $f(x^2+xf(y))=xf(x+y)$ for all $x,y\in \mathbb{R}$. $P(0,0)$ easily gives us $f(0)=0$ and then $P(x,0)$ gives $f(x^2)=xf(x)$. Suppose there exists non-zero real number $a$ that $f(a)=0$. $P(x,a)$ gives us $f(x^2)=xf(x+a)=xf(x)$ for all $x\in \mathbb{R}$. So, $f(x)=f(x+a)$ for all $x\in \mathbb{R} -\{ 0\}$. Since $f(0)=f(a)=0$, we conclude that $f(x)=f(x+a)$ for all $x\in \mathbb{R}$. We can easily prove by induction that $f(x)=f(x+na)$ for all $x\in \mathbb{R}$ and $n\in \mathbb{Z}^+$. $P(x+a,y)$ gives us $f(x^2+2xa+a^2+xf(y)+af(y))=(x+a)f(x+a+y)=(x+a)f(x+y)$ for all $x,y \in \mathbb{R}$. So, $f\Big( x^2+xf(y) +a(2x+a+f(y))\Big) =xf(x+y)+af(x+y)$ for all $x,y \in \mathbb{R}$. Plugging in $x$ by $t=\frac{n-a-f(y)}{2}$ where $n$ is a positive integer gives us $$f(t^2+tf(y)+na)=f(t^2+tf(y))=tf(t+y)=tf(t+y)+af(t+y)\implies f(t+y)=0.$$Hence, we get that for any $a\neq 0$ that $f(a)=0$, $f\Big( \frac{n-a-f(y)}{2}+y\Big) =0$ for all positive integer $n$ and real number $y$. Suppose there exists non-zero real number $a$ that $f(a)=0$. We've $f\Big( \frac{1-a-f(1)}{2}+1\Big) =0$ and $f\Big( \frac{2-a-f(1)}{2}+1\Big) =0$. From the preceding paragraph, we've $f(x)=f(x+\frac{1-a-f(1)}{2}+1 )=f(x+\frac{2-a-f(1)}{2}+1)$ for all $x\in \mathbb{R}$. So, $f(x)=f(x+\frac{1}{2})$ for all real number $x$. From $f(0)=0$, we can easily prove that $f(m)=0$ for all integer $m$. Hence, we've $f(x)=f(x+n)$ for all $x\in \mathbb{R}$ and $n\in \mathbb{Z}$. Using the preceding paragraph again, we've $f\Big( \frac{n-m-f(y)}{2}+y\Big) =0$ for all $n\in \mathbb{Z}^+,m\in \mathbb{Z},y\in \mathbb{R}$. So, $f(y-\frac{f(y)}{2})=0$ for all real number $y$. This gives $f(n^2+nf(y)-\frac{nf(n+y)}{2})=f(n^2+\frac{nf(y)}{2})=0$ for all $n\in \mathbb{Z}^+$ and $y\in \mathbb{R}$. So, $f(4+f(y))=f(f(y))=0\implies f(x+f(y))=f(x)$ for all $x,y\in \mathbb{R}$. And so $f(x+f(y))=f(x)=f(x+f(z))\implies f(f(y)-f(z))=0$ for all $x,y,z\in \mathbb{R}$. If there exists $k\in \mathbb{R}$ that $f(k)\neq 0$. For any $r\in \mathbb{R}$, not hard to show that there exists $a,b,c\in \mathbb{R}$ that $b-a=\frac{r}{f(k)}$ and $a+b+c=k$. Also, there exists $u,v\in \mathbb{R}$ that $f(u)=af(a+b+c)$ and $f(v)=bf(a+b+c)$. From the preceding paragraph, we get $f\Big( (b-a)f(a+b+c)\Big) =f(r)=0$. This means $f(x)=0$ for all $x\in \mathbb{R}$, contradiction with the existence of $k$. Hence, we conclude that if there exists non-zero real number $a$ that $f(a)=0$ then $f(x)=0$ for all $x\in \mathbb{R}$. If there not exists such $a$, $P(x,-x)$ gives $f(x^2+xf(-x))=0\implies x^2+xf(-x)=0$ for all $x\in \mathbb{R}$, this easily gives $f(x)=x$ for all $x\in \mathbb{R}$.

