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Source: RMMSL-19 A1

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#1 h May 27, 2020, 9:40 AM • 5 Y

Y by son7, tiendung2006, Mango247, ItsBesi, farhad.fritl

Determine all the functions $f:\mathbb R\mapsto\mathbb R$ satisfies the equation
$f(a^2 +ab+ f(b^2))=af(b)+b^2+ f(a^2)\,\forall a,b\in\mathbb R$

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TLP.39

#2 h May 27, 2020, 10:24 AM • 2 Y

Y by son7, farhad.fritl

$P(a,b)$ and $P(-a-b,b)$ imply that $f((a+b)^2)-f(a^2)=(2a+b)f(b).$

Let $Q(a,b)$ denotes the above equation. $Q(a,-2a-b)$ implies that $f((-a-b)^2)-f(a^2)=-bf(-2a-b).$

Hence, $(-2a-b)f(b)=bf(-2a-b)$ for all $a,b\in\mathbb{R}.$ This means that $af(b)=bf(a)$ for all $a,b\in\mathbb{R}.$ In particular, $f(x)\equiv cx$ for some constant $c.$

Plugging back in the equation, we get $c\in\{-1,1\}.$ Thus, $f(x)\equiv x$ or $f(x)\equiv -x,$ both of which are solutions.

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#3 h Jun 19, 2020, 2:32 PM • 1 Y

Y by son7

a nice one
let $P(x,y)$ be the assertion
$f(a^2 +ab+ f(b^2))=af(b)+b^2+ f(a^2)$
$P(-\frac{f(b^2)}{b},b) \implies f(b^2)f(b)=b^3$ for $b \neq 0$
so $f(1)=1$ (if $f(1)=0$ then $P(0,1) \implies f(0)=f(0)+1$ contradiction)
$P(0,b) \implies f(f(b^2))=b^2+f(0)$ plugging $b=1 \implies f(0)=0$
so $f(f(b))=b \forall b \ge 0$
$P(-a,a)\implies f(a^2)=af(a)$
so $f(a)^2=a^2$
so $f(a)=a \forall a \in \mathbb{R}$ or $f(a)=-a \forall a \in \mathbb{R}$ (if we have $f(a)=a ,f(b)=-b$ for some $a,b$ it's easy to get contradiction just by $P(a,b)$ )

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Atpar

#4 h Jun 19, 2020, 2:52 PM

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Another solution

This post has been edited 2 times. Last edited by Atpar, Jun 21, 2020, 11:05 AM

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#5 h Jul 22, 2020, 5:22 PM • 1 Y

Y by AlastorMoody

This FE is very suitable for a introductory exercise

. It captures all the basic tricks .

Sketch

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#6 h Jul 22, 2020, 6:38 PM

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TLP.39 wrote:

$P(a,b)$ and $P(-a-b,b)$ imply that $f((a+b)^2)-f(a^2)=(2a+b)f(b).$

Let $Q(a,b)$ denotes the above equation. $Q(a,-2a-b)$ implies that $f((-a-b)^2)-f(a^2)=-bf(-2a-b).$

Hence, $(-2a-b)f(b)=bf(-2a-b)$ for all $a,b\in\mathbb{R}.$ This means that $af(b)=bf(a)$ for all $a,b\in\mathbb{R}.$ In particular, $f(x)\equiv cx$ for some constant $c.$

Plugging back in the equation, we get $c\in\{-1,1\}.$ Thus, $f(x)\equiv x$ or $f(x)\equiv -x,$ both of which are solutions.

General question, why do you ( and many others) write $f(x)\equiv x$ instead of $f(x)=x$ ?

