Functional equation on the set of reals

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Source: MEMO 2017 I1

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abeker

#1 h Aug 25, 2017, 2:35 PM • 3 Y

Y by Amir Hossein, Adventure10, ItsBesi

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 $y$ .

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pco

#3 h Aug 25, 2017, 3:08 PM • 5 Y

Y by FC_YangGuifei, vsathiam, Amir Hossein, Adventure10, Mango247

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

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407 posts

DerJan

#4 h Aug 25, 2017, 4:36 PM • 2 Y

Y by Adventure10, Mango247

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.

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#5 h Aug 26, 2017, 9:43 AM • 1 Y

Y by Adventure10

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.

This post has been edited 3 times. Last edited by MilosMilicev, May 15, 2018, 1:12 PM

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#6 h Sep 1, 2017, 7:26 AM • 2 Y

Y by Adventure10, Mango247

How can I enter submitter information on the problems?

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#7 h Oct 13, 2017, 5:18 PM • 1 Y

Y by Adventure10

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 $f(x)=\pm x$

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pco

#8 h Oct 13, 2017, 5:41 PM • 2 Y

Y by Adventure10, Mango247

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 ...

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#9 h Oct 13, 2017, 6:10 PM • 1 Y

Y by Adventure10

But I think that we need to prove the injectivity ?

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pco

#10 h Oct 14, 2017, 7:56 AM • 2 Y

Y by Adventure10, Mango247

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.

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#11 h Oct 16, 2017, 7:57 PM • 3 Y

Y by mistakesinsolutions, Adventure10, Mango247

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

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#13 h Feb 19, 2018, 1:14 PM • 2 Y

Y by Adventure10, Mango247

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.

This post has been edited 3 times. Last edited by rkm0959, Feb 19, 2018, 1:15 PM

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Mathematicsislovely

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Mathematicsislovely

#14 h Sep 21, 2020, 7:08 PM • 1 Y

Y by ismayilzadei1387

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. $\blacksquare$

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#15 h May 4, 2021, 12:59 PM

Y by

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 :blush:

:blush:

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#16 h May 4, 2021, 1:19 PM • 1 Y

Y by Mango247

$\boxed{f(x)=0}$ 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 $\boxed{f(x)=-x}$ .

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hakN

#17 h May 4, 2021, 2:41 PM

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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.

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#18 h May 9, 2021, 7:02 PM

Y by

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 $f(x)=-x$

This post has been edited 1 time. Last edited by logrange, May 9, 2021, 7:03 PM

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#19 h May 9, 2021, 7:04 PM

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Quote:

$f(0)f(y)$ can be any real number

What if $f(x)=x^2$ or $f(x)=\operatorname{sgn}(x)$ ?

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#20 h May 9, 2021, 7:06 PM • 1 Y

Y by Mango247

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.

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#21 h May 9, 2021, 7:09 PM

Y by

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

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#22 h May 9, 2021, 8:41 PM • 1 Y

Y by Mango247

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 $f\not\equiv0$ ).

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llplp

#23 h Jul 27, 2021, 11:32 PM

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strange solution

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Fibonacci_11235

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Fibonacci_11235

#24 h Jun 1, 2023, 7:09 PM

Y by

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.

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ezpotd

#25 h Jun 2, 2023, 3:42 AM

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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 $f = 0$

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#26 h Feb 19, 2024, 10:22 AM

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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 $x$ .

This post has been edited 2 times. Last edited by rafigamath, Feb 26, 2024, 1:00 PM
Reason: typo

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#27 h Feb 22, 2024, 3:13 PM

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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.

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#30 h Jan 24, 2025, 10:33 AM

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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

This post has been edited 1 time. Last edited by ItsBesi, Jan 24, 2025, 11:44 AM

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#31 h Apr 29, 2025, 6:11 PM

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Let $P(x,y)$ denote the given assertion.
$f \equiv 0$ is a solution, we want to find the other solutions, so we can assume there exists $k$ such that $f(k) \neq 0$
$P(x, k - x): f(x^2 + f(x)f(k-x)) = xf(k)$ so $f$ is surjective.
$(0, y): f(f(0)f(y)) = 0$ . If $f(0) \neq 0$ , then by surjectivity there exists $y$ such that $f(y) = \frac{k}{f(0)}$ so $f(k) = 0$ , contradiction. Therefore, $f(0) = 0$ .
$P(x, 0): f(x^2) = xf(x) \quad x:= -x \to f(x^2) = xf(x) = -xf(-x) \to f(-x) = -f(x)$ .
$P(x, -x): f(x^2 + f(x)f(-x)) = 0 \to f(x^2 - f(x^2)) = 0$ .
Suppose $f(t) = 0$ . Then $P(x, k): f(x^2) = xf(x+t) = xf(x) \to \forall x \neq 0: f(x) = f(x+t), f(0) = f(t) \to \forall x: f(x) = f(x+a)$
$P(t, y): f(t^2) = tf(t+y) = tf(y)$ . Now if $t \neq 0$ then $f$ is constant, but $f$ is surjective, contradiction. As a result, $t = 0$ .
Now, $f(x^2 - f(x^2)) = 0 \to f(x^2) = x^2 \to f(x) = \pm x$
Now, as we don't fall into the pointwise trap

, we assume there exists $a, b \neq 0$ such that $f(a) = a, f(b) = -b$ .
$P(a, b) \pm(a^2 - ab) = a. \pm(a+b)$ .
We write all the four possibilities:
- $a^2 - ab = a^2 + ab \to 2ab = 0$ , contradition.
- $a^2 - ab = -a^2 - ab \to 2a^2 = 0$ , contradition.
- $-a^2 + ab = a^2 + ab \to 2a^2 = 0$ , contradition.
- $-a^2 + ab = -a^2 - ab \to 2ab = 0$ , contradition.
So, no such $a, b$ exists. This means $f(x) = -x \forall x$ or $f(x) = x \quad \forall x$ , which we can check and see that both are indeed solutions.
To sum up, we proved the only solutions are $\boxed{f(x) = 0 \quad \forall x \in \mathbb{R}}$ , $\boxed{f(x) = x \quad \forall x \in \mathbb{R}}$ , and $\boxed{f(x) = -x \quad \forall x \in \mathbb{R}}$ . $\blacksquare$

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