Good Problem. Ankoganit wrote: Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y)$$for all reals $x,y$. We shall show that $f(x) \equiv x \ \forall x \in \mathbb{R}$ and $f \equiv 0$ are the only solutions to the given functional equation and both of them clearly work. $ $ Now assume that $f$ is not identically $0$ and hence there exists at least one $\alpha \in \mathbb{R}$ such that $f(\alpha) \neq 0$. Then, $P(x, \alpha-x) : f(x^2+x \cdot f(\alpha)) = \underbrace{x \cdot f(\alpha)}_{\text{Spans all of} \ \mathbb{R}} \implies f$ is surjective. $ $ Note that $P(0,0) \implies f(0) = 0$ and $P(x,0) : f(x^2) = x \cdot f(x) \forall x \ \in \mathbb{R} \dots (\star)$. Now, suppose there exists $\lambda \in \mathbb{R}$ such that $f(\lambda) = 0 , \lambda \neq 0$. Also by surjectivity of $f$ there must exist $\mu \in \mathbb{R}$ such that $f(\mu) = 1$. $ $ $P(x , \lambda) : f(x^2) = \underbrace{x \cdot f(x + \lambda) =f(x)}_{\bigstar} \implies f$ is periodic. $P(\lambda , \mu): f(\lambda^2 + \lambda) = \underbrace{\lambda \cdot f(\lambda + \mu) =\lambda \cdot f(\mu)}_{\text{Periodic}} = \lambda$. $ $ But, $f(\lambda^2 + \lambda ) = f(\lambda^2)= \lambda \cdot f(\lambda) = 0 \iff \lambda = 0$. So , $f$ is injective at $0$. Now, the problem easily breaks down as: $ $ $P(-f(y), y) : f(y) \cdot f(y-f(y))=0 \implies f(y - f(y))=0 \forall y \neq 0 \iff y=f(y) \forall y \neq 0$. As $f(0) = 0$, we conclude that $\boxed{f(x) \equiv x \forall x \ \in \mathbb{R}}$

my short solution: claim(1):$f $ is surjective proof: $P(\frac{x}{f(y)},y-x) \implies f(something)=x$ $\blacksquare$ claim(2): $f$ is injective proof: suppose there's $a \neq b : f(a)=f(b)$ note that $P(x,a) \& P(x,b) \implies f(x+a)=f(x+b)$ define $\mathbb{S}=\{a-b | f(a)=f(b) \}$ then $f(x)=f(x+s) \forall s \in \mathbb{S}$ now let $f(y_1)=1 ,f(y_2)=2$ $P(s,y_1) \& P(s,y_2) \implies f(s^2)=s=2s \implies s=0$ $\blacksquare$ now just put $P(1,y) \implies f(y)=y$ and we win

Ankoganit wrote: Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y)$$for all reals $x,y$. Nice actually and pritty short Let $P(x,y)$ be the assertion $$f\left( x^2+xf(y)\right)=xf(x+y)$$Notice that if $f \equiv c$ then $c=0$. Suppose we are searching for a non-constant solution. In other words, there exist $t \in \mathbb{R}$ with $f(t) \neq 0$, then $P(x,t-x)$ gives $$ f\left( x^2+xf(t-x)\right)=xf(t) \ \forall x \in \mathbb{R} $$Thus $f$ is surjective, and $P(0,0)$ gives $f(0)=0$, and by $P(x,0)$ we get $$f\left( x^2 \right)=xf(x) \quad (1)$$Now for each $z \in \mathbb{R}^*$, there exist $x_0 \in \mathbb{R}^*$ such that $f(x_0)=z \neq 0$, thus $P(-f(x_0),x_0)$ gives $$0=f(0)=-f(x_0)f\left(x_0-f(x_0)\right)$$because $f(x_0) \neq 0$, it is forced that $f\left(x_0-f(x_0)\right)=0$. Finally, $P(f(x_0),x_0-f(x_0)$ gives $$f\left( f(x_0)^2 \right)=f(x_0)^2 $$from $(1)$, we get $$f(x_0)f\left(f(x_0)\right) = f\left( f(x_0)^2 \right)=f(x_0)^2 $$$$ \Rightarrow f(z) = f\left(f(x_0)\right) = f(x_0) = z \ \forall z \in \mathbb{R}^* $$Hence $f \equiv x$ for all $x \in \mathbb{R}$ and $f \equiv 0$ are the only solutions. $\blacksquare$

