Nice and easy

Let $P(x,y)$ be the assertion $(f(x)+y)(f(y)+x)=f(x^2)+f(y^2)+2f(xy)$. $P(0,0)\Rightarrow f(0)^2=4f(0)\Rightarrow f(0)\in\{0,4\}$ $\textbf{Case 1: }f(0)=0$ $P(x,0)\Rightarrow xf(x)=f(x^2)$ $P(x,x)\Rightarrow(f(x)+x)^2=4xf(x)\Rightarrow(f(x)-x)^2=0\Rightarrow\boxed{f(x)=x}$, which fits. $\textbf{Case 2: }f(0)=4$ $P(1,0)\Rightarrow f(1)=3$ Comparing $P(x,0)$ with $P(x,1)$, we have $f(x)=\frac13x+4$, which doesn't work. $\square$

anyone__42 wrote: Find all functions $f$ $:$ $\mathbb{R} \rightarrow \mathbb{R}$ such that $\forall x,y \in \mathbb{R}$ : $$(f(x)+y)(f(y)+x)=f(x^2)+f(y^2)+2f(xy)$$ Let $P(x,y)$ the assertion of the given FE. $P(0,0)$ $$f(0)^2=4f(0) \implies f(0)=\{0,4\}$$If $f(0)=4$ $P(0,-4)$ $$4+f(16)+8=0 \implies f(16)=-12$$$P(4,4)$ $$(4+f(4))^2=-48 \implies \; \text{contradiction!!}$$So $f(0)=0$ $P(x,0)$ $$xf(x)=f(x^2)$$$P(x,x)$ $$(f(x)+x)^2=4xf(x) \implies (f(x)-x)^2=0 \implies f(x)=x$$Thus the only solution is: $\boxed{f(x)=x \; \forall x \in \mathbb R}$ Thus we are done

Let $P(x,y)$ be the assertion of $(f(x)+y)(f(y)+x)=f(x^2)+f(y^2)+2f(xy)$. $P(0,0)\Rightarrow f(0)^2=4f(0)\Rightarrow f(0)\in\{0,4\}$ Now we have $2$ cases, Case 1: $f(0)=4$ If $f(0)=4$ $P(0,-4)$ $$4+f(16)+8=0 \implies f(16)=-12$$$P(4,4)$ $$(4+f(4))^2=-48$$so we have a contradiction. Case 2: $f(0)=0$ $P(x,0)$ $$xf(x)=f(x^2)$$$P(x, x)$ gives $f(x)=x$ and that works, so the only solution is $\boxed{f(x)=x}.$

Dang went a little overboard when trying to get rid of the $f(0)=4$ case The only answer is $f(x)=x$. It is easy to see that this works. We now prove that this is the only solution. As per usual, denote the assertion by $P(x,y)$. $P(0,0)$ gives that $f(0)=0,4$. We take cases based on whether $f(0)=0$ or $f(0)=4$. Case 1: $f(0)=4$ The idea is to just bash a bunch of values until you get some time of contradiction. The intuition is to try to make things cancels, so notice $1^2=1$ so plugging in $P(1,0)$ give that $f(1)=3$. $P(x,0)$ gives us that $f(x)(x+4)=f(x^2)+12,$ and $f(-x)(-x+4)=f(x^2)+12,$ so $f(x)(x+4)=f(-x)(-x+4)$. Plugging in $x=4$ gives that $f(4)=0$. Plugging in $P(2,2)$ gives that $f(2)=-2$ and from here it is enough to get the contradiction: plugging in $P(1,2)$ gives that $5(-1)=3+0-4,$ an obvious contradiction. Therefore, this case yields no solutions. Case 2: $f(0)=0$ $P(x,0)$ gives that $yf(y)=f(y^2),$ and using the same tactic as in case 1, we have that $-yf(-y)=f(y^2),$ and so $f(-y)=-f(y)$, so we have that $f$ is odd. From here, it is visceral to plug in $P(x,-x)$ and this indeed finishes the problem. We get that $(f(x)-x)(f(-x)+x)=f(x^2)+f(x^2)+2f(-x^2)=0\implies f(x)=x$ or $f(-x)=-x$, but these are the same function, so therefore this case yields $\boxed{f(x)=x}$ as the solution.

$P(0,0) : f(0)^2 = 4f(0) \implies f(0) = 0,4$ Case $1 : f(0) = 0$ $P(x,0) : xf(x) = f(x^2)$. $P(x,x) : f(x)^2 + x^2 + 2xf(x) = 4f(x^2) = 4xf(x) \implies f(x)^2 + x^2 - 2xf(x) = 0 \implies (f(x) - x)^2 = 0 \implies f(x) = x$. Case $2 : f(0) = 4$ $P(1,0) : 5f(1) = f(1) + 12 \implies f(1) = 3$. $P(x,0) : f(x)(4+x) = f(x^2) + 12$. $(1)$ $P(x,1) : (x+1)f(x) + x = f(x^2)$. $(2)$ From $(1),(2)$ we have $3f(x) = 12 + x \implies f(x) = \frac{x}{3} + 4$ But it gives contradiction with main equation. Answers : $f(x) = x$.

