Does this work? Let, $P(x,y)$ be the assertion, now $$P(0,0) \implies f(0)=0,1$$Take, $f(0)=1$, now, we have, $$P(x,0) \implies x^2+2 = f(x)(f(x)+1)$$$$P(-x,0) \implies x^2+2 = f(-x)(f(-x)+1)$$So, we have, $f(x)=f(-x)$. $$P(x,-x) \implies 2x^2+2f(-x^2) =f(x)+f(-x) \implies x^2 + f(x^2) = 2f(x)$$$$P(x,x) \implies 2x^2+2f(x^2) = 2f(2x)f(x)$$So, $$2f(x) = 2f(2x) f(x) \implies f(2x)=1, f(x)=0$$Both do not work, so we have, $f(0)=0$, now, $$P(x,0) \implies x^2 = f(x)^2 \implies f(x) = \pm x$$Verifying only $f(x)=x$ works, so we are done!

very funny... $P(0,0):f(0)=f(0)(f(0)+f(0))\Rightarrow f(0)=1$ or $f(0)=0$. If $f(0)=1$: $P(x,0):x^2+2=f(x)(f(x)+1)\Rightarrow 4x^2+9=4f(x)^2+4f(x)+1=(2f(x)+1)^2 \Rightarrow f(x)=\frac{\sqrt{4x^2+9}-1}{2}$ or $f(x)=\frac{-\sqrt{4x^2+9}-1}{2}$. $P(1,-1): 2+2f(-1)=f(1)+f(-1)\Rightarrow f(1)-f(-1)=2$ Since $f(x)=\frac{\sqrt{4x^2+9}-1}{2}$ or $f(x)=\frac{-\sqrt{4x^2+9}-1}{2}$ we get that $f(1), f(-1)$ either equals $\frac{\sqrt{13}-1}2$ or $\frac{-\sqrt{13}-1}2$ which is impossible. If $f(0)=0$: $P(x,0): x^2=f(x)^2$ $P(x,-x):2x^2+2f(-x^2)=0\Rightarrow f(-x^2)=-x^2$ $\Rightarrow f(x)=x \forall x \in \mathbb{R}$.

khan.academy wrote: Verifying both of these solutions work, so we are done! no...

hectorleo123 wrote: Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for any real numbers $x$ and $y$ the following is true: $$x^2+y^2+2f(xy)=f(x+y)(f(x)+f(y))$$ $\color{blue}\boxed{\textbf{Answer:f(x)}\equiv \textbf{x}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$x^2+y^2+2f(xy)=f(x+y)(f(x)+f(y))...(\alpha)$$In $(\alpha) x=0, y=0:$ $$\Rightarrow 2f(0)=2f(0)^2$$$$\Rightarrow f(0)\in \{ 0,1\}$$$\color{red}\boxed{\textbf{If f(0)=0}}$ $\color{red}\rule{24cm}{0.3pt}$ In $(\alpha) y=-x:$ $$\Rightarrow 2x^2+2f(-x^2)=0$$$$\Rightarrow f(-x^2)=-x^2$$$$\Rightarrow f(x)=x, \forall x\in \mathbb{R}^-_0...(I)$$In $(\alpha) x,y\in\mathbb{R}^+_0, x=-x, y=-y$ $$\Rightarrow 2x^2+2f(x^2)=4x^2$$$$\Rightarrow f(x^2)=x^2$$$$\Rightarrow f(x)=x, \forall x\in\mathbb{R}^+_0...(II)$$By $(I)$ and $(II):$ $$\Rightarrow f(x)\equiv x$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If f(0)=1}}$ $\color{red}\rule{24cm}{0.3pt}$ In $(\alpha) y=0:$ $$\Rightarrow x^2+2=f(x)(f(x)+1)$$$$\Rightarrow f(x)^2+f(x)-x^2-2=0$$$$\Rightarrow f(x)=\frac{-1\pm \sqrt{4x^2+9}}{2}...(\beta)$$In $(\alpha) y=-x:$ $$\Rightarrow 2x^2+2f(-x^2)=f(x)+f(-x)$$By $(\beta):$ $$\Rightarrow 2x^2+2f(-x^2)=-1\pm\frac{\sqrt{4x^2+9}}{2}\pm\frac{\sqrt{4x^2+9}}{2}$$$$\Rightarrow f(-x^2)\in\{ -2x^2-1, -2x^2-1\pm\sqrt{4x^2+9} \}$$By $(\beta) (\Rightarrow \Leftarrow)$ $\color{red}\rule{24cm}{0.3pt}$ $\Rightarrow f(x)\equiv x_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

