Seems daunting, but actually not that bad. Substituting $y=x^2$ yields $f(f(x)+x^2)=f(0)+4(x^2-2)(f(x)+2)$ which means $$f(0)=f(f(x)+x^2)-4(x^2-2)(f(x)+2)$$ Now substituting $y=-f(x)$ yields $f(0)=f(x^2+f(x))-4(f(x)+2)(f(x)+2)$ Now note that the LHS of both equations are the same, so the RHS must have the same value this means $$f(f(x)+x^2)-4(x^2-2)(f(x)+2)=f(x^2+f(x))-4(f(x)+2)(f(x)+2)$$ Subtracting the first term, which is the same, on both sides and then dividing by $-4$ yields $$(x^2-2)(f(x)+2)=(f(x)+2)(f(x)+2)$$ If $f(x)+2=0$, then $\boxed{f(x)=-2}$, which works if we plug it into the original equation. Otherwise, we can divide both sides by $f(x)+2$ to get $f(x)+2=x^2-2$, or $\boxed{f(x)=x^2-4}$, which also works if we plug it in $\blacksquare$

Note that you can get the conclusion $\forall x$, either $f(x)=-2$, either $f(x)=x^2-4$ much quicker : Just set $y=\frac{x^2-f(x)}2$ in the original equation. But you can not, as you wrongly did, skip from the sentence : $\forall x$, either $f(x)=-2$, either $f(x)=x^2-4$ To the sentence : Either $f(x)=-2\quad\forall x$, either $f(x)=x^2-4\quad\forall x$ You must consider the case where $f(x)=-2$ for some $x$ and $f(x)=x^2-4$ for some other ....

JustPostTaiwanTST wrote: Find all functions $ f: \mathbb{R} \to \mathbb{R} $ such that $$ f\left(f\left(x\right)+y\right) = f\left(x^2-y\right)+4\left(y-2\right)\left(f\left(x\right)+2\right) $$holds for all $ x, y \in \mathbb{R} $ Let $P(x,y)$ be the assertion $f(f(x)+y)=f(x^2-y)+4(y-2)(f(x)+2)$ $P(x,\frac{x^2-f(x)}2)$ $\implies$ $(f(x)-(x^2-4))(f(x)+2)=0$ and so : $\forall x$, either $f(x)=-2$, either $f(x)=x^2-4$ This implies $f(\sqrt 2)=f(-\sqrt 2)=-2$ If $f(u)=-2$ for some $u\notin\{-\sqrt 2,\sqrt 2\}$ : $P(\sqrt 2,2-x)$ $\implies$ $f(-x)=f(x)$ $P(u,2-x)$ $\implies$ $f(-x)=f(x+u^2-2)$ And so $f(x)=f(x+\Delta)$ $\forall x$ and for some $\Delta\ne 0$ Subtracting then $P(x,y)$ from $P(x,y+\Delta)$, we get $\boxed{\text{S1 : }f(x)=-2\quad\forall x}$ Which indeed is a solution If $f(x)\ne -2$ $\forall x\notin\{-\sqrt 2,\sqrt 2\}$ Then $f(x)=x^2-4$ $\forall x\notin\{-\sqrt 2,\sqrt 2\}$ And so $\boxed{\text{S2 : }f(x)=x^2-4\quad\forall x}$ Which indeed is a solution

pco wrote: JustPostTaiwanTST wrote: Find all functions $ f: \mathbb{R} \to \mathbb{R} $ such that $$ f\left(f\left(x\right)+y\right) = f\left(x^2-y\right)+4\left(y-2\right)\left(f\left(x\right)+2\right) $$holds for all $ x, y \in \mathbb{R} $ Let $P(x,y)$ be the assertion $f(f(x)+y)=f(x^2-y)+4(y-2)(f(x)+2)$ $P(x,\frac{x^2-f(x)}2)$ $\implies$ $(f(x)-(x^2-4))(f(x)+2)=0$ and so : $\forall x$, either $f(x)=-2$, either $f(x)=x^2-4$ This implies $f(\sqrt 2)=f(-\sqrt 2)=-2$ This implication is only a $P (a,b) $?

Al3jandro0000 wrote: This implication is only a $P (a,b) $? I dont understand your question. I'm sorry.

pco wrote: This implies $f(\sqrt 2)=f(-\sqrt 2)=-2$ This implication, is only a yield of a assertion $P (a,b) $ (basic algebra) or is another concept?

