See here.

oVlad wrote: Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that all real numbers $x$ and $y$ satisfy \[f(f(x)+y)=f(x^2-y)+4f(x)y.\] Cool! Let \(P(x,y)\) denote the assertion of the given functional equation. I claim that \(f(a)=f(b)\) if and only if \(a=\pm b\) or \(f\equiv 0\) First of all, \(P(0,0)\) gives \(f(f(0))=f(0)\). Next, \(P(0,-f(0))\) gives us that \(f(0)=0\). So \(P(0,x)\) gives us \(f\) even. Now, let \(f(a)=f(b)\). Then, \(P(a,x)\) and \(P(b,x)\) give us \(f(a^2-x)=f(b^2-x)\), so if \(a\neq\pm b\), then \(f\) is periodic with period \(t\). Now \(P(x,y)\) and \(P(x,y+t)\) give us that \[f(f(x)+y+t)=f(x^2-y-t)+4f(x)(y+t)\]or \(4f(x)y=4f(x)(y+t)\) for all reals \(x,y\). Therefore, \(f\equiv 0\) or \(a=\pm b\) as claimed. Now, \(P(x,0)\) gives us that \(f(f(x))=f(x^2)\), implying that \(f(x)=x^2\) or \(f(x)=x^2\). Clearly \(f(x)=-x^2\) is not a solution, therefore \(f(x)=x^2\) for all reals \(x\) which indeed fits. In conclusion, the only solution are \(f\equiv0\) or \(f(x)=x^2\)

sarjinius wrote: See here. So this problem is unoriginal...

Let $P(x,y)$ be the assertion. Claim 1:$f(0)=0$ Proof: $P(0,0)$ gives $f(f(0))=f(0)$ $P(0,-f(0))$ and using above fact gives that $f(0)=0$.$\blacksquare$ Claim 2 : $\forall x$,$f(x)=0$ or $x^2$ Proof: $P\left(x,\frac{x^2-f(x)}{2}\right)$ gives $f(x) \left(x^2-f(x)\right) =0 $ As we can vary $x$ over all real numbers, we get that $\forall x$, $f(x)=0$ or $f(x)=x^2$. $\blacksquare$ Hence, $f(1)=0$ or $f(1)=1$. Case 1 : $f(1)=0$ We prove that $f(x)=0 \forall x$ is the only function in this case. Assume that there exists some $m$ such that $f(m) \neq 0 $ Using claim 2, we get that $f(m)=m^2$ $P(1,m)$ gives $m^2=f(1-m)$ Using claim 2, $m^2= \left(1-m \right)^2 $ Solving, we get that $m=\frac{1}{2}$ Hence, $f(m) \neq 0 \Rightarrow m=\frac{1}{2} ....(1)$ $P\left(\frac{1}{2},0\right)$ gives $f\left(\frac{1}{4}\right)=0$ contradicting (1). Hence, no such $m$ exists. Hence,$\boxed{f(x)=0 \forall x}$ which indeed works. Case 2 : $f(1)=1$ We prove that $f(x)=x^2 \forall x$ is the only function in this case. Assume that there exists some $m \neq 0$ such that $f(m) \neq m^2 $ Using claim 2, we get that $f(m)=0$ $P(m,1)$ gives $1=f(m^2-1)$. Using claim 2, we get that $1=\left(m^2-1\right)^2$ Solving we get that $m=0$ or $+-\sqrt2$. As $m \neq 0$ , we get that $f(m) \neq m^2 \Rightarrow m=+-\sqrt2 ....(2)$ But both $P(\sqrt2,0)$ and $P(-\sqrt2,0)$ give $P(2)=0$Contradicting (2). Hence, no such $m$ exists. Using claim 2, we get that $\boxed{f(x)=x^2 \forall x}$ ,which indeed works

