Ankoganit wrote: Determine all functions $f : \mathbb R \to \mathbb R$ such that $$f(f(x)- f(y)) = f(f(x)) - 2x^2f(y) + f\left(y^2\right),$$for all reals $x, y$. Posted (and solved) many many times. Dont hesitate to use the search function (see here ). Set (for example copy/paste) in the "search term" field the exact following string : +"f(f(x)-f(y))" +"2x^2f(y)" You'll get in the five first results (excluded your own post and this post itself) all the help you are requesting for.

Japanese MO finals 2011

The solution I submitted: $\textbf{Answer:}$ 1. $f(x)=x^2 \ \forall x \in \mathbb{R}$ 2. $f(x)=-x^2 \ \forall x \in \mathbb{R}$ 3. $f(x)=x^2-1 \ \forall x \in \mathbb{R}$ 4. $f(x)=1-x^2 \ \forall x \in \mathbb{R}$ 5. $f(x) \equiv 0 \ \forall x \in \mathbb{R}$ $\textbf{Solution:}$ We begin by noting that an easy check shows that these are indeed solutions. Now we proceed to prove that these are the only ones. Let $P(x,y)$ denote the statement $f(f(x)-f(y))=f(f(x))-2x^2f(y)+f(y^2)$. Let $f(0)=c$. Suppose $\exists x_1, x_2$ such that $f(x_1)=f(x_2)$ and $x_1^2 \neq x_2^2$. Then $P(x_1,x)$ and $P(x_2,x)$ for arbitrary $x$ and subtracting, along with the assumption that $x_1^2 \neq x_2^2$ yields that $f(x) \equiv 0$ for all real $x$. Now we assume that $\forall x_1, x_2$ such that $f(x_1)=f(x_2)$, $x_1^2=x_2^2$. $P(0,0) \Longrightarrow f(f(0))=0$ $P(0,x) \Longrightarrow f(c-f(x))=f(x^2) \Longrightarrow (c-f(x))^2=x^4 \Longrightarrow f(x)= c+x^2$ or $c-x^2$. $P(x,0) \Longrightarrow f(f(x)-c)=f(f(x))-2x^2c+c$. $P(x,x) \Longrightarrow c=f(f(x))-2x^2f(x)+f(x^2) $. Let $f(x)=c\pm x^2$, with regards to sign. Subtracting the last 2 equations gives us that $c-f(\pm x^2)=\mp 2x^4+f(x^2)-c$ $\Longrightarrow c-c \mp_1 x^4 = \mp 2x^4 \pm_2 x^4$, where the $\pm$ signs with subscripts denote the coefficients of $x^4$ in $f(\pm x^2), f(x^2)$ respectively, which gives us that unless x=0, coefficient of $x^4$ in $f(\pm x^2)$ is the same as the coefficient of $x^4$ in $f(x^2)$ and the coefficient of $x^2$ in $f(x)$, and since any power of $0$ is $0$, this holds true for $x=0$ as well. Now we use this in $P(x,0)$ to get that $f(\pm x^2)=f(c\pm x^2)-2x^2c+c \Longrightarrow \pm x^4=c\pm_3 (c\pm x^2)^2-2x^2c+c \Longrightarrow \pm x^4= c \pm_3 c^2 \pm_3 x^4 \pm_3 2(\pm x^2)c-2x^2c$. If $\pm_3$ is opposite to $\pm$, we get a quartic equation with some plus or minus signs, which has a finite number of solutions in $\mathbb{R}$, so for all other (infinite number of) $x$, $\pm_3$ is same as $\pm$. Call all such numbers with the signs $\pm_3 = \pm$ good and all others bad. For the good numbers, by plugging in the values of the signs $\pm_3 = \pm$ in $P(x,0)$, we get that $\pm x^4= c\pm c^2 + 2x^2c - 2x^2c$, which shows that $c\pm c^2 = 0$, yielding that $c=0, 1$ or $-1$. We handle the case $c= \pm1$ first. Case 1: $c=1$ In this case note that since $c+c^2 \neq 0, c-c^2=0$, the good numbers all have the coefficient of $x^2$ in $f(x)$ as $-$ because we used the sign $\pm$ for coefficient of $x^2$ in $f(x)$, and thus $f(x) = 1-x^2$ for all the good numbers. For the bad numbers, the equation becomes (with $\pm_3$ opposite to $\pm$, the latter of which is now $-$) $-2x^4 +4x^2-1-1 = 0 \Longrightarrow x^2-1=0 \Longrightarrow x=\pm1$. Since $f(1-(\pm1)^2)=1+(1-(\pm1)^2) = 1- (1-(\pm1)^2)$, the possibility of $\pm_3$ being either + or - does not get ruled out and hence $\pm$ can be either of + or - for $\pm 1$. If $\pm$ is - for both -1 and +1, there is no problem so we focus on the case where $\pm$ is + for any $s \in \{1, -1\}$. This would imply that $f(s)=2$, and $P(t,s), t\neq0,1,-1 \Longrightarrow f(1)=0 $, which shows that $s=-1$, so we just need to check the case when $f(-1)=2, f(1)=0$. To this end, $P(-1,0)$ gives that f(2) is 1, which is a contradiction. So the only solution in this case is $f(x)=1-x^2 \ \ \ \ \ \ \ \ \forall x \in \mathbb{R}$ Case 2: $c=-1$ In this case note that since $c-c^2 \neq 0, c+c^2=0$, the good numbers all have the coefficient of $x^2$ in $f(x)$ as $+$ because we used the sign $\pm$ for coefficient of $x^2$ in $f(x)$, and thus $f(x) =x^2-1$ for all the good numbers. For the bad numbers, the equation becomes (with $\pm_3$ opposite to $\pm$, the latter of which is now $+$) $2x^4 -4x^2+1+1 = 0 \Longrightarrow x^2-1=0 \Longrightarrow x=\pm1$. Since $f(1-(\pm1)^2)=1+(1-(\pm1)^2) = 1- (1-(\pm1)^2)$, the possibility of $\pm_3$ being either + or - does not get ruled out and hence $\pm$ can be either of + or - for $\pm 1$. If $\pm$ is + for both -1 and +1, there is no problem so we focus on the case where $\pm$ is - for any $s \in \{1, -1\}$. This would imply that $f(s)=-2$, and $P(t,s), t\neq0,1,-1, \Longrightarrow f(1)=0$, so it suffices to check that $f(-1)$ can't happen to be -2. To prove this we note that $P(-1,0) \Longrightarrow f(2)=-1$ which is a contradiction, so in this case, $f(x)=1-x^2 \ \ \forall x \in \mathbb{R}$ Case 3: $c=0$ Suppose that $\exists x, y$ such that $f(x)=x^2$ and $f(y)=-y^2$. We show that either of x, y is 0. Note that by our observation, $f(t)$ and $f(t^2)$ have the same sign, so $f(x^2)=x^4$ and $f(y^2)=-y^4$ a) If $f(x^2+y^2)=(x^2+y^2)^2$ $P(x,y) \Longrightarrow (x^2+y^2)^2=x^4+2x^2y^2-y^4 \Longrightarrow y=0$ b) If $f(x^2+y^2)=-(x^2+y^2)^2$ $P(x,y) \Longrightarrow -(x^2+y^2)^2=x^4+2x^2y^2-y^4 \Longrightarrow x=0$ or $x^2+2y^2=0 \Longrightarrow x=0$ So in this case we get that the signs of $f(x)$ and $f(y)$ are the same for all x, y, thus the solutions in this case are $f(x)=x^2 \ \ \ \forall x \in \mathbb{R}$ and $f(x)=-x^2 \ \ \ \forall x \in \mathbb{R}$. \ \ $ \blacksquare$ Thus we have shown that the only solutions of the functional equations are those we mentioned before, thereby completing the solution.