Let $P(x,y):=x^2+y^2+2f(xy)=f(x+y)(f(x)+f(y))$ $P(0,0)$ yields $2f(0)=2f(0)^2\Longrightarrow f(0)(f(0)-1)=0$ thus either $f(0)=0\text{ or }1$ Case 1: $f(0)=0$ $P(x,0)$ yields $f(x)^2=x^2$ thus $f(x)=\pm x$ If $f(x)=x$, plugging this in $P(x,y)$ yields $x^2+y^2+2xy=(x+y)^2$ which is clearly true. Thus $f(x)=x$ is a solution If $f(x)=-x$, plugging this in $P(x,y)$ yields $x^2+y^2-2xy=(x+y)^2$ which is a contradiction. So $f(x)=x, \forall x\in\mathbb{R}$ is the only solution in this case. Case 2: $f(0)=1$ $P(x,0)$ yields $f(x)^2+f(x)=x^2+2$ $P(-x,0)$ yields $f(-x)^2+f(-x)=x^2+2$ Thus from $P(x,0)\text{ and }P(-x,0)$ we obtain $f(x)^2+f(x)=f(-x)^2+f(-x)\Longrightarrow f(x)^2-f(-x)^2=-(f(x)-f(-x))$ Furthermore $(f(x)-f(-x))(f(x)+f(-x))=-(f(x)-f(-x))$ If $f(x)\neq f(-x)$ we can cancel out some terms which yields $f(x)+f(-x)=-1$, however notice that letting $x\to0$ yields $2f(0)=1\Longrightarrow 2=1$ which is a contradiction. So $f(x)=f(-x)$ which implies that $f$ is even $P(x,-x)$ yields $2(x^2+f(-x^2))=f(x)+f(-x)=2f(x)\Longrightarrow x^2+f(-x^2)=f(x)$ however since $f(x^2)=f(-x^2)$ the $\text{LHS}$ can be transformed into $x^2+f(x^2)$ Thus $f(x)=x^2+f(x^2)$ however notice that letting $x\to1$ in our last equation yields $f(1)=1+f(1)\Longrightarrow 1=0$ which is clearly a contradiction. Therefore there exists no solution to the functional equation when $f(0)=1$ In conclusion $\boxed{f(x)=x, \forall x\in\mathbb{R}}$ $\blacksquare$.

Let $P(x,y)$ denote the assertion $P(0,0)\implies 2f(0)=2f(0)^2\implies f(0)=0\text{ or }1$ $P(x,0)\implies x^2+2f(0)=f(x)^2+f(x)f(0)$ Case 1: $f(0)=1$ $\implies f(x)^2+f(x)-x^2-2=0$ $\implies f(x)=\frac{-1\pm\sqrt{4x^2+9}}{2}$ $P(1,1)\implies 1+1+2\cdot\left(\frac{-1\pm\sqrt{13}}{2}\right)=\left(\frac{-1\pm5}{2}\right)\cdot2\cdot\left(\frac{-1\pm\sqrt{13}}{2}\right)$ With some arithmetic we can see that this isn't true Case 2: $f(0)=0$ $\implies f(x)^2=x^2$ $\implies f(x)=\pm x$ $P(x,x)\implies x^2+f(x^2)=f(2x)f(x)$ If $f(x^2)=-x^2\implies f(2x)f(x)=0\implies x=0$ contradiction $\implies f(x^2)=x^2\implies f(a)=a$ for positive a $P(x,-x)\implies x^2+f(-x^2)=0$ $\implies f(-x^2)=-x^2\implies f(a)=a$ for negative a $\implies f(x)=x$ is the only solution

Question: how does $f(x)(f(x)+1)=f(-x)(f(-x)+1)?$ imply $f(x)=f(-x)?$ What if $f(-x)=-1-f(x)?$

@above $P(x,x)+P(-x,-x)$ $f(x)f(2x)=f(-x)f(-2x)$ if we assume $f(x) \ne f(-x)$ and $f(-x)=-1-f(x)$ so $f(-2x)=-1-f(2x)$ $f(x)f(2x)=(f(x)+1)(f(2x)+1)$ which is obviously contradicition ...

Sorry, I was talking about this part: Quote: Take, $f(0)=1$, now, we have, $$P(x,0) \implies x^2+2 = f(x)(f(x)+1)$$$$P(-x,0) \implies x^2+2 = f(-x)(f(-x)+1)$$So, we have, $f(x)=f(-x)$.

DouDragon wrote: Sorry, I was talking about this part: Quote: Take, $f(0)=1$, now, we have, $$P(x,0) \implies x^2+2 = f(x)(f(x)+1)$$$$P(-x,0) \implies x^2+2 = f(-x)(f(-x)+1)$$So, we have, $f(x)=f(-x)$. I was talking too I have been showed how to get contradicition $f(x) \ne f(-x)$ case

ismayilzadei1387 wrote: DouDragon wrote: Sorry, I was talking about this part: Quote: Take, $f(0)=1$, now, we have, $$P(x,0) \implies x^2+2 = f(x)(f(x)+1)$$$$P(-x,0) \implies x^2+2 = f(-x)(f(-x)+1)$$So, we have, $f(x)=f(-x)$. I was talking too I have been showed how to get contradicition $f(x) \ne f(-x)$ case Oh, thank you. That completes the proof