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Find all functions $f: \Bbb{R} \to \Bbb{R}$ which satisfy:

$xf(x+y)+f(y)=f(y)f(x+1)+f(x^2)$

\begin{align} & \text{let } P(x, y): xf(x+y)+f(y)=f(y)f(x+1)+f(x^2). \\ P(0, y): \; & f(y)=f(y)f(1)+f(0). \\ \Rightarrow \; & f(y)(1-f(1))=f(0). \\ \text{ if } \; & f(1) \neq 1: \\ & f(y)=c, xc+c=c^2+c, c=0 \rightarrow \boxed{f \equiv 0}. \\ \text{if } \; & f(1)=1: f(0)=0.\\ P(x, 0): \; & xf(x)=f(x^2). \end{align}

Here's my attempt and effort, but can't do somethin' else. Can anyone help me solve this F.E.?

(This problem is from Olympiad similarities collection.)

Kenny Wong
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RDK
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2 Answers2

3

The case $f(1) \neq 0$ already grants $f \equiv 0$ with what OP already did, so let's look at the case $f(1) = 1$.

Thanks to $f(0) = 0$ showed by OP, we have: $$P(-1, y):\quad -f(y - 1) + f(y) = f((-1)^2) = f(1) = 1$$ As such, by taking $y+1$ instead of $y$: $$\forall y \in \mathbb{R},\quad f(y +1) = f(y) + 1 \tag{$1$}$$ Therefore $P(x,y)$ is equivalent to: $xf(x+y) = f(y)f(x) + f(x^2)$.
Afterwards: $$P(x,0):\quad xf(x) = f(x^2)$$ Since $x^2 = (-x)^2$, this gives: $xf(x) = f(x^2) = -xf(-x)$, and thus, as we already have $f(0) = 0$: $$\forall x \in \mathbb{R},\quad f(x) = -f(-x) \tag{2}$$ Using $(2)$, we get: $$P(x, -x):\quad 0 = -f(x)^2 + f(x^2)$$ Hence: $$\forall x \in \mathbb{R},\quad f(x^2) = f(x)^2 \tag{3}$$ But thanks to the post $f(x+1)=f(x)+1$ and $f(x^2)=f(x)^2$ we can conclude that this would imply $f = \operatorname{id}_\mathbb{R}$ since it satisfies $(1)$ and $(3)$.

Finally, $0$ and $\operatorname{id}_\mathbb{R}$ are solutions, and as such we have found all the solutions of OP's equation.

Bruno B
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$$xf(x+y) + f(y) = f(y) f(x+1) + f(x^2)$$

  1. Put, $x = 0$, $$f(y) = f(y) f(1) + f(0) \implies f(y) = f(0)/(1-f(1))$$ OR $f(1) = 1$ and $f(0) = 0$,

  2. Put $y = 0$, $xf(x) = f(x^2) \implies$ Putting $x \rightarrow -x$ in the this equation, $-xf(-x) = f(x^2) = xf(x) \implies f(x) = -f(-x)$. So $f$ is an odd function.

  3. Look at $$f(y^2) = yf(y) \rightarrow f(y^2) = y f((\sqrt{y})^2) = y^{1+0.5} f(y^{0.5}) \implies$$ $$ f(y^2) = y^{1+\sum_i \frac{1}{2^i}} \lim_{i \rightarrow \infty} f(y^{\frac{1}{2^i}}) = y^2 f(1) = y^2$$,

Hence $f(y^2) = y^2$ and coupled with odd property of function, we have $f(y) = y$.

In the above we have assumed continuity of $f(x)$ at $x = 1$.

  1. Only Solution, is $f(x) = x$ OR $f(0) = 0$.
Balaji sb
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    This would have been a cool solution if continuity had been assumed, so I'll still give a +1, but as shown in my answer you don't actually need to assume continuity. – Bruno B May 20 '23 at 10:26