Let $P(x,y)$ denote the given assertion. $P(0,0): f(0)^2=4f(0)\implies f(0)\in \{0,4\}$. Case 1: $f(0)=0$. $P(x,0): xf(x)=f(x^2)$. $P(x,x): (f(x)+x)^2=4f(x^2)=4xf(x)\implies f(x)^2+2xf(x)+x^2=4xf(x)\implies f(x)^2+x^2=2xf(x)$. So $f(x)^2-2xf(x)+x^2=(f(x)-x)^2=0\implies \boxed{f(x)=x}$, which works. Case 2: $f(0)=4$. $P(0,-4): 0=12+f(16)\implies f(16)=-12$. $P(4,4): (f(4)+4)^2=4f(16)=-48$, a contradiction.

anyone__42 wrote: Find all functions $f$ $:$ $\mathbb{R} \rightarrow \mathbb{R}$ such that $\forall x,y \in \mathbb{R}$ : $$(f(x)+y)(f(y)+x)=f(x^2)+f(y^2)+2f(xy)$$ $$P(0,0)\implies f^2(0)=4f(0)\implies f(0)=0 or f(0)=4$$ $$P(x,0)\implies f(x^2)=xf(x)$$ $$P(x,x)\implies (f(x)-x)^2=0 \implies \boxed{f(x)=x}$$now we want to see the solution if $f(0)=4$ $$P(0,-4)\implies 0=12+f(16)\implies f(16)=-12$$$$P(4,4)\implies (f(4)+4)^2=4f(16)=-48$$, a contradiction.

$(f(x) + y)(f(y) + x) = f(x^2) + f(y^2) + 2f(xy) (1), \forall x, y \in \mathbb{R}$ Let $x = y = 0,$ we have $(f(0))^2 = 4f(0)$. Then $f(0) = 0$ or $f(0) = 4$ Case 1: $f(0) = 0$ Let $y = 0,$ we have $xf(x) = f(x^2), \forall x \in \mathbb{R}$ Then $(1)$ becomes $(f(x) + y)(f(y) + x) = xf(x) + yf(y) + 2f(xy), \forall x, y \in \mathbb{R}$ Hence $f(x)f(y) + xy = 2f(xy), \forall x, y \in \mathbb{R}$ In here, let $y = 1,$ we have $f(x)f(1) + x = 2f(x)$ If $f(1) = 2,$ let $x = y = 1$ in $(1),$ we have $9 = 8,$ which is contradiction So $f(1) \ne 2,$ this leads to $f(x) = \dfrac{x}{2 - f(1)}, \forall x \in \mathbb{R}$ or $f(x) = ax, \forall x \in \mathbb{R}, a \in \mathbb{R}$ Case 2: $f(0) = 4$ Let $y = 0,$ we have $(x + 4)f(x) = f(x^2) + 12, \forall x \in \mathbb{R}$ Then $(1)$ becomes $(f(x) + y)(f(y) + x) = (x + 4)f(x) + (y + 4)f(y) - 24 + 2f(xy), \forall x, y \in \mathbb{R}$ Hence $f(x)f(y) + xy = 4f(x) + 4f(y) - 24 + 2f(xy), \forall x, y \in \mathbb{R}$ In here, let $y = 1,$ we have $f(x)f(1) + x = 4f(x) + 4f(1) - 24 + 2f(x)$ If $f(1) = 6,$ let $x = y = 1$ in $(1),$ we have $49 = 24,$ which is contradiction So $f(1) \ne 6,$ this leads to $f(x) = \dfrac{x + 24 - 4f(1)}{6 - f(1)}, \forall x \in \mathbb{R}$ or $f(x) = ax + b, \forall x \in \mathbb{R}, a, b \in \mathbb{R}$ But $f(0) = 4$ then $b = 4$ or $f(x) = ax + 4, \forall x \in \mathbb{R}, a \in \mathbb{R}$ With $f(x) = ax, \forall x \in \mathbb{R}, a \in \mathbb{R},$ $(1)$ becomes $(ax + y)(ay + x) = ax^2 + ay^2 + 2axy$ From this, we see that $a = 1$ or $f(x) = x, \forall x \in \mathbb{R}$ With $f(x) = ax + 4, \forall x \in \mathbb{R}, a \in \mathbb{R},$ $(1)$ becomes $(ax + 4 + y)(ay + 4 + x) = ax^2 + ay^2 + 2axy + 16$ From this, we see that no such $a \in \mathbb{R}$ satisfies In conclusion, we have $f(x) = x, \forall x \in \mathbb{R}$