yaron235 wrote: Find all $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying (for all $x,y \in \mathbb{R}$): $f(x+y)^2 - f(2x^2) = f(y-x)f(y+x) + 2x\cdot f(y)$. Let $P(x,y)$ be the assertion $f(x+y)^2-f(2x^2)=f(y-x)f(y+x)+2xf(y)$ $P(0,0)$ $\implies$ $f(0)=0$ Comparing $P(-x,x)$ with $P(x,-x)$, we get $f(-x)=-f(x)$ $\forall x$ $P(x,0)$ $\implies$ $f(2x^2)=2f(x)^2$ $P(-x,x)$ $\implies$ $f(2x^2)=2xf(x)$ And so $f(x)(f(x)-x)=0$ And so $\forall x$; either $f(x)=0$, either $f(x)=x$ If $\exists u\ne 0$ such that $f(u)=0$, then : $P(x,u-x)$ $\implies$ $f(2x^2)=-2xf(u-x)$ $P(-x,x)$ $\implies$ $f(2x^2)=2xf(x)$ And so $f(x)+f(u-x)=0$ $\forall x\ne 0$ Which implies $f(x)=0$ (else $f(x)=x$ and so $f(u-x)=-x$, which neither $0$, neither $u-x$) And so two solutions $\boxed{\text{S1 : }f(x)=0\text{ }\forall x}$ which indeed is a solution $\boxed{\text{S2 : }f(x)=x\text{ }\forall x}$ which indeed is a solution

yaron235 wrote: Find all $f:\mathbb{R}\rightarrow \mathbb{R}$ satisfying (for all $x,y \in \mathbb{R}$): $f(x+y)^2 - f(2x^2) = f(y-x)f(y+x) + 2x\cdot f(y)$.

$\blacklozenge \color{green}{\textit{\textbf{ANS:}}}$ $f(x)=0 \quad \textrm{and} \quad f(x)=x \quad \forall x\in \mathbb{R}.$ $\spadesuit \color{blue}{\textit{\textbf{Proof:}}}$ It is easy to see that these are indeed solutions to given FE. Let $P(x,y)$ denote the given assertion, we have \[P(0,0): f(0)=0.\]$f$ is an odd function, indeed \[P(x,-x) \quad \textrm{and} \quad P(-x,x): f(-x)=-f(x)\]for all $x\ne 0$ but since $f(0)=0$, $f(-x)=-f(x) \quad \forall x\in \mathbb{R}$. Therefore, \[P(x,0): 2f(x)^2=f(2x^2)\]and since \[P(x,-x): f(2x^2)=2xf(x),\]we have $f(x)(f(x)-x)=0$. Also, \[P(x,x):f(2x)^2-f(2x^2)=2xf(x) \implies 2xf(2x)=f(2x)^2=4xf(x) \implies f(2x)=2f(x).\]This means $2f(x)^2=f(2x^2)=2f(x^2) \implies f(x)^2=f(x^2)$. Finally, let $f(a)=a, f(b)=0$ for some reals $a,b \ne 0$. We know that $f(2a^2)=2f(a)^2=2a^2, f(2b^2)=2f(b)^2=0$, so by comparing \[P(a,b): f(a+b)^2-2a^2=f(b-a)f(b+a)\]\[P(b,a): f(a+b)^2=-f(b-a)f(b+a)+2ab\]we get \[f(b-a)f(b+a)+a^2=ab.\]If $f(b+a)$ or $f(b-a)$ is equal to $0$, then $a=b$ which is a contradiction as it yields $0\ne a=f(a)=f(b)=0$. If $f(b+a)=b+a, f(b-a)=b-a$, then $b=a$ which is also a contradiction by the same previous argument. Hence, The solutions are \[\boxed{f(x)=0 \quad \textrm{and} \quad f(x)=x \quad \forall x\in \mathbb{R}}. \quad \blacksquare\]