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#7 h Jan 16, 2021, 4:55 PM

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The answers are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$ .
It is easy to see that they satisfy the given functional equation, we now show that these are the only solutions.
Let $P(a,b)$ denote the given assertion, we prove that $f$ is odd, let $a\ne 0$ , from $P(-a,b)$ and $P(a,-b)$ , we have $P(-a,b) : f(a^2-ab+f(b^2))=-af(b)+b^2+f(a^2)$ $P(a,-b) : f(a^2-ab+f(b^2))=af(-b)+b^2+f(a^2)$ and by comparing these two equations, we get $f(b)=-f(-b)$ for all $b\in \mathbb{R}$ . So, we can deduce that $f(0)=0$ by letting $b=0$ .
We now show that $f$ is an involution, $P(0,a) : f(f(a^2))=a^2 \ \ \textrm{and} \ \ -f(f(a^2))=f(f(-a^2))=-a^2 \ \ \forall a\in \mathbb{R}$ as desired. Since $f$ is an involution, it is also bijective.
Then, we show that $f(a^2)=af(a)$ , we have $P(a,-a) : a^2=f(f(a^2))=-af(a)+a^2+f(a^2) \implies f(a^2)=af(a) \ \ \forall a\in \mathbb{R}$ as desired. Thus, $f(f(a)^2)=f(a)f(f(a))=af(a)=f(a^2) \implies f(a)^2=a^2 \implies f(a)=\pm a$ .
Now, assume that there exist $a,b$ such that $f(a)=a$ and $f(b)=-b$ where $a,b\ne 0$ , we have $f(a^2)=af(a)=a^2$ and $f(b^2)=bf(b)=-b^2$ , therefore, $f(a^2+ab-b^2)=f(a^2+ab+f(b^2))=af(b)+b^2+f(a^2)=-ab+b^2+a^2.$ If $f(a^2+ab-b^2)=a^2+ab-b^2$ , then $a=b$ , so $-a=-b=f(b)=f(a)=a \implies a=0$ which is impossible. If $f(a^2+ab-b^2)=-a^2-ab+b^2$ , then $a=0$ which is also impossible.
Hence, $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$ are indeed the only solutions. $\blacksquare$

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#8 h Mar 2, 2021, 1:29 PM • 2 Y

Y by aritrads, centslordm

The answer is $f(x)=x$ and $f(x)=-x$ . This can be easily checked to work. Denote $P(a,b)$ the given assertion.
Let $f(0)=c$ . From $P(a,0)$ we get:
$f(a^2+c)=ac+f(a^2).$ But by $P(-a,0)$ we get:
$f(a^2+c)=-ac+f(a^2).$ Comparing these two equations gives us $c=0$ , so $f(0)=0$ .
Now, comparing $P(1,b)$ and $P(-1,-b)$ we find that $f(b)=-f(-b)$ , so $f$ is odd. Then, by $P(0,b)$ we derive:
$f(f(b^2))=b^2.$ Combining this with $f$ odd, we get that $f(f(x))=x$ for all $x$ . Therefore $f$ is involutive and also bijective, etc.
WIth $P(-b,b)$ we get:
$f(f(b^2))=-bf(b)+b^2+f(b^2).$ Comparing this with $P(0,b)$ we get $f(b^2)=bf(b)$ . But if we replace $b$ with $f(b)$ and use $f(f(x))=x$ , we get:
$f(b^2)=f(f(b)^2) \implies f(b)^2=b^2 \implies f(x) \in \{x,-x\},$ where we use the injectivity of $f$ here.
Now suppose there exist some $r,s \neq 0$ such that $f(r)=r$ and $f(s)=-s$ . Since $f$ is odd, WLOG $r,s>0$ . From $P(\sqrt{s},\sqrt{r})$ we get:
$f(s+\sqrt{rs}+r)=\sqrt{s}f(\sqrt{r})+r-s.$ Now taking a total of four cases on the value of $f(s+\sqrt{rs}+r)$ and $f(\sqrt{r})$ will give $r=0$ , $s=0$ , or $r=s$ in each of them, none of which are possible. This leaves $f(x)=x$ and $f(x)=-x$ as the only solutions. $\blacksquare$

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#9 h Apr 10, 2021, 6:41 PM • 1 Y

