abeker wrote: Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying $$f(x^2 + f(x)f(y)) = xf(x + y)$$for all real numbers $x$ and $y$. The only constant solution is $\boxed{\text{S1 : }f(x)=0\quad\forall x}$. So let us from now look only for nonconstant solutions. Let $P(x,y)$ be the assertion $f(x^2+f(x)f(y))=xf(x+y)$ If $\exists u$ such that $f(u)=0$ $P(u,x-u)$ $\implies$ $f(u^2)=uf(x)$ And so $u=0$, else $f(x)=\frac{f(u^2)}u$ is constant. And since $P(0,0)$ implies $f(f(0)^2)=0$, we get that $f(x)=0$ $\iff$ $x=0$ Comparing $P(x,0)$ with $P(-x,0)$, we get $f(-x)=-f(x)$ $\forall x$ $P(x,-x)$ $\implies$ $f(x^2-f(x)^2)=0$ and so $f(x)=\pm x$ $\forall x$ Suppose now that $\exists x,y$ such that $f(x)=x$ and $f(y)=-y$ $P(x,y)$ $\implies$ $x^2-xy=\pm(x^2+xy)$ which implies $xy=0$ and so Either $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ which indeed is a solution Either $\boxed{\text{S3 : }f(x)=-x\quad\forall x}$ which indeed is a solution

Let $P(x,y)$ be the assertion $f(x^2+f(x)f(y))=xf(x+y)$. $P(0,0) \Rightarrow f(f(0)^2)=0$ $P(0,f(0)^2) \Rightarrow f(0)=0$ Now, let $u,v$ satisfy $f(u)=f(v)$ and $v\neq 0$. We have $$-vf(-v+u)=f(v^2+f(-v)f(u))=f(v^2+f(-v)f(v))=-vf(-v+v) \Rightarrow f(u-v)=f(0)=0$$Note that $f(x)=0\quad\forall x$ is a solution. Assume there exists an $a$, such that $f(a) \neq 0$. We get $P(u-v,0) \Rightarrow f((u-v)^2)=(u-v)f(u-v)=0$ $P(u-v,a-(u-v)) \Rightarrow 0=(u-v)f(a) \Rightarrow u=v$ Thus, $f$ is injective. $P(1,y) \Rightarrow f(1+f(1)f(y))=f(1+y) \Rightarrow f(y)=\frac{y}{f(1)}$ From here we easily get the other two solutions.

For $x=0$, we get that $f$ has a zero point. Let $f(c)=0$. Plugging in $x=0, y=c$, we get $f(0)=0$. Now for $y=0$, for all $x$ we have $f(x^2)=xf(x)$, and by changing $x$ by $-x$ we get that $f$ is odd. 1) There exists t different from $0$ s. t. $f(t)=0$. For $x=t$, for all $y$, we have $f(t^2)=t*f(y+t)=t*f(t)=0$. So for all $y, f(y+t)=0$, but $y+t$ can get every real value while $y$ varying, so $f$ is a zero function. 2) $f(x)=0 => x=0$: Take $x,-x$: $f(x^2+f(x)f(-x))=0$, so $x^2=-f(x)*f(-x)=f(x)^2$, so for all $x, f(x)$ is either $x$ or $-x$. We can easily check that $f(x)=x, f(y)= -y$ implies $xy=0$. So $f$ is zero, fixed or $f(x)=-x$ for all $x$. These three functions satisfy the condition.