rama1728 wrote: oVlad wrote: Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that all real numbers $x$ and $y$ satisfy \[f(f(x)+y)=f(x^2-y)+4f(x)y.\] Cool! Let \(P(x,y)\) denote the assertion of the given functional equation. I claim that \(f(a)=f(b)\) if and only if \(a=\pm b\) or \(f\equiv 0\) First of all, \(P(0,0)\) gives \(f(f(0))=f(0)\). Next, \(P(0,-f(0))\) gives us that \(f(0)=0\). So \(P(0,x)\) gives us \(f\) even. Now, let \(f(a)=f(b)\). Then, \(P(a,x)\) and \(P(b,x)\) give us \(f(a^2-x)=f(b^2-x)\), so if \(a\neq\pm b\), then \(f\) is periodic with period \(t\). Now \(P(x,y)\) and \(P(x,y+t)\) give us that \[f(f(x)+y+t)=f(x^2-y-t)+4f(x)(y+t)\]or \(4f(x)y=4f(x)(y+t)\) for all reals \(x,y\). Therefore, \(f\equiv 0\) or \(a=\pm b\) as claimed. Now, \(P(x,0)\) gives us that \(f(f(x))=f(x^2)\), implying that \(f(x)=x^2\) or \(f(x)=-x^2\). Clearly \(f(x)=-x^2\) is not a solution, therefore \(f(x)=x^2\) for all reals \(x\) which indeed fits. In conclusion, the only solution are \(f\equiv0\) or \(f(x)=x^2\) Well, aren't you missing the pointwise trap ($f(a)=a^2, f(b)=-b^2$)? Otherwise, everything else is fine.

a_507_bc wrote: rama1728 wrote: oVlad wrote: Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that all real numbers $x$ and $y$ satisfy \[f(f(x)+y)=f(x^2-y)+4f(x)y.\] Cool! Let \(P(x,y)\) denote the assertion of the given functional equation. I claim that \(f(a)=f(b)\) if and only if \(a=\pm b\) or \(f\equiv 0\) First of all, \(P(0,0)\) gives \(f(f(0))=f(0)\). Next, \(P(0,-f(0))\) gives us that \(f(0)=0\). So \(P(0,x)\) gives us \(f\) even. Now, let \(f(a)=f(b)\). Then, \(P(a,x)\) and \(P(b,x)\) give us \(f(a^2-x)=f(b^2-x)\), so if \(a\neq\pm b\), then \(f\) is periodic with period \(t\). Now \(P(x,y)\) and \(P(x,y+t)\) give us that \[f(f(x)+y+t)=f(x^2-y-t)+4f(x)(y+t)\]or \(4f(x)y=4f(x)(y+t)\) for all reals \(x,y\). Therefore, \(f\equiv 0\) or \(a=\pm b\) as claimed. Now, \(P(x,0)\) gives us that \(f(f(x))=f(x^2)\), implying that \(f(x)=x^2\) or \(f(x)=-x^2\). Clearly \(f(x)=-x^2\) is not a solution, therefore \(f(x)=x^2\) for all reals \(x\) which indeed fits. In conclusion, the only solution are \(f\equiv0\) or \(f(x)=x^2\) Well, aren't you missing the pointwise trap ($f(a)=a^2, f(b)=-b^2$)? Otherwise, everything else is fine. Oh to end this we could even make the substitution \(P(x,\frac{x^2-f(x)}{2})\) to see \(f(x)=x^2\) or \(f(x)=0\), so \(f(x)=-x^2\) isn't in the picture here. PS: \(f(x)=-x^2\) isn't even a solution anyways, but just for the sake of completion