Y by mkomisarova

Let $P(x,y)$ denotes assertion of given functional equation. Note that $P(1,0)$ and $P(-1,0)$ gives us:
$f(1+f(0)) =f(0)+f(1)=-f(0)+f(1) \implies f(0) = 0$ Consequently from $P(0,b)$ it follows $f(f(b^2)) =b^2$ or in other words $f(f(x)) = x$ for all positive reals $x$ .
Now we seek to gain some more useful identities, thus we take look at $P(a,-a)$ :
$f(f(a^2)) =af(-a) + a^2 +f(a^2) \implies f(a^2) = -af(-a)$ Replacing $a$ by $-a$ gives in last equality yields to:
$f(a^2) = af(a) = -af(-a) \implies f(-a) = -f(a)$ This helps us since:
$-f(f(x)) = f(-f(x)) = f(f(-x)) =-x$ Last identity simply implies that we $f(f(x)) =x$ for all negative reals $x$ , therefore we conclude that $f(f(x)) =x$ for all real numbers $x$ . Now we take $f$ from both sides of $f(a^2) = af(a)$ :
$f(f(a^2)) = f(af(a)) = a^2$ Finally we replace $a$ by $f(a)$ in the last identity:
$f(af(a)) = f(a)^2 =a^2$ This implies that $f(a) =a$ or $f(a) =-a$ . Now we are left to escape pointwise trap. Assume that for some non zero real numbers $x,y$ we have $f(x) =x$ and $f(y)=-y$ . Consider $P(x,y)$ :
$f(x^2 + xy + yf(y)) = -xy +y^2 + xf(x)$ This transforms into:
$f(x^2+ xy -y^2) =-xy+y^2 +x^2$ If $f(x^2+ xy -y^2) = x^2+xy-y^2$ , then $2y^2 = 2xy \implies x=y$ - contradiction, otherwise:
$-x^2- xy+y^2 = -xy+y^2 +x^2 \implies x=0$ Therefore functions $f(x) =x$ and $f(x) =-x$ are the only ones that work.

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#10 h May 13, 2021, 7:27 PM

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Let $P(x,y)$ be the assertion $f(x^2+xy+f(y^2))=xf(y)+y^2+f(x^2)$ .

$\textbf{1. }f(x^2)=xf(x),f(f(x))=x$
$P(-x,x)\Rightarrow f(f(x^2))=-xf(x)+x^2+f(x^2)$
$P(0,x)\Rightarrow f(f(x^2))=x^2+f(0)$
So $f(x^2)=xf(x)+f(0)$ . Plugging in $x=1$ yields $f(0)=0$ , hence $f(x^2)=xf(x)$ and $f$ is odd. We immediately get $f(f(x^2))=x^2$ and then $f(f(x))=x$ by oddness.

$\textbf{2. }f(x)^2=x^2$
Since $f(f(x^2))=x^2$ , we have $f(xf(x))=x^2$ , and taking $x\mapsto f(x)$ and comparing yields $f(x)^2=x^2$ .

$\textbf{3. }f(x)=x\forall x\text{ or }f(x)=-x\forall x$
Assume FTSOC that $\exists a,b$ such that $f(a)=a$ , $f(b)=-b$ , and $ab\ne0$ . Note that $f(a^2)=af(a)=a^2$ , and we similarly have $f(b^2)=-b^2$ .
$P(a,b)\Rightarrow f(a^2+ab-b^2)=-ab+b^2+a^2\Rightarrow (a^2+ab-b^2)^2=(-ab+b^2+a^2)^2\Rightarrow 2(a-b)^2=0$ , so $a=b$ . This means that $a=f(a)=f(b)=-b=-a$ , so $a=0$ , contradiction. Then the only solutions are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ , which both work.

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554183

#11 h Jun 16, 2021, 3:44 PM • 1 Y

Y by Mango247

Nice

Choose $b \ne 0$ . $P(\frac{-f(b^2)}{b},b) \implies f(b^2)f(b)=b^3$ . Now put it $b=1 \implies f(1)^2=1 \implies f(1)= \pm 1$ . Now put in $b=-1$ to get $f(1)f(-1)=-1$ . So $f(-1) = \mp 1$ . Now $P(1,0) \implies f(f(1))=1+f(0)$ . In any case, $f(f(1))=1 \implies f(0)=0$ . Now $P(0,b) \implies f(f(b^2))=b^2$ . Also by putting in $a=-b$ , we get $f(x^2)=xf(x) \forall x$ . Recall that we had obtained $f(b^2)f(b)=bf(b)^2=b^3 \implies f(b) = \pm b$ . If $f(a)=a, f(b)=-b$ for some $a,b \ne 0$ , we get a contradiction by $P(a,b)$ .
So, $f(x)=+x$ and $f(x)=-x$ are the solutions, which clearly work $\Box$

