Find all functions $f:\mathbb{R}\to \mathbb{R}$ so that for any reals $x,y$ the following holds: \[f(x\cdot f(x+y))+f(f(y)\cdot f(x+y))=(x+y)^2\]
Problem
Source: 2021 Israel TST 8 P2
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01.06.2022 00:40
Phorphyrion wrote: Find all functions $f:\mathbb{R}\to \mathbb{R}$ so that for any reals $x,y$ the following holds: \[f(x\cdot f(x+y))+f(f(y)\cdot f(x+y))=(x+y)^2\] Let $P(x, y)$ be the assertion to given FE Then $P(0,x)\implies f(0)+f(f(x)^2)=x^2\cdots (1)$ assume there exist $x_1>x_2\geq 0$ Such that $f(x_1)=f(x_2)$ So $P(0, x_1), P(0, x_2)$ are equal hence $(x_1-x_2)(x_1+x_2)=0$ So $x_1=x_2$ which is contradiction hence $f(x)$ is Injective over $R_+=\{x| x\geq 0\}$ Assume there is $x_1, x_2$ such that $f(x_1)=-f(x_2)$ then by $(1)$ we have $x_1^2=x_2^2$ $P(0, 0)\implies f(0)=-f(f(0)^2) \implies f(0)=0$ So $(1)$ becomes $f(f(x)^2) =x^2$ this gives $f(x)$ is onto over $R_+$ and $f(x)\geq 0 \forall x\geq 0$ and $P(x, 0)\implies f(xf(x))=x^2$ So as $f$ is Injective over $R_+$ we have $xf(x)=f(x)^2$ for all $x\geq 0$ and hence because of Injectivity $f(x)=x \forall x\geq 0$ Now as for any $x_1>0$ we have $f(-x_1f(-x_1))=f(x_1^2)$ So By $(1)$ we have $f(-x_1)=x_1$ or $-x_1$ If $f(-x_1)=x_1$ then $P(2x_1, -x_1)\implies f(2x_1f(x_1))+f(f(-x_1)f(x_1))=x_1^2 \implies 3x_1^2=x_1^2$ contradiction. So $f(-x_1)=-x_1 \forall x_1>0$ Hence $f(x)=x \forall x\in R$ is the only solution.
01.06.2022 01:00
Denote the given assertion by $P(x,y)$. First, $P(0,y)$ yields \[ f(0) + f\bigl(f(y)^2\bigr) = y^2. \]In particular, if $y_1$ and $y_2$ are such that $f(y_1)^2=f(y_2)^2$, then we conclude that $y_1^2=y_2^2$. Now, we use this observation with the previous display: setting $y=0$, we get \[ f(0)^2 = f\left(f(0)^2\right)^2 \implies f(0)^4 = 0 \implies f(0)=0. \]Now, $P(x,0)$ yields $f(xf(x)) = x^2$. In particular, if $x_0$ is such that $f(x_0)=0$, we get that $x_0=0$. Now, comparing this with $f(f(y)^2)=y^2$ established above, we get \[ f(y)^4 = y^2f(y)^2 \implies f(y)^2\bigl(f(y)-y\bigr)\bigl(f(y)+y\bigr)=0. \]Namely, for any $y\ne 0$, $f(y)=\pm y$. Moreover, using $f(x)^2 = x^2$, we get that $f(x)=x$ for all $x\ge 0$. Finally, assume there is a $-x_0<0$ such that $f(-x_0)=x_0$. Inspecting $P(2x_0,-x_0)$, we immediately get a contradiction. Hence, $f(x)=x$ for all $x$.
23.06.2022 01:43
This F.E. is kinda nice tbh. $P(0,x)$ $$f(f(x)^2)=x^2-f(0) \implies \; \text{if} \; f(a)^2=f(b)^2 \; \text{then} \; a^2=b^2$$$P(0,0)$ $$f(f(0)^2)^2=f(0)^2 \implies f(0)^4=0 \implies f(0)=0 \implies f(f(x)^2)=x^2 \implies f \; \text{injective at 0 and surjective in} \; \mathbb R_{\ge 0}$$$P(x,0)$ $$f(xf(x))=x^2=f(f(x)^2) \implies x^2f(x)^2=f(x)^4 \implies f(x)=x \; \text{or} \; f(x)=-x$$Assume that there exists $b \ne 0$ with $f(b)=-b$, then clearly $b$ has to be negative, becuase if it was positive then there would exists $t$ such that $f(t)^2=b$ and then $f(b)$ would be positive. Now this tells us that $f(x)=x$ holds for all $x \ge 0$, now set $x \ge 0$ and do $P(x,b)-P(b,x)$ $$f(-bf(x+b))=f(bf(x+b)) \implies f \; \text{even} \; \implies f(x)=x \; \forall x \ge 0 \; \text{and} \; f(x)=-x \; \forall x \le 0$$Now set $P(x,-y)$ with $x>y>0$ $$x^2-y^2=x^2+y^2-2xy \implies x=y \; \text{contradiction!l}$$Hence the only function that work is $\boxed{f(x)=x \; \forall x \in \mathbb R}$ thus we are done
05.08.2022 18:37
Denote the assertion by $P(x,y).$ Let $|f(u)|=|f(v)|$ then $P(0,u)$ and $P(0,v)$ implies $|u|=|v|.$ With this, $P(0,0)$ implies $f(0)=0.$ $P(x,0)$ implies $f(xf(x))=x^2=f(f(x)^2)$ hence using the result, $f(x)=\pm x$ since $f\not \equiv 0.$ Moreover $f(f(x)^2)=f(x^2)=x^2$ so $f(x)=x$ for $x\geq 0.$ If $f(y)=-y$ for $y\leq 0$ and $x\geq 0$ then $P(x,y)$ gives $y=0$ for $f(x+y)=x+y$ or $x=0$ for $f(x+y)=-(x+y).$ Hence $f\equiv \text{Id}.$