How can I enter submitter information on the problems?

pco wrote: abeker wrote: Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying $$f(x^2 + f(x)f(y)) = xf(x + y)$$for all real numbers $x$ and $y$. The only constant solution is $\boxed{\text{S1 : }f(x)=0\quad\forall x}$. So let us from now look only for nonconstant solutions. Let $P(x,y)$ be the assertion $f(x^2+f(x)f(y))=xf(x+y)$ If $\exists u$ such that $f(u)=0$ $P(u,x-u)$ $\implies$ $f(u^2)=uf(x)$ And so $u=0$, else $f(x)=\frac{f(u^2)}u$ is constant. And since $P(0,0)$ implies $f(f(0)^2)=0$, we get that $f(x)=0$ $\iff$ $x=0$ Comparing $P(x,0)$ with $P(-x,0)$, we get $f(-x)=-f(x)$ $\forall x$ $P(x,-x)$ $\implies$ $f(x^2-f(x)^2)=0$ and so $f(x)=\pm x$ $\forall x$ Suppose now that $\exists x,y$ such that $f(x)=x$ and $f(y)=-y$ $P(x,y)$ $\implies$ $x^2-xy=\pm(x^2+xy)$ which implies $xy=0$ and so Either $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ which indeed is a solution Either $\boxed{\text{S3 : }f(x)=-x\quad\forall x}$ which indeed is a solution I didn't understand the passage from $f(x^2-f(x)^2)=0$ to $f(x)=\pm x$

NathalieShwarz wrote: I didn't understand the passage from $f(x^2-f(x)^2)=0$ to $f(x)=\pm x$ I first proved that $f(u)=0$ implies $u=0$ I then proved that $f(x^2-f(x)^2)=0$ So ...

But I think that we need to prove the injectivity ?

NathalieShwarz wrote: But I think that we need to prove the injectivity ? Have you just read me ? : (1) I proved first that $f(u)=0$ implies $u=0$ Do you understand this phrase ? Do you agree the proof ? If so : (2) I proved then that $f(x^2-f(x)^2)=0$ $\forall x$ Do you understand this phrase ? Do you agree the proof ? If so: Let $x\in\mathbb R$ (3) Let $u=x^2-f(x)^2$ Then, using (2) above : $f(u)=0$ Then, using (1) above : $u=0$ Then, using (3) above : $f(x)=\pm x$ Is it OK now ? And no need for injectivity.

pco wrote: NathalieShwarz wrote: But I think that we need to prove the injectivity ? Have you just read me ? : (1) I proved first that $f(u)=0$ implies $u=0$ Do you understand this phrase ? Do you agree the proof ? If so : (2) I proved then that $f(x^2-f(x)^2)=0$ $\forall x$ Do you understand this phrase ? Do you agree the proof ? If so: Let $x\in\mathbb R$ (3) Let $u=x^2-f(x)^2$ Then, using (2) above : $f(u)=0$ Then, using (1) above : $u=0$ Then, using (3) above : $f(x)=\pm x$ Is it OK now ? And no need for injectivity. hhhhh thank you, I'm kind of stupid

Denote $P(x,y)$ as the assertion $f(x^2+f(x)f(y)) = xf(x+y)$ and assume that such function $f$ exists. $f \equiv 0$ is a solution, and it is the only constant solution. Let's look at nonconstant solutions now. Since $f$ is nonconstant, we may take $\alpha$ such that $f(\alpha)$ is nonzero. We will use this to show that $f$ is surjective. For any $T \in \mathbb{R}$, take $P \left( \frac{T}{f(\alpha)} , \alpha - \frac{T}{f(\alpha)}. \right)$. Next, we will show that $f$ is injective. First, we will show that if $f(u)=0$ if and only if $u=0$. First, take $P(0,y)$ to have $f(f(0)f(y))=0$ for all $y$. If $f(0) \neq 0$, by the fact that $f$ is surjective we get $f(x) \equiv 0$ for all $x$. This is an obvious contradiction, so $f(0)=0$. Now if $f(u)=0$ and $u \neq 0$, by $P(u,y)$ we have $f(u^2) = uf(u+y)$, so again, contradiction to surjective condition. We now have $f(u)=0 \iff u=0$. To improve this to injectivity, assume $f(u)=f(v)$ and $u \neq v$. Then by $P(x,u)$ and $P(x,v)$, it is easy to note that $f$ is a periodic function with period $|u-v|$. Since $f(|u-v|)$ is nonzero, this cannot hold. Therefore, $f$ is injective. Now $P(x,-x)$ gives $f(x^2+f(x)f(-x))=0$ and $P(x,0)$ gives $f(x^2)=xf(x)$. So by injectivity we have $x^2+f(x)f(-x)=0$, and $-f(-x)=f(x)$. This gives us that $f(x)$ is equal to either $x$ or $-x$. Now we take care of the "point-wise trap". Assume nonzero reals $u, v$ exist such that $f(u)=u$ and $f(v)=-v$. $P(u,v)$ gives the desired contradiction. So in the end we have three solutions, $f \equiv 0$, $f \equiv x$, and $f \equiv -x$. These clearly work, done.