$P(x,y)$ denotes the assertion $f(x+y)^2 - f(2x^2) = f(y-x)f(y+x) + 2xf(y)$. $P(0,0)\Rightarrow f(0)=0$ $P(-x,x)\Rightarrow f(2x^2)=2xf(x)$ $P(x,0)\Rightarrow 2f(x)^2=f(2x^2)\Rightarrow f(x)\in\{0,x\}$ Assume that $\exists a,b\ne0:f(a)=0,f(b)=b$. $P(-a,a)\Rightarrow f(2a^2)=0$ $P(-b,b)\Rightarrow f(2b^2)=2b^2$ $P(a,b)+P(b,a)\Rightarrow f(a+b)^2=ab+b^2$ If $f(a+b)=0$, then $ab+b^2=0$, so $a+b=0$. Then $P(a,b)\Rightarrow ab=0$, contradiction. If $f(a+b)=a+b$, then $a^2+ab=0$, so $a+b=0$ again, and we have a contradiction. Then the only solutions are $\boxed{f(x)=0}$ and $\boxed{f(x)=x}$ which both work.

We claim that only $\boxed{f(x)=x}$ and $\boxed{f(x)=0} \ \forall x \in \mathbb{R}$ are the only solutions. Let $P(x,y)$ be our assertion then, $P(0,0) \rightarrow$ $$f(0)=0$$$P(-x,x) \rightarrow $ $$f(2x^2)=2xf(x) \cdots \quad{(1)}$$$P(x,-x) \rightarrow $ $$f(2x^2)=-2xf(-x) \cdots \quad{(2)}$$Comparing eqution $(1)$ and $(2)$ we get $f(x)=-f(-x)$ $P(x,0) \rightarrow $ $$f(x)^2-f(2x^2)=f(-x)f(x) \Rightarrow 2f(x^2)=f(2x^2)=2xf(x) \Rightarrow f(x)(f(x)-x)=0$$$\Rightarrow f(x)=0 \ \text{or} f(x)=x$ To resolve point wise trap Let us assume that $\exists \ a,b \ne0$ such that $f(a)=0$ and $f(b)=b$ $P(-a,a)\rightarrow$ $$f(2a^2)=0$$$P(-b,b)\rightarrow$ $$f(2b^2)=2b^2$$$P(a,b) \rightarrow$ $$f(a+b)^2-2b^2=f(b-a)f(b+a)+2ab$$$P(b,a) \rightarrow $ $$f(a+b)^2=f(b-a)f(b+a)$$Adding this two we get, $$f(a+b)^2=ab+b^2$$If $f(a+b)=0 \Rightarrow a+b=0$ which is a contradiction. If $f(a+b)=a+b \Rightarrow a+b=0$ which is again a contradiction. Hence,we are done.

For storage Denote the assertion by $P(x,y).$ Clearly $f(0)=0.$ Subtracting $P(x,0)$ from $P(-x,x)$ gives $f(x)=x,0.$ Take $f(u)=0$ and $f(v)=v$; $u,v\neq 0.$ Then $P(u-v,v)$ gives $u=0$ since $f(2(v-u)^2)\neq 0.$

$f(x+y)^2-f(2x^2)=f(x+y)f(y-x)+2xf(y)$ 1.$P(0,0)$ $f(0)=0$ 2.$P(x,0)$ $f(x)^2-f(2x^2)=f(x)f(-x)$ 3.$P(x,-x)$ $-f(2x^2)=2xf(-x)$ 4.$P(-x,x)$ $-f(2x^2)=-2xf(x)$ From $3$ and $4$ we get $f(x)=-f(-x)$ Let's use this Claim into $2$ $2f(x)^2=f(2x^2)$ From $4$ we get $f(x)(f(x)-x)=0$ Case 1: $f(x)=0$ Case 2: $f(x)=x$ But we have Point wise trap Assume $f(a)=0$ $f(b)=b$ $a,b \ne 0$ $P(a,b)$ $f(a+b)^2-f(2a^2)=f(b-a)f(a+b)+2af(b)$ which is easy contradicition.