aloski1687 wrote: Find all the functions $f:R\to R$ such that \[f(x^2) + 4y^2f(y) = (f(x-y) + y^2)(f(x+y) + f(y))\] for every real $x,y$. Let $P(x,y)$ be the assertion $f(x^2)+4y^2f(y)=(f(x-y)+y^2)(f(x+y)+f(y))$ Subtracting $P(0,-x)$ from $P(0,x)$, we get $f(-x)=f(x)$ $\forall x$ Subtracting then $P(x,-y)$ from $P(x,y)$, we get $(f(x+y)-f(x-y))(f(y)-y^2)=0$ And so three cases : 1) $\boxed{\text{S1 : }f(x)=x^2\text{ }\forall x}$ which indeed is a solution. 2) $f(x)=x^2$ $\forall x\ne 0$ and $f0)\ne 0$, and then $P(1,0)$ $\implies$ contradiction 3) $\exists u\ne 0$ such that $f(u)\ne u^2$, then this implies $f(x+2u)=f(x)$ $\forall u$ Then subtracting $P(0,x)$ from $P(0,x+2u)$, we get $(x+u)f(x)=0$ and so $f(x)=0$ $\forall x\ne -u$ But this implies $f(u)=0$ and so $f(-u)=0$ too and so $\boxed{\text{S2 : }f(x)=0\text{ }\forall x}$ which indeed is a solution.

pco wrote: aloski1687 wrote: Find all the functions $f:R\to R$ such that \[f(x^2) + 4y^2f(y) = (f(x-y) + y^2)(f(x+y) + f(y))\]for every real $x,y$. Let $P(x,y)$ be the assertion $f(x^2)+4y^2f(y)=(f(x-y)+y^2)(f(x+y)+f(y))$ Subtracting $P(0,-x)$ from $P(0,x)$, we get $f(-x)=f(x)$ $\forall x$ Subtracting then $P(x,-y)$ from $P(x,y)$, we get $(f(x+y)-f(x-y))(f(y)-y^2)=0$ And so three cases : 1) $\boxed{\text{S1 : }f(x)=x^2\text{ }\forall x}$ which indeed is a solution. 2) $f(x)=x^2$ $\forall x\ne 0$ and $f0)\ne 0$, and then $P(1,0)$ $\implies$ contradiction 3) $\exists u\ne 0$ such that $f(u)\ne u^2$, then this implies $f(x+2u)=f(x)$ $\forall u$ Then subtracting $P(0,x)$ from $P(0,x+2u)$, we get $(x+u)f(x)=0$ and so $f(x)=0$ $\forall x\ne -u$ But this implies $f(u)=0$ and so $f(-u)=0$ too and so $\boxed{\text{S2 : }f(x)=0\text{ }\forall x}$ which indeed is a solution. PCO , I have a real concern here .. $P(0,0) \implies 2f(0)^2=f(0)$ Then , we have : $2f(0)^2-f(0)=0$ ; $f(0)[2f(0)-1]=0$ ; thus we get : $f(0)=0$ and $f(0)=\frac{1}{2}$ Does this help , or it gets us nowhere ?

JustinHerchel wrote: $P(0,0) \implies 2f(0)^2=f(0)$ Then , we have : $2f(0)^2-f(0)=0$ ; $f(0)[2f(0)-1]=0$ ; thus we get : $f(0)=0$ and $f(0)=\frac{1}{2}$ Does this help , or it gets us nowhere ? According to me, this does not help really to get the final solutions.

My solution : Let $P(x,y)$ be the assertion of this equation. $P(0,0)$ $f(0)=0, \frac{1}{2}$ 1.If $f(0)=\frac{1}{2}$ If we subtract $P(0,-x)$ from $P(0,x)$ we ll get $f(y)=f(-y)$ $\Longrightarrow$ $1+4y^2f(y)=4f(y)^2$. $P(1,0)$ gives $f(1)=0,\frac{1}{2}$ (since $f(0)=\frac{1}{2}$ ) But these two values of $f(1)$ gives us contradiction. So $f(0)=0$ Let's again subtract $P(0,-x)$ from $P(0,x)$ we ll get $f(y)=f(-y)$ again. So $f(y)(f(y)-y^2)=0$. so Solutions are $\boxed{\text{S1 : }f(x)=x^2\text{ }\forall x}$ and $\boxed{\text{S2 : }f(x)=0\text{ }\forall x}$. Murad Aghazade.