Denote $P(x,y)$ the assertion of the question.Observe that $f\equiv 0$ is the only constant solution of $f$.Now assume that $f$ is non-constant. $P(0,0)\implies f(f(0)^2)=0$.So assume for some $t_0, f(t_0)=0$. $P(t_0,x-t_0)\implies f(t_0^2)=t_0f(x)$.If $t_0\ne 0$ then $f(x)=\frac{f(t_0^2)}{t_0}=\text {constant}$.Which is a contradiction. Hence we get $f(x)=0\iff x=0$. Now we claim that $f$ is injective function.Suppose for some $a,b\in \mathbb R,f(a)=f(b)$. Then $P(x,a)$ and $P(x,b)$ implies, $f(x^2+f(x)f(a))=f(x^2+f(x)f(b))\\ \iff xf(x+a)=xf(x+b)\\ \iff f(x+a)=f(x+b)$. Putting, $x=-a$ in the last equation we get $f(b-a)=0\iff b=a$. Now, $P(1,y)$ implies $f(1+f(1)f(y))=f(1+y)\iff f(y)=cy$ where $c=\frac{1}{f(1)}$. Putting it in the main equation we get $c=1$ or $c=-1$. Hence (1)$f\equiv 0$, (2)$f(x)=x\forall x\in \mathbb R$ and (3)$f(x)=-x\forall x\in \mathbb R$ are all the solutions.$\blacksquare$

abeker wrote: Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying $$f(x^2 + f(x)f(y)) = xf(x + y)$$for all real numbers $x$ and $y$. Assume that $f(x)=c$ for $c$ any constant. $$c=cx \; \forall x \in \mathbb R \implies c=0 \implies f(x) \equiv 0 \; \forall x \in \mathbb R$$Now take that $f$ is not constant. Let $P(x,y)$ the assertion of the given FE. $P(0,x)$ $$f(f(x)f(0))=0 \; \forall x \in \mathbb R \implies f(0)=0$$$P(x,0)$ $$f(x^2)=xf(x) \; \forall x \in \mathbb R$$$P(x,-x)$ $$f(x^2+f(x)f(-x))=0 \; \forall x \in \mathbb R$$Now assume that exists more than 1 cero no the function. Let $\mathbb C$ the set of the ceros of $f$ and take $c \in \mathbb C$ then $f(c)=0$. $P(x,c)$ $$f(x^2)=xf(x+c) \; \forall x \in \mathbb R$$Then $f$ is periodic at $c$. Note that $c^2$ is also a cero by $f(x^2)=xf(x)$. Then: $$cf(2c)=0 \implies c=0 \; \text{or} \; f(2c)=0$$Now take $f(2c)=0$ then $f(4c^2)=0$ In fact if we skip the case when $c=0$ then $f(n \cdot c)=0 \; \forall n \in \mathbb Z$ $P(-x,0)$ $$f(x^2)=xf(x)=-xf(-x) \implies f(-x)=-f(x)$$Assume that exists $a,b \ne n \cdot c$ such that $f(a)=f(b)$ then i will prove that $a=b$. $P(a,-b)+P(-b,a)$ $$f(a^2-f(a)^2)+f(b^2-f(b)^2)=(a-b)f(a-b) \implies a=b$$Then $f$ is injective. That means exists an only $0$ and hence $c=0$ (contradiction!!) (We assumed that exists more than 1 cero, thats why contradiction) Hence $f(0)=0$ has an unique $0$ The same proof for get $f$ injective. Then: $$x^2-f(x)^2=0 \implies f(x)= \pm x$$Then the solutions are: $\boxed{f(x)=x \; \forall x \in \mathbb R}$ $\boxed{f(x)=-x \; \forall x \in \mathbb R}$ $\boxed{f(x) \equiv 0 \; \forall x \in \mathbb R}$ Thus we are done