Very cool problem! Let $P(x,y)$ be the given assertion. Observe that $P(0,y)$ yields $f(0)=0.$ Thud, $P(x,0)$ implies $f(x^2)=xf(x).$ Note that $f\equiv 0$ is a solution; assume this sin't the case. Then, there exists $c$ such that $f(c)\neq 0$ then by varying $x$ in $P(x,c-x)$ we get that $f$ is surjective. Claim: If $f(a)=f(b)$ then $f$ is periodic with period $|a-b|.$ Proof: If $f(a)=f(b)$ then by combining $P(x,a)$ and $P(x,b)$ we get \[xf(x+a)=f(x^2+xf(a))=f(x^2+xf(b))=xf(x+b).\]Therefore, for all $x\neq 0,$ we have $f(x+a)=f(x+b).$ But since $f(a)=f(b)$ then $f(x+a)=f(x+b)$ for $x=0$ as well, so $f$ is periodic with period $|a-b|. \ \square$ Case 1: Assume that $f$ is not injective. Then there exist $a\neq b$ such that $f(a)=f(b).$ According to our claim, $f$ is periodic with period $c:=|a-b|\neq 0.$ Now, recall that $f(x^2)=xf(x).$ Combining this with the periodicity of $f$, we can infer that for any integer $n$ we have \begin{align*}\frac{n-c}{2}\cdot f\bigg(\frac{n-c}{2}\bigg)&=f\bigg(\frac{n^2+c^2-2nc}{4}\bigg)=f\bigg(\frac{n^2+c^2-2nc}{4}+n\cdot c\bigg)=f\bigg(\frac{n^2+c^2+2nc}{4}\bigg) \\ &=\frac{n+c}{2}\cdot f\bigg(\frac{n+c}{2}\bigg) =\frac{n+c}{2}\cdot f\bigg(\frac{n+c}{2}-c\bigg)=\frac{n+c}{2}\cdot f\bigg(\frac{n-c}{2}\bigg)\end{align*}Since $c\neq 0$ then $(n-c)/2\neq(n+c)/2$ so we must have $f((n-c)/2)=0.$ But using our claim, since $f(0)=0=f((n-c)/2)$ then $f$ is periodic with period $(n-c)/2.$ In particular, this implies that $f$ is periodic with period $n-c.$ Recall that $n$ could be any integer, so $f$ has periods $1-c$ and $2-c,$ resulting in the fact that $f$ has period $1.$ However, $P(1,y)$ gives us $f(1+f(y))=f(1+y).$ Using the fact that $f$ has period $1,$ we actually have $$f(f(y))=f(1+f(y))=f(1+y)=f(y).$$But since $f$ is surjective, $f(y)$ can take any real value, so we get that $f(x)=x$ for all $x$, contradicting the fact that $f$ is not injective! Case 2: Assume that $f$ is injective. Well $P(1,y)$ gives us $f(1+f(y))=f(1+y)$ resulting in $f(y)=y$ for all $y.$ Therefore, the only functions are $f\equiv 0$ and $f\equiv\text{id}.$

This is one of the best FE's I've seen in a while. Let $P(x,y)$ be the assertion. $P(0,0)$ implies $f(0)=0$. First, notice $f$ is surjective by taking $y=k-x$ and $x=L/f(k)$ (as we vary $L$, $f$ goes through all reals). Then, $y=0$ implies $f(x^2)=xf(x)=-xf(-x)$ and thus $f$ is odd. Now notice that $y=-x$ implies $f(x^2-f(x^2))=0$, and hence if we prove injectivity at 0 we're done, as we'd prove that $f(x^2)=x^2$ and to extend to the negative reals just use it is odd. To prove injectivity at 0, let $S=\{a_1,...\}$ the set of all real numbers that satisfy $f(a_i)=0$. But notice that if $a \in S$, then $a^2 \in S$, because of $x=a,y=0$. Now, to finish, just notice that: $P(x,a_i)$ implies $f(x^2)=xf(x)=xf(x+a_i)$. Thus the function has period $a_i$ ($f(x)=f(x+a_i)$. Now, by surjectivity, take $c$ so that $f(c)=1$, and $P(a_i,c) \implies f(a_i^2+(1.a_i))=f(a_i^2)=a_i.f(a_i+c)=a_i$, because of $f(k+a_i)=f(k)$. Thus $f(a_i^2)=0=a_i$, and we're done. $\blacksquare$