Murad.Aghazade wrote: ... So $f(y)(f(y)-y^2)=0$. so Solutions are $\boxed{\text{S1 : }f(x)=x^2\text{ }\forall x}$ and $\boxed{\text{S2 : }f(x)=0\text{ }\forall x}$. Certainly not You can only conclude : $\forall x$, either $f(x)=0$, either $f(x)=x^2$ And not, as you wrongly said : Either $f(x)=0$ $\forall x$, either $f(x)=x^2$ $\forall x$

pco wrote: Murad.Aghazade wrote: ... So $f(y)(f(y)-y^2)=0$. so Solutions are $\boxed{\text{S1 : }f(x)=x^2\text{ }\forall x}$ and $\boxed{\text{S2 : }f(x)=0\text{ }\forall x}$. Certainly not You can only conclude : $\forall x$, either $f(x)=0$, either $f(x)=x^2$ And not, as you wrongly said : Either $f(x)=0$ $\forall x$, either $f(x)=x^2$ $\forall x$ From here must I assume that there can exist $a,b$ such that $f(a)=0$ and $f(b)=b^2$. Please answer. Thanks!!!!

$f(x)=\left(\frac{x+|x|}2\right)^2$ is neither $\equiv 0$, neither $\equiv x^2$ but is such that $f(x)(f(x)-x^2)=0$ $\forall x$

pco wrote: Murad.Aghazade wrote: ... So $f(y)(f(y)-y^2)=0$. so Solutions are $\boxed{\text{S1 : }f(x)=x^2\text{ }\forall x}$ and $\boxed{\text{S2 : }f(x)=0\text{ }\forall x}$. Certainly not You can only conclude : $\forall x$, either $f(x)=0$, either $f(x)=x^2$ And not, as you wrongly said : Either $f(x)=0$ $\forall x$, either $f(x)=x^2$ $\forall x$ Is there a way to correctly solve from here? Thanks! btw, is this the pointwise trap?

@winnertakeover Normally, we will suppose that there exist a,b such that f(a)=0 and f(b)=b^2 then we will get contradiction. But I think to prove this we must have f(0)=0..

winnertakeover wrote: pco wrote: Murad.Aghazade wrote: ... So $f(y)(f(y)-y^2)=0$. so Solutions are $\boxed{\text{S1 : }f(x)=x^2\text{ }\forall x}$ and $\boxed{\text{S2 : }f(x)=0\text{ }\forall x}$. Certainly not You can only conclude : $\forall x$, either $f(x)=0$, either $f(x)=x^2$ And not, as you wrongly said : Either $f(x)=0$ $\forall x$, either $f(x)=x^2$ $\forall x$ Is there a way to correctly solve from here? Thanks! btw, is this the pointwise trap? Yes, this is exactly the dreaded pointwise trap.

My argument (a lot longer than pco's argument), but analyzing both cases $f(0)=0$ and $f(0)=1/2$, and showing that the second is absurd. Let $P(x,y)$ be the given assertion. $P(0,y)$ gives us, $$ f(0)+4y^2f(y)=(f(-y)+y^2)2f(y)\implies f(0)+2y^2f(y)=2f(y)f(-y). $$Now, arguing similarly, and considering $P(0,-y)$, we get, $f(0)+2y^2f(-y)=2f(-y)f(y)=f(0)+2y^2f(y)$. Hence, $f(y)=f(-y)$. Plugging this in what was just found, we get, $$ 2f(y)^2 = 2y^2f(y)+f(0). $$Therefore, letting $y=0$, $f(0)=0$ or $f(0)=1/2$. $a)$ We first claim that $f(0)\neq 1/2$. Suppose that $f(0)=1/2$. Taking $y=0$, we get, $$ f(x^2)=f(x)\cdot (f(x)+\frac{1}{2}) \implies f(1)=f(1)(f(1)+\frac{1}{2}). $$Now, if $f(1)=0$, then, using $2f(y)^2=2y^2f(y)+f(0)$, with $y=1$, we get a contradiction. If $f(1) \neq 0$, then, $f(1)=f(1)(f(1)+1/2)$ gives us $f(1)=1/2$. However, with this, using $2f(1)^2=2f(1)+f(0)$ one more time, we get a contradiction, as left-hand-side is $1/2$; while the right-hand-side is $3/2$. $b)$ Therefore, $f(0)=0$. Namely, $f(y)^2 = f(y)y^2 \implies f(y)=0 \text{ or } f(y)=y^2$. In the assertion above, consider $P(x,-y)$, and notice that, $P(-x,y)$ gives, $$ f(x^2)+4y^2f(y)=(f(-x-y)+y^2)(f(-x+y)+f(y))=(f(x+y)+y^2)(f(x-y)+f(y)), $$using the fact that $f(\cdot)$ is even. On the other hand, the left-hand-side above is also equal to $(f(x-y)+y^2)(f(x+y)+f(y))$, hence, $$ f(x-y)f(x+y) + f(x-y)f(y)+y^2f(x+y)+y^2f(y) = f(x+y)f(x-y) + f(x+y)f(y)+y^2f(x-y)+y^2f(y), $$which implies, $$ f(x-y)(f(y)-y^2)=f(x+y)(f(y)-y^2). $$From here, we recover our first solution. If, for every $y$, $f(y)=y^2$, the entire assertion holds. Thus, $\boxed{f(x)=x^2,\forall x\in\mathbb{R}}$ is a solution. Next, suppose that, $\exists y_0$ such that $f(y_0)\neq y_0^2$. We clearly have, $y_0\neq 0$, since, $f(0)=0$. For this choice of $y_0$, and for every $x$, we get (keeping $x$ intact, and after letting $y=y_0$ above), $f(x-x_0)=f(x+x_0)$. Now, if $f$ is identically $0$, we recover a second solution, namely, $\boxed{f(x)=0,\forall x\in\mathbb{R}}$ is also a solution. Now, assume that, $\exists t$ such that $f(t)\neq 0$. In particular, since $f(y_0)\neq y_0^2\implies f(y_0)=0$, $t\neq y_0$ and $t\neq 0$. Taking $x=t-y_0$ above, we get, $$ f(t)=f(t-2y_0)\implies t^2 = (t-2y_0)^2 \implies t=t-2y_0 \text{ or } t=2y_0-t, $$where, $f(t)$ is necessarily $t^2$ (as it is not zero), hence, $f(t-2y_0)$ cannot be $0$, and thus, should be $(t-2y_0)^2$. If $t=t-2y_0$, then we get $y_0=0$, a contradiction. If $t=2y_0-t$, we get, $t=y_0$, again, a contradiction. Hence, no further solutions exists. Therefore, the answer is $\boxed{f(x)=x^2, \forall x\in\mathbb{R}}$ and $\boxed{f(x)=0,\forall x\in\mathbb{R}}$.