$\boxed{f(x)=0}$ works, assume now that $\exists j:f(j)\ne0$. Let $P(x,y)$ denote the given assertion. $P(0,x)\Rightarrow f(f(0)f(x))=0$ $P(0,f(0)f(x))\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x^2)=xf(x)\Rightarrow f$ is odd If $\exists k:f(k)=0$: $P(k,j-k)\Rightarrow kf(j)=f(k^2)=kf(k)=0\Rightarrow k=0$ $P(x,-x)\Rightarrow f(x^2-f(x)^2)=0\Rightarrow f(x)^2=x^2$ The assertion becomes $f(x^2+f(x)f(y))^2=x^2f(x+y)^2$, or $2x^2f(x)f(y)=2x^3y$. Then $f(x)f(y)=xy\forall x\ne0$, but since it also holds for $x=0$, we set $y=1$ to get that $f(x)=cx$ for some constant $c$. Testing, we have that either $\boxed{f(x)=x}$ or $\boxed{f(x)=-x}$.

Note that $\boxed{f(x) = 0 \ \forall x \in \mathbb{R}}$ is the only constant solution. Now we look for non-constant solutions. Let $P(x,y)$ be the assertion. If there exist a $u\neq 0$ such that $f(u) = 0$, then $P(u,y-u) \implies \frac{f(u^2)}{u} = f(y)$ and so $f$ is constant, contradiction. $P(0,x) \implies f(f(0)f(x)) = 0$ so $f(0)f(x) = 0 \implies f(0) = 0$. $P(x,0) \implies f(x^2) = xf(x) \implies f$ is odd. $P(x,-x) \implies f(x^2 - f(x)^2) = 0 \implies f(x)^2 = x^2$ so $f(x) \in \{x,-x\}$ for all $x\in \mathbb{R}$. So we have $\boxed{f(x) = x \ \forall x\in \mathbb{R}}$ and $\boxed{f(x) = -x \ \forall x\in \mathbb{R}}$ as solutions. Let for some $a , b \neq 0$ we have $f(a) = a$ and $f(b) = -b$. $P(a,b) \implies f(a^2 - ab) = af(a+b) \in \{a^2 + ab , -a^2 - ab\} \implies ab = 0$, contradiction. So we have our three solutions.

I have a much simpler solution but I don't know that it is correct or not, kindly check. Substituting $x=0$ in the original equation gives $f(f(0)f(y))=0$ $f(0)f(y)$ can be any real number (except if either $f(0)=0$ or $f(y)=0$) and it is not possible that $f(R)=0$ (R represents every real number/$f(0)f(y)$) and so the above equation is true if and only if either $f(0)=0$ or $f(y)=0$. Case 1 - $f(x)=0$, it is indeed a solution. Case 2 - $f(0)=0$ $P(x,-x) - x^2+f(x)f(-x)=0$ $f(x^2)=xf(x)=-xf(-x)$ which gives $f(-x)=-f(x)$ Combining, we get $f(x)=x$ and $f(x)=-x$

Quote: $f(0)f(y)$ can be any real number What if $f(x)=x^2$ or $f(x)=\operatorname{sgn}(x)$?

jasperE3 wrote: Quote: $f(0)f(y)$ can be any real number What if $f(x)=x^2$ or $f(x)=\operatorname{sgn}(x)$? Nice! Thank you for checking.

What if I write finitely many, will my solution become wrong?