Note that $f(x)\equiv 0$ and $f(x)\equiv x$ are both solutions. From now on, we can assume there exist (not necessarily distinct) $u$ and $v$ such that $f(u)\neq 0$ and $f(v)\neq v$. Let $c = |v| - f(|v|)$. $P(x,u-x)\implies f$ is surjective. $P(0,0)\implies f(0) = 0$. $P(x,0)\implies f(x^2) = xf(x)$. In particular $f$ is odd, so $f(|v|)\neq |v|$. $P\left(-\sqrt{|v|},\sqrt{|v|}\right)\implies f\left(|v| - \sqrt{|v|}f\left(\sqrt{|v|}\right)\right) = 0\implies f(c) = 0$. $P(x,c)\implies f(x^2) = xf(x+c)\implies f(x) = f(x + c)$ for $x\neq 0$. But $0 = f(0) = f(c)$, so $f(x) = f(x+c)$ for all $x$. $P(c,y)\implies f(c^2 + cf(y)) = cf(c+y) = cf(y)$. Since $c\neq 0$ and $f$ is surjective the funtion $cf(y)$ is surjective, so $f(c^2 + x) = x$ for all $x$. In particular $f$ is linear with slope $1$, but plugging back in yields no new solutions.

Let $P(x,y)$ denote the given assertion. $P(0,0): f(0)=0$. $P(x,0): f(x^2)=xf(x)$. Clearly $\boxed{f\equiv 0}$ is a solution. Henceforth assume there exists a $k$ with $f(k)\ne 0$. Claim: $f$ is surjective. Proof: $P(x,k-x): f(x^2+xf(k-x))=xf(k)$. Since $f(k)\ne 0$, $xf(k)$ can take on any value, so $f$ is surjective $\blacksquare$. Claim: If $f(k)=0$, then $k=0$. Proof: Suppose there exists a $k\ne 0$ with $f(k)=0$. $P(x,k): f(x^2)=xf(x+k)$. So $xf(x)=xf(x+k)\implies f(x)=f(x+k)$. Let $a$ satisfy $f(a)=-k$. $P(k,a): f(k^2-k^2)=0=kf(a+k)=kf(a)=-k^2$, a contradiction. $\blacksquare$ $P(x,-x): f(x^2+xf(-x))=0\implies x^2=-xf(-x)$. If $x\ne 0$, then $-f(-x)=x\implies \boxed{f(x)=x}$.

We prove that $f(x)=x$ and $f\equiv 0$ are the only solutions. Let $P(x,y)$ be the assertion in $f\left( x^2+xf(y)\right)=xf(x+y)$. It can be easily checked that the only constant function is $\boxed{f\equiv0}$, which indeed works and is our first solution. Let $f$ be non-constant. Claim : $f$ is injective. Proof : Assume possible $f(y_1)=f(y_2)$ and $y_1\neq y_2$. $P(x,y_1)-P(x,y_2) : f(x+y_1)=f(x+y_2) \forall x$. Hence, $f$ is periodic with period $|y_2-y_1|$. Hence, $f$ is bounded above. But, we can choose very large $x$ and $y$ such that $f(x+y)\neq0$(Note we can select such $y$ because $f\not\equiv 0$) : $f\left( x^2+xf(y)\right)<xf(x+y)$. Contradiction! Hence, $f(y_1)=f(y_2) \implies y_1=y_2$. Hence proved claim! $P(0,y) : f(0)=0$. $P(-x,x) : f(x^2-xf(x))=0$. Hence, $f(x^2-xf(x))=f(0) \overset{\text{injective}}{\implies} x^2-xf(x)=0 \implies f(x)=x \forall x\neq0$. Combining with $f(0)=0$, we get $\boxed{f(x)=x },\forall x$, which indeed satisfies given equation and is our second solution. Hence, we are done