Al3jandro0000 wrote: pco wrote: This implies $f(\sqrt 2)=f(-\sqrt 2)=-2$ This implication, is only a yield of a assertion $P (a,b) $ (basic algebra) or is another concept? Oh, I understand ! We just established : $\forall x$, either $f(x)=-2$, either $f(x)=x^2-4$ Plug there $x=\sqrt 2$ and this becomes "either $f(\sqrt 2)=-2$, either $f(\sqrt 2)=(\sqrt 2)^2-4=-2$ " And same for $x=-\sqrt 2$

Here is another way to finish the problem off without any roots of $2$: As in pco's solution, considering $y=\frac{x^2-f(x)}{2}$ we immediately find that for each $x$ we either have $f(x)=-2$ or $f(x)=x^2-4$. In particular $f(0)=-2$ or $f(0)=-4$. Case 1: $f(0)=-4$. Suppose that $f(x_0) \ne x_0^2-4$ for some $x_0$. Then $f(x_0)=-2$ and $P(x_0,2)$ shows that $f(x_0^2-2)=-4$. So $x_0^2=2$ and hence $f(x_0)=x_0^2-4$. Contradiction! So in this case we must have $f(x)=x^2-4$ for all $x$ which is indeed a solution. Case 2: $f(0)=-2$. Then $P(0,-y)$ shows that $f(y)=f(-y-2)$. Suppose that $f(y) \ne -2$ for some $y$. Then $f(-y-2) \ne -2$ and hence $f(y)=y^2-4$ and $f(-y-2)=(y+2)^2-4$ so that $y^2=(y+2)^2$ and hence $y=-1$. So $f(x)=-2$ for all $x \ne -1$ and $f(-1)=-3$. But then $P(-1,42)$ is a contradiction. So in this case we must have $f(x)=-2$ for all $x$ which is indeed another solution.

Can you somehow show injectivity?

itslumi wrote: Can you somehow show injectivity? This is going to be difficult since $x^2-4$ is a solution which is not injective.

It’s nearly injective Claim: If $f(a)=f(b), a^2 \neq b^2$, then $f(x)=-2 \forall x \in \mathbb{R}$. Proof.: Assume that there exist $a,b$ such that $f(a)=f(b)$ but $a^2=b^2$. Compare the two equations, $P(a,y)$ and $P(b,y)$, we have $f(a^2-y)=f(b^2-y)$. So $f$ is periodic with period $t=a^2-b^2$. For any $x$, pick an integer $N$ such that $y=\frac{x^2-f(x)+Nt}{2} \neq 2$. Plug in $P(x,y)$ then $(y-2)(f(x)+2)=0$, therefore $f(x)=-2 \forall x \in \mathbb{R}$. As desired. Now we assume that $f$ is not a constant function. $P(x,2) \rightarrow f(f(x)+2)=f(x^2-2)$ By our claim we get $|f(x)+2|=|x^2-2|$, that is $f(x)=-x^2$ or $f(x)=x^2-4$. If $f(a)=-a^2$ $ P(a,a^2) \rightarrow f(0)=f(0)+(a^2-2)(f(a)+2)$ Both we get $a=\pm\sqrt{2}$. Conclude that $f(x)=x^2-4 \forall \mathbb{R}$. Done.

The only solutions are $\boxed{f\equiv -2} $ and $\boxed{f(x) = x^2 - 4}$. These clearly work. Let $P(x,y)$ be the given assertion. $P\left(x, \frac{x^2 - f(x)}{2}\right)$ gives that either $x^2 - f(x) = 4$ or $f(x) = -2$. So for each $x$, we have $f(x)\in \{x^2 - 4, -2\}$. Suppose there existed $a$ such that $f(a) = -2$ and $a^2 \ne 2$. We claim that $f\equiv -2$, and indeed this will prove the desired. Notice that $f\left(\sqrt{2}\right) = -2$, so \[P(\sqrt{2},x ): f(x-2) = f(2-x)\]so $f$ is even. $P(a,x): f(x-2) = f(x-a^2)$ so $f$ is periodic. Let $f$ have period $t$, where $t\ne 0$. Supposing that $f$ is not identically $-2$, comparing $P(x,y)$ and $P(x,y+t)$ for some $x$ with $f(x)\ne -2$ gives $4(y-2) = 4(y+t-2)$, which is not possible as $t\ne 0$. Done.

Solved with Taco12. I claim that the solutions are $f(x)=-2$ and $f(x)=x^2-4$, easy to check. Using $P\left(x,\frac{x^2-f(x)}{2}\right)$, we get $f(x)=x^2-4$ or $-2$ at each individual $x$. If no point besides $\pm\sqrt2$ goes to $-2$, we get $f(x)=x^2-4$ and are done. Otherwise, let that other point that goes to $-2$ be $x_1$. Now, $P(\sqrt2,x)$ gives $f(x-2)=f(2-x)$ so $f$ is even. $P(x_1,y)$ gives $$f(y-2)=f(x_1^2-y)=f(y-x_1^2)$$so $f$ is periodic mod $x_1^2-2$. If there exists $x_2$ such that $f(x_2)\ne -2$, then $P(x_2,y)$ and $P(x_2,y+x_1^2-2)$ gives a contradiction, so in this case $f$ is identically $-2$, which finishes.

Let $P(x,y)$ denote the assertion $f(f(x)+y)=f(x^2-y)+4(y-2)(f(x)+2)$. $P(x,\tfrac{x^2-f(x)}2)$ yields $f(x)\in \{x^2-4,-2\}$. Assume $f(m)=-2$ then comparing $P(m,2-x)$ and $P(\sqrt 2,2-x)$ gives $f(x)=f(x+p)$ for some $p=m^2-2$. Now comparing $P(x,0)$ and $P(x,p)$ gives $f\equiv -2$, which works, or $m=\pm \sqrt 2$ whence $f(x)\equiv x^2-4$, which works.