For $f(x)=x^2+1$, there are infinitely many possible values for $f(0)f(y)$ and in the case of $f(x)=1+(-1)^{|x|+2}$, there are finitely many (both cases have $f(0)\ne0$ and $f\not\equiv0$).

I denote by $P(x, y)$ plugging in some $x$ and $y$ into the given equation. Firstly, suppose that there exists a real number $a$ such that $f(a) \ne 0$, otherwise $f(x) = 0$ for all $x \in \mathbb{R}$ $P(\frac{x}{f(a)}, a-\frac{x}{f(a)}):f(...) = x$ and thus f is bijective. $P(1, y): f(1 + f(1)f(y)) = f(1+y)$ since f is bijective and therefore injective, we know: $1+f(1)f(y) = 1+y$ Case 1: $f(1) = 0$ then we have $1 + 0 = 1 + y \implies y = 0$ but that does not hold true for all $y \in \mathbb{R}$ so $f(1) \ne 0$ $1 + f(1)f(y) = 1+y \implies f(y) = \frac{y}{f(1)} = cy$ for some $c \ne 0$ Plugging back in the original equation, we get $c=1$ or $c=-1$ So... $f(x) = 0$, $f(x) = x$, $f(x)=-x$ are all solutions.

Consider $P(x,-x)$ and $P(-x,x)$, this gives $f(x^2 + f(x)f(-x)) = xf(0)= -xf(0)$, thus $f(0) = 0$. Then $P(x,0)$ gives $f(x^2) = xf(x)$, and $f(x) = -f(-x)$. Now consider $a$ with $f(a) = 0$. $P(a,y)$ gives $f(a^2 + f(a)f(y)) = af(a + y)$. The left hand side evaluates to $f(a^2 + f(a)f(y)) = f(a^2) = af(a) =0$. Thus we are either forced $a = 0$ or $f$ is all zero. Assume the former. Now we show $f$ injective, consider $a,b$ with $a < b$, $f(a) = f(b)$, then $P(x,a), P(x,b)$ forces $xf(x + a) = xf(x+b)$, so either $f(x + a) = f(x + b)$ or $x = 0$(this second case doesn't matter because $f(a) = f(b)$ by definition). Thus we can conclude $f(x + a) = f(x + b)$ for all $x$. Then consider $x = -a$, this gives $f(0)= f(b)$, which is a contradiction. Thus $f$ is injective. Then take $P(1,a)$ to get $f(1)f(a) = a$, so $f$ is linear going thru the origin, testing functions we get $f(x) = x,-x$, along with our original solution of $f = 0$

First assume that there exists $u\ne 0$ such that $f(u)=0$.$P(u,0)$ implies $f(u^2)=0$.So either 1.$f(x)=0$ for all $x$ or 2.$u=0$.So we proved that $f(0)=0$ and $f$ is injective at $0$(since $f(u^2)=uf(u+y)$ and $f(u^2)=0$ at $y=0$ we can say this).$P(x,0)$ implies $f(x^2)=xf(x)$ so $f$ is odd,and $P(x,-x)$ gives $f(x^2+f(x)f(-x))=0$ $\implies$ $f(x)=x$ or $f(x)=-x$.Plug in $f(x)=x,f(y)=-y$ and this will cause a contradiction so $f(x)=0$;or $f(x)=x$; or $f(x)=-x$ for all $x$.