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hakN

#12 h Jun 16, 2021, 4:52 PM

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Let $P(a,b)$ denote the assertion.
$P(a,0)$ and $P(-a,0)$ gives $f(0) = 0$ .
Comparing $P(a,b)$ and $P(-a,-b)$ gives $f$ is odd.
$P(0,b) \implies f(f(b^2)) = b^2$ and so $f(f(x)) = x$ for all $x\geq 0$ .
But since $f$ is odd, we have $f(f(-x)) = f(-f(x)) = -f(f(x)) = -x$ and so $f(f(x)) = x$ for all $x\in \mathbb{R}$ , so $f$ is bijective.
$P(-b,b) \implies b^2 = f(f(b^2)) = -bf(b) + b^2 + f(b^2) \implies f(b^2) = bf(b)$ for all $b \in \mathbb{R}$ .
So, $f(a^2) = af(a) \implies f(f(a)^2) = af(a) = f(a^2)$ and since injective, we have $f(a)^2 = a^2 \implies f(a) \in \{a, -a\}$ for all $a\in \mathbb{R}$ .
Clearly, $f(a) = a \ \forall a \in \mathbb{R}$ and $f(a) = -a \ \forall a \in \mathbb{R}$ are solutions.
Lets assume for $a , b \neq 0$ , $f(a) = a$ and $f(b) = -b$ . Then $P(a,b)$ gives us $a=b$ or $a=0$ , contradiction. So we have our 2 solutions
$\boxed{f(a) = a \ \forall a \in \mathbb{R}}$ and $\boxed{f(a) = -a \ \forall a \in \mathbb{R}}$ .

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DakuMangalSingh

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DakuMangalSingh

#13 h Oct 4, 2021, 4:44 PM

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Answer
Solution

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#15 h Oct 21, 2021, 3:35 PM

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Physicsknight wrote:

Determine all the functions $f:\mathbb R\mapsto\mathbb R$ satisfies the equation
$f(a^2 +ab+ f(b^2))=af(b)+b^2+ f(a^2)\,\forall a,b\in\mathbb R$

Cute

Let $P(a,b)$ be the assertion of the functional equation. Then. $P(0,b)$ gives us that $f(f(b^2))=b^2+f(0)$ . $P(-b,b)$ also gives us that $f(f(b^2))=-bf(b)+b^2+f(b^2)$ so $f(b^2)=bf(b)+f(0)$ . Put $b=1$ in this to get $f(0)=0$ . Thus, $f(b^2)=bf(b)$ . Now, $P(\frac{-f(b^2)}{b},b)$ gives us that $f(b)f(b^2)=b^3$ or $f(b)^2=b^2$ , implying that $f(b)=\pm b$ . If $f(n)=n$ for some $n$ and $f(m)=-m$ for some $m$ , $P(m,n)$ gives us the desired contradiction.

This post has been edited 1 time. Last edited by rama1728, Oct 21, 2021, 3:35 PM
Reason: .

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#16 h Mar 6, 2022, 4:50 PM

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Physicsknight wrote:

Determine all the functions $f:\mathbb R\mapsto\mathbb R$ satisfies the equation
$f(x^2 +xy+ f(y^2))=xf(y)+y^2+ f(x^2)\,\forall x,y\in\mathbb R$

19:01

Let $P(x,y)$ denote the given assertion.

$P(0,0): f(f(0))=f(0)$ .

$P(x,0): f(x^2+f(0))=xf(0)+f(x^2)$ .

$P(0,x): f(f(x^2))=x^2+f(0)$ .

Thus, $f(f(x))=x+f(0)$ for all nonnegative reals $x$ .

Case 1: $f(0)\ge 0$ .
Then $f^3(0)=2f(0)\implies f(0)=2f(0)\implies f(0)=0$ . Thus, $f(f(x))=x$ for all nonnegative reals $x$ .

$P(x,-x): f(f(x^2))=x^2=xf(-x)+x^2+f(x^2)\implies -xf(-x)=f(x^2)$ .

$P(-x,x): f(f(x^2))=x^2=-xf(x)+x^2+f(x^2)\implies xf(x)=f(x^2)$ .

So $xf(x)=-xf(-x)$ . Since $f(0)=0$ , $f$ is odd and an involution.

The FE can be rewritten as $f(x^2+xy+yf(y))=xf(y)+y^2+xf(x)$ . Let $Q(x,y)$ denote this new assertion.

$Q(-f(x),x): f((-f(x))^2)=-f(x)^2+x^2+(-f(x))f(-f(x))=-f(x)^2+x^2+f((-f(x))^2)\implies x^2=f(x)^2$ . So $f(x)=x$ or $f(x)=-x$ for each $x$ .