Ankoganit wrote: Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y)$$for all reals $x,y$. My solution, without using the injectivity. Let $P(x,y)$ be the assertion that $f\left( x^2+xf(y)\right)=xf(x+y)$. $P(0;y) \implies f(0)=0$ $P(x;0) \implies f(x^2)=xf(x)$, which leads to $f$ is odd. Clearly $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ is a solution, hence assume that there exist $k$ such that $f(k)\ne 0$. $P(x;k-x) \implies f(\cdots)=xf(k)$, it follows that $f$ is surjective, so there exists $a \in \mathbb R$ such that $f(a)=0$. $P(x;a) \implies f(x+a)=f(x) \quad \forall x \in \mathbb R \implies f(x+na)=f(x) \quad \forall n \in \mathbb Z$ $P(x;-x) \implies f(x^2-xf(x))=0 \implies f(a^2)=0 \quad (*)$ $P(x;a^2) \implies f(x+a^2)=f(x) \quad \forall x \in \mathbb R$ Put $x \longrightarrow x+a$ into $f(x^2)=xf(x)$, we have: $f(x^2+2ax+a^2)=(x+a)f(x+a) \implies f(n^2)=(n+a)f(n)=nf(n) \implies af(n)=0 \quad \forall n \in \mathbb Z$ If $a=0$, from $(*)$ we have $\boxed{f(x)=x,\forall x\in\mathbb{R}}$ is also a solution. If $f(n)=0 \quad \forall n \in \mathbb Z$: $P(x;n) \implies f(x+n)=f(x) \quad \forall n \in \mathbb Z$ $P(x+n;y) \implies f(x^2+2nx+n^2+(x+n)f(y))=(x+n)f(x+y+n)$ $\implies f(x^2+2nx+(x+n)f(y))=(x+n)f(x+y) \quad \forall n \in \mathbb Z$ Because of the surjectivity, for all $x \in \mathbb R, n \in \mathbb Z$, there exists $t$ such that $f(t)=n-x$. Put $y \longrightarrow t$ into the above equation, we have: $f(2nx)=(x+n)f(x+t) \quad \forall n \in \mathbb Z$ Put $x \longrightarrow x-n$ into the above equation: $xf(x+t)=f(2nx) \implies nf(x+t)=0 \implies f(x+t)=0 \quad \forall n \in \mathbb Z$ So all the solutions are $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ or $\boxed{f(x)=x,\forall x\in\mathbb{R}}$.

let $P(x,y)$ be the natural assertion. $P(0,0) : f(0)=0$ $p(x,0) : f(x^2)=xf(x)$ if $\exists k \neq 0$ such that $f(k)=0$ then by $P(x,k) : f(x^2)=xf(x+k) \implies f(x)=f(x+k)$ then $f$ is $k-$periodic . and we have by $P(x+k,-k ) : f((x+k)^2)=(x+k)f(x) \implies f(x^2)=xf(x)+kf(x) \implies k=0$ or $\boxed{\forall x : f(x)=0}$ is a solution if $k=0$ then $\forall z\neq 0 : f(z) \neq 0 $ : $P(-x,x) : f(x^2-xf(x))=0 \implies xf(x)=x^2 \implies f(x)=x$ so we have the second solution $\boxed{\forall x : f(x)=x}$

Oly grind back maybe? Let $P(x,y)$ the assertion of the F.E. clearly if $f$ is constant then $f(x)=0$ works so lets work when $f$ is not constant. $P(0,x)$ $$f(0)=0$$$P(x,0)$ $$f(x^2)=xf(x) \implies f \; \text{odd}$$$P(f(x),-x)$ $$f(f(x)-x)=0$$$P(x,f(x)-x)$ $$xf(x)=f(x^2)=xf(f(x)) \implies f(x)=f(f(x))$$As $f$ is non-cero there exists $d$ with $f(d) \ne 0$ so by $P(x,d-x)$ $$f(x^2+xf(d-x))=xf(d) \implies f \; \text{surjective}$$Since $f$ is surjective we use that on the last equation by setting $t=f(x)$ to get $$f(t)=t \; \forall t \in \mathbb R$$Hence $\boxed{f(x)=0,x \; \forall x \in \mathbb R}$ work thus we are done