Let $P(x,y)$ denote the given assertion. $P(0,0): f(f(0))=f(0)$. $P(0,-f(0)): f(0)=f(0)-4f(0)^2\implies f(0)=0$. $P(0,x): f(x)=f(-x)$. $P(x,0): f(f(x))=f(x^2)$. Suppose $f(a)=f(b)$. $P(a,x): f(f(a)+x)=f(a^2-x)+4f(a)x$. $P(b,x): f(f(a)+x)=f(b^2-x)+4f(a)x$. So $f(a^2-x)=f(b^2-x)$ for all reals $x$. Now suppose $a\ne \pm b$. Then $f$ is periodic, and call it's period $n$. $P(x,y+n): f(f(x)+y+n)=f(x^2-y-n)+4f(x)(y+n)\implies f(f(x)+y)=f(x^2-y)+4f(x)(y+n)$. Comparing with the original FE gives $4f(x)y=4f(x)(y+n)$, so $\boxed{f\equiv0}$, which works. Henceforth assume that $f$ is not identically zero and that if $f(a)=f(b)$, then $a=\pm b$. Since $f(f(x))=f(x^2)$, either $f(x)=-x^2$ or $f(x)=x^2$. Suppose $f(x)=-x^2$. $P(x,-x): f(-x^2-x)=f(x^2+x)+4x^3\implies 4x^3=0\implies x=0$. So $\boxed{f(x)=x^2}$, which works.

we want to make $f(f(x)+y) = f(x^2-y)$ so $P(x,\frac{x^2-f(x)}{2}) : 4f(x)(x^2-f(x)) = 0 \implies f(x) = 0$ or $f(x) = x^2$. case $1$ : $f(1) = 0$. Assume there exists $a$ such that $f(a) = a^2$. $P(1,a) : f(a) = f(1-a) \implies a^2 = (1-a)^2 \implies a = \frac{1}{2}$. $P(\frac{1}{2},0) : f(0) = f(\frac{1}{4}) + y \implies f(\frac{1}{4}) \neq 0$ which gives contradiction so no such $a$ exists so $f(x) = 0$. case $2$ : $f(1) = 1$. Let assume there exists $a$ such that $f(a) = 0$. $P(a,1) : 1 = f(a^2-1) \implies a^2-1 = \pm 1 \implies a = \sqrt{2}$ $P(\sqrt{2},0) : 0 = f(2)$ which gives contradiction so no such $a$ exists so $f(x) = x^2$. so we have the answers and the cases. we're Done.

Copied here also Solved with Ritwin. $P\left(x,\frac{f(x)-x^2}{2}\right): 4f(x)\frac{f(x)-x^2}{2}=0$. Thus, either $f(x)=0$ or $f(x)=x^2$ for each $x$. Suppose FTSOC there exist $a,b\ne 0$ with $f(a)=a$ and $f(b)=b^2$. $P(a,-b): f(a^2-b)=b^2$. Since $b\ne 0$, we have $f(a^2-b)=(a^2-b)^2$. Thus, either $a^2-b=b$ or $a^2-b=-b$. The latter gives $a=0$, a contradiction. So the former must hold true, which implies $a^2=2b$. So $a^2=2b$, if $f(a)=0$ and $f(b)=b^2$. $P(a,b): f(3b)=f(-b)$. Clearly $(3b)^2\ne (-b)^2$, so $f(3b)=f(-b)=0$. Since $f(-b)=0$, we have $(-b)^2=2b\implies b=2$. Since $f(3b)=0$, we have $(3b)^2=9b^2=2b\implies b=\frac{2}{9}$. These contradict each other. Thus, $\boxed{f\equiv 0}$ or $\boxed{f(x)=x^2}$, which both fit.

here solution

Attachments:

Denote the given assertion by $P(x,y).$ We claim that $f\equiv 0$ and $f(x)\equiv x^2$ are the only solutions. They clearly work. $P(x,\tfrac{x^2-f(x)}{2})$ implies $f(x)\in \{0,x^2\}.$ Indeed if $f(u)=u^2$ and $f(v)=0$ then $P(v,u)$ gives $f(v^2-u)=u^2\neq 0$ hence $2u=v^2$ absurd. As desired.

megarnie wrote: So $a^2=2b$, if $f(a)=0$ and $f(b)=b^2$. $P(a,b): f(3b)=f(-b)$. Is it my blindness or there's a little error here?

P(x,(x²-f(x))/2) implies f(x)×(x²-f(x))/2=0 then we get f(x)=0 or f(x)=x².We are done.