We claim the only solutions are $f(x) \equiv 0 \forall x$, $f(x) = x \forall x$ and $f(x) = -x \forall x$. Clearly all of these functions work. Now we prove these are the only such functions. Let the above assertion be represented by $P(x, y)$. Clearly the only constant solution is $f(x) \equiv 0$. Now assume $f$ to be nonconstant. Claim 1: $\boxed{f(0) = 0}$ Proof: Note that $P(0,0)$ gives us $f(f(0)^2) = 0$. Now $P(0, f(0)^2)$ gives us $f(0) = 0 \square$. Claim 2: $\boxed{f \text{ is odd}}$ Proof: Observe that $P(x, 0)$ gives us $f(x^2) = xf(x) = -xf(-x) \implies f(-x) = -f(x) \square$ Claim 3: $\boxed{\text{ There exists no nonzero } a \text{ such that } f(a) = 0}$ Proof: Assume FtSoC that such $a$ exists. Then $P(a, y-a)$ gives us $f(a^2) = af(y) \implies f(y) \equiv \frac{f(a^2)}{a}$. However we have assumed $f$ to be nonconstant. Contradiction. $\square$ Claim 4: $\boxed{f(x) = x \text{ or } -x \forall x}$ Proof: $P(x, -x)$ yields $f(x^2 + f(x)f(-x)) = 0 \implies x^2 + f(x)f(-x) = x^2 - f(x)^2 = 0 \implies f(x) = x \text{ or } -x \forall x \square$ Claim 5: $\boxed{ f(x) = x \forall x \text{ or } f(x) = -x \forall x}$ Proof: Assume FtSoC that for some nonzero $m$, $n$ we have $f(m) = m, f(n) = -n$. Then $f(m^2 - mn) = mf(m+n)$. If $f(m^2 - mn) = m^2 - mn, f(m+n) = m-n, \text{else} f(m+n) = n-m$, both of which lead to a contradiction. $\square$ Therefore we are done. Q. E. D.

Nice one Answer: $f \equiv 0 , f(x)=\pm x \forall x \in \mathbb{R}$ It's easy to see that these work hence we move on. Solution: Let $P(x,y)$-denote the given assertion. Note that the only constant solution is $f \equiv 0$ hence assume $f$-is not constant Claim: $f(0)=0$ Proof: $P(0,0) \implies f(f(0)^2)=0 ...(1)$ $P(0,f(0)^2) \implies f(0)=0$ $\square$ Claim: $f(x^2)=xf(x)$ Proof: $P(x,0) \implies f(x^2)=xf(x) \forall x \in \mathbb{R} ...(*)$ $\square$ Claim: $f-\text{odd}$ Proof: From $(*) \implies f(x^2)=xf(x) , x \rightarrow -x \implies xf(x) \stackrel{(*)}{=} f(x)^2=-xf(x) \implies f(x)=-f(-x) \forall x \in \mathbb{R} / \{0\}.$ Combining with $f(0)=0 \implies f(x)=-f(x) \forall x \in \mathbb{R} \square$. Claim: $\exists \alpha \in \mathbb{R} : f(\alpha)=0 \implies \alpha=0$ or $f-\text{injective on} 0$ Proof: FTSOC assume $\alpha \neq 0$ From $(*) \implies f(x^2)=xf(x) , x \rightarrow \alpha \implies f(\alpha^2)=\alpha \cdot f(\alpha)=0 \implies f(\alpha^2)=0$ $P(\alpha,1) \implies \alpha (\alpha+1)=0 \implies \alpha=0 \vee f(\alpha+1)=0$. Since we assumed $\alpha \neq 0 \implies f(\alpha+1)=0$ $P(1,\alpha) \implies f(1)=0$ $P(1,x) \implies f(f(x)f(1)+1)=f(x+1) \implies f(1)=f(x+1) \implies f(x+1)=0 , x \rightarrow x-1 \implies f(x)=0 \iff f(x) \equiv 0$. which is a contradition since we assumed $f$-is not constant. Hence our assumption is wrong so $\alpha=0$ $\square$ Claim: $f(x)=\pm x$ Proof: $P(x,-x) \implies f(x^2+f(x)f(-x))=0=f(0) \implies f(x^2+f(x)f(-x)=f(0)$ combining with our previous claim we have: $x^2+f(x)f(-x)=0 \implies f(x)f(-x)=-x^2 \stackrel{f-\text{odd}}{\implies} -f(x)f(x)=-x^2 \implies f(x)^2=x^2 \implies f(x)=\pm x$ DON'T FORGET POINTWISE TRAP