Let $P(x,y)$ be the assertion. $P(0,0)$: $f(0)=0$ $P(x,0)$: $f(x^2)=xf(x)$ Case 1: $f(x)=0 \iff x=0$ $P(-f(x),x)$: $f(x)f(x-f(x))=0 \implies f(x)=x, \forall x \in\mathbb{R}$, which is a solution. Case 2: There exists $t\neq0$ such that $f(t)=0$ $P(x,t)$: $f(x^2)=xf(x+t) \implies xf(x+t)=xf(x) \implies f(x)=f(x+t), \forall x \in\mathbb{R}$. This means $f(d)=0 \implies f$ has period $d$. $f(t^2)=tf(t)=0$, so $f$ has period $t^2$. Now, $f(x^2+2tx+t^2)=f((x+t)^2)=(x+t)f(x+t)=xf(x)+tf(x)=f(x^2)+tf(x)$. Let $x$ be an integer, then $f(x^2+2tx+t^2)=f(x^2) \implies tf(x)=0 \implies\forall x \in\mathbb{Z}, f(x)=0$, so $f$ has period $1$. Take a positive integer $k$ that is not a perfect square, $f(k)=\sqrt{k}f(\sqrt{k})=0 \implies f(\sqrt{k})=0$, where $\sqrt{k}$ is irrational. Thus, $f$ has periods $1$ and $\sqrt{k}$. We aim to prove $f(x) \equiv 0$. Only need to prove $f(x)=0$ for all $x \in [0,1)$. For fixed irrational number $\alpha$, let $n$ span all integers, $\{n\alpha\}$ is dense in $[0,1)$. Hence $0=f(0)=f(n\sqrt{k}-[n\sqrt{k}])$, and $n\sqrt{k}-[n\sqrt{k}]$ can be any number in $[0,1)$ $\implies f(x)=0, \forall x \in [0,1)$. So $f(x) \equiv 0$, which indeed is a solution. Conclusion: Solutions are $f(x)=x, \forall x \in\mathbb{R}$ or $f(x) \equiv 0$. Weird solution I just came up with (any mistakes?).

Let $P(x,y)$ be the given assertion. $P(0,0)$ gives $f(0)=0.$ If constant then $f\equiv 0.$ If not then $P(x,y-x)$ gives $f$ is surjective. Now let $f(u)=f(v).$ Comparing $P(z,u)$ and $P(z,v)$ gives $f(z)=f(z+w)$ where $w=u-v.$ Take $f(m)=0$ and $f(n)=1.$ Comparing $P(w,m)$ and $P(w,n)$ forces $w=0.$ So $f$ is injective and $P(1,x)$ gives $f(x)=x.$ Both work.

Ali3085 wrote: my short solution: claim(1):$f $ is surjective proof: $P(\frac{x}{f(y)},y-x) \implies f(something)=x$ $\blacksquare$ claim(2): $f$ is injective proof: suppose there's $a \neq b : f(a)=f(b)$ note that $P(x,a) \& P(x,b) \implies f(x+a)=f(x+b)$ define $\mathbb{S}=\{a-b | f(a)=f(b) \}$ then $f(x)=f(x+s) \forall s \in \mathbb{S}$ now let $f(y_1)=1 ,f(y_2)=2$ $P(s,y_1) \& P(s,y_2) \implies f(s^2)=s=2s \implies s=0$ $\blacksquare$ now just put $P(1,y) \implies f(y)=y$ and we win Your solution needs a fix since surjectivity holds $\forall f(x)\neq 0.$

2016 India IMO Training T2P2 wrote: Find all functions $f:\mathbb R\to\mathbb R$ such that $$f\left( x^2+xf(y)\right)=xf(x+y),\forall x,y\in \mathbb{R}^{(1)}$$ When $f$ is a constant function, we have $\boxed{f(x)=0,\forall x\in \mathbb{R}}$ Now consider the case when $f$ is not a constant function, then $\exists a \in \mathbb{R}$ such that $f(a)\ne 0$ $(1). P(x,a-x)\Rightarrow f(x^2+xf(a-x))=xf(a),$ subtitute $x\rightarrow \dfrac{x}{f(a)}\Rightarrow f(\text{something})=x\Rightarrow f$ is surjective $(1).P(0,0)\Rightarrow f(0)=0; (1).P(x,0)\Rightarrow f(x^2)=xf(x)$ Assume that there is $c\ne 0$ such that $f(c)=0$ $(1).P(x,c)\Rightarrow f(x^2)=xf(x+c)\Rightarrow f(x)=f(x+c),\forall x\in \mathbb{R}$ (as we already have $f(x^2)=xf(x),\forall x\in \mathbb{R}$) $(1).P(c,y)\Rightarrow f(c^2+cf(y))=cf(y+c)=cf(y)\Rightarrow f(xc+c^2)=xc\Rightarrow f(x+c^2)=x$ Subtitute $x\rightarrow c-c^2\Rightarrow c-c^2=0\Rightarrow c=1\Rightarrow f(x+1)=x\Rightarrow f(0)=-1,$ a contradiction. Hence $f(x)=0\Leftrightarrow x=0; (1).P(x,-x)\Rightarrow f(x^2+xf(-x))=0\Rightarrow x^2=-xf(-x)\Rightarrow \boxed{f(x)=x,\forall x\in \mathbb{R}}$