@winnertakeover Substituting $y=0$, we get $f(x^2)=f(x)^2$. Suppose $x,y \ne 0$ are the zeroes of $f$. Since $f(x-y)^2+y^2 \ge y^2 > 0$, $x+y$ is also the zero of $f$. Assume that for some $a, b \ne 0$, $f(a)=0$ and $f(b)=b^2$. Then $f(a-b), f(a+b) \ne 0$ else $f((b \pm a) \mp a) = f(b) = 0$. Since $f(x)=x^2$ is a solution for $P(x,y)$, $P(a, b)$ implies $f(a^2)=a^2$ and contradicts the assumption. Therefore, $\boxed{\text{S1 : }f(x)=x^2\text{ }\forall x}$ and $\boxed{\text{S2 : }f(x)=0\text{ }\forall x}$ This is a similar technique used at the end of USAMO 2016 #4 (https://artofproblemsolving.com/community/c5h1231010p6220308)

Ohhh, I get it now. Here's my solution. Let $P(x,y)$ be the given assertion. Observe that $P(0,0)\implies f(0)=0$. $P(x,0)$ gives $f(x^2)=f(x)^2$ and $P(0,y)$ implies $f(y)(f(-y)-y^2)=0$. We claim $f$ is even. Indeed, if $f(a)=0$, then we have $0=f(a)^2=f(a^2)=f(-a)^2$, which shows $f(-a)=0=f(a)$. Otherwise, $f(a)\neq 0$, so the equation $f(a)(f(-a)-a^2)=0$ forces $f(-a)=a^2$. Similarly, $f(-a)(f(a)-a^2)=0$ implies $f(a)=a^2$. Thus, $f(a)=a^2=f(-a)$, and we see that $f$ is even in either case. Therefore $P(0,y)$ is equivalent to $f(y)(f(y)-y^2)=0$. So, $f(x)\in \{0,x^2\}$ for every point $x$. Assume there are $a,b\in\mathbb{R}^\times$, such that $f(a)=a^2$ and $f(b)=0$. Since $f$ is even, we have $f(-b)=0$. Note that $P(x,b)$ is $$f(x^2)=(f(x-b)+b^2)f(x+b)$$and $P(x,-b)$ gives $$f(x^2)=(f(x+b)+b^2)f(x-b).$$Therefore $(f(x-b)+b^2)f(x+b)=(f(x+b)+b^2)f(x-b)$, or $f(x+b)=f(x-b)$. Taking $x=a+b$ and $x=a-b$, we obtain $f(a+2b)=f(a)=f(a-2b)$. But $f(a)=a^2>0$, so we must have $f(a+2b)=(a+2b)^2$ and $f(a-2b)=(a-2b)^2$. However, this implies $(a+2b)^2=(a-2b)^2$, which gives $ab=0$, impossible for $a,b\in\mathbb{R}^\times$. Hence, the only solutions are $\boxed{f=x^2}$ and $\boxed{f=0}$.