If $f(a)=a$ and $f(b)=-b$ with $a,b\ne 0$ , then $Q(a,b): f(a^2+ab-b^2)=a^2-ab+b^2$ . If $a^2+ab-b^2=a^2-ab+b^2$ , then $a-b=b-a\implies a=b$ , a contradiction. If $-a^2-ab+b^2=a^2-ab+b^2$ , then $a^2=0$ , a contradiction.

So the only solutions here are $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ , which work.

Case 2: $f(0)<0$ .
$P(-x,x): f(f(x^2))=-xf(x)+x^2+f(x^2)$ . Now we know $f(f(x^2))=x^2+f(0)$ , so $f(0)=-xf(x)+f(x^2)$ . Set $x=1$ here to achieve $f(0)=0$ , a contradiction.

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#17 h Apr 2, 2022, 2:55 PM

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$P(a,-a) : f(f(a^2)) = af(-a) + a^2 + f(a^2)$
$P(a,0) : f(f(a^2)) = a^2 + f(0)$
so $f(a^2) + af(-a) = f(0)$ .
$P(-a,a),P(-a,0) : f(a^2) - af(a) = f(0)$ so now we have $f(a) = -f(-a)$ .
$a = 1 : f(1) - 1f(1) = f(0) \implies f(0) = 0$ so $P(a,0) : f(f(a^2)) = a^2 + f(0) = a^2$ and $f(a^2) = af(a)$ .
Note that $f(f(a^2)) = a^2$ covers all nonnegative numbers so for all $n \ge 0$ we have $f(f(n)) = n$ .
we had $f(a^2) = af(a)$ now put $a = f(a)$ so we have $f(f(a)^2) = f(a)f(f(a)) = af(a) = f(a^2) \implies f(f(f(a)^2)) = f(f(a^2))$ . Note that both $f(a)^2 , a^2$ are nonnegative so $f(a)^2 = a^2 \implies f(a) = \pm a$ for all nonnegative $a$ but back we proved $f(a) = -f(-a)$ . so we have $f$ for negative values as well.
Answers : $f(a) = \pm a$ .

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khina

#18 h Apr 2, 2022, 4:35 PM • 2 Y

Y by megarnie, IAmTheHazard

Can we change the title of this post? Having the solution set (however obvious) of a functional equation as the title seems like a major and completely unnecessary giveaway.

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ZETA_in_olympiad

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ZETA_in_olympiad

#19 h May 30, 2022, 2:02 PM

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Let $P(a,b)$ denote the given assertion.

$P(1,0)-P(-1,0)\implies f(0)=0.$
$P(0,x)\implies f(f(x^2))=x^2.$
$P(x,-x)\implies f(x^2)=xf(x).$

So $f(f(x))=x~\forall x$ by oddness. Also $f(xf(x))=x^2.$ Setting $x=f(x) \implies f(x)=\pm x.$ Note that existence of $u,v:f(u)=u, f(v)=-v$ is absurd by $P(u,v).$ Thus $f\equiv \text{Id}$ and $f\equiv -\text{Id}$ , indeed work.

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#20 h Mar 6, 2023, 5:42 PM

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Here is my solution: https://calimath.org/pdf/RMM-SL2019-A1.pdf
And I uploaded the solution with motivation to: https://youtu.be/3RjpZHIIVEo

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Aoxz

#21 h Mar 20, 2024, 9:01 PM

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Eazy
P(0,b) + P(a,0)+P(-f(b^2)/b ,b)+P(a,-a)+P(1) kill

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#22 h Apr 13, 2025, 5:31 PM

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Put $a=-b$ to get $f(f(b^2))=b^2+f(b^2)-bf(b)$ Put $a=0$ to get $f(f(b^2))=b^2+f(0)$ Thus $f(b^2)=bf(b)-f(0)$ . From $b=0$ we get $f(0)=0$ , and so $f(b^2)=bf(b)$ .
Put $a=-a-b$ into the initial equation. We get $f(a^2+ab+f(b^2))=-(a+b)f(b)+b^2+f((a+b)^2)$ And so $f((a+b)^2)-f(a^2)=(2a+b)f(b).$ Rewrite it as $f((a+b)^2)-f(a^2)-f(b^2)=2af(b)$ The left side is symmetric, thus $af(b)=bf(a)$ , from which $f(x)=cx$ . Substituting it into the original equation we get $c=\pm 1$ , which means our solutions are $x$ and $-x$ .

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