\[f(x^2) + 4y^2f(y) = (f(x-y) + y^2)(f(x+y) + f(y))\]$P(0,0)$ gives $f(0)=2f(0)^2\iff f(0)(f(0)-\frac{1}{2})=0\iff f(0)=0,\frac{1}{2}$ $i)f(0)=0$, $P(x,0)\implies f(x^2)=f(x)^2\geq 0\implies f(x^2)\geq 0$ $P(-x,0)\implies f(x^2)=f(-x)^2$ These two give that $|f(x)|=|f(-x)|$ $P(0,y)\implies 4y^2f(y)=(f(-y)+y^2)(2f(y))=2f(y)f(-y)+2y^2f(y)$ $2y^2f(y)=2f(y)f(-y)\iff f(y)(f(-y)-y^2)=0$ Hence $f(y)=0$ or $f(-y)=y^2$ $\implies f(x)=x^2,-x^2,0$ If there exists $c\neq 0$, $f(c)=c^2$ and $f(-c)=-c^2$ $P(0,c)\implies 4c^2.c^2=(-c^2+c^2)(2c^2)=0\implies c=0$ which gives a contradiction. This gives that $f(c)=-f(-c)\iff f(c)=0$ Also this means that $f(x)=f(-x)$ By replacing $-x$ with $x$, $LHS$ doesn't change so $(f(x-y)+y^2)(f(x+y)+f(y))=(f(x+y)+y^2)(f(x-y)+f(y))$ $f(x-y)f(y)+y^2f(x+y)=f(x+y)f(y)+y^2f(x-y)$ $f(y)(f(x+y)-f(x-y))-y^2(f(x+y)-f(x-y))=0\iff (f(y)-y^2)(f(x+y)-f(x-y))=0$ $f(y)=y^2$ or $f(x+y)=f(x-y)$ If there exists some $a\neq 0, f(x+a)=f(x-a)$ then $x\rightarrow x+a$ gives that $f(x+2a)=f(x)\implies f$ is periodic. Let the period be $T$. $P(0,y)\implies 4y^2f(y)=(f(y)+y^2).2f(y)=2f(y)^2+2y^2f(y)\iff f(y)^2=y^2f(y)$ Replacing $y+T$ with $y$ doesn't change $LHS$ but changes $RHS$. Assume that $T>0$. $\implies y^2f(y)=(y+T)^2f(y+T)=(y+T)^2f(y)=y^2f(y)+2yTf(y)+T^2f(y)\implies f(y)(T+2y)=0$ $T$ must be positive so $y\geq 0 \implies f(y)=0$ Thus $f(x^2)=0$ also $|f(-x^2)|=|f(x^2)|=0$ gives that $\forall x, f(x)=0$ So $T=0$ and this gives that $\boxed{f\equiv 0}$ If there doesn't exist any $a\neq 0, f(x+a)=f(x-a),$ then $f(y)=y^2$ So $\boxed{f(x)=x^2}$ $ii)f(0)=\frac{1}{2}$ $P(0,y)\implies 4y^2f(y)+\frac{1}{2}=2f(y)f(-y)+2y^2f(y)\iff 2f(y)f(-y)=2y^2f(y)+\frac{1}{2}$ $\iff f(y)(f(-y)-y^2)=\frac{1}{4}$ Replacing $-y$ with $y$ gives $f(-y)(f(y)-y^2)=\frac{1}{4}$ So $f(y)\neq 0$ and $f(-y)f(y)-y^2f(-y)=f(-y)f(y)-y^2f(y)\iff f(y)=f(-y)$ $LHS$ stays same if we replace $-x$ with $x$. So $(f(x-y)+y^2)(f(x+y)+f(y))=(f(x+y)+y^2)(f(x-y)+f(y))$ $\iff f(x-y)f(y)+y^2f(x+y)=f(x+y)f(y)+y^2f(x-y)$ $\iff (f(x+y)-f(x-y))(y^2-f(y))=0$ This gives that $f(y)=y^2$ or $f(x+y)=f(x-y)\forall x$. If $f(x+a)=f(x-a)$ for some $a\neq 0$, taking $x=a$ gives $f(2a)=\frac{1}{2}$ $P(0,a)\implies \frac{1}{2}=f(0)=(f(a)+a^2)f(a)=f(a)^2+a^2f(a)=0$ which gives a contradiction. So $f(x)=x^2$ for $x\neq 0$. $P(x,x)\implies 5x^4=f(x^2)+4x^2f(x)=(\frac{1}{2}+x^2)(f(2x)+f(x))=(\frac{1}{2}+x^2)(5x^2)=5x^4+\frac{5}{2}x^2$ and this gives a contradiction again.