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A functional equation from MEMO

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Source: Middle European Mathematical Olympiad 2022, problem I-1

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square_root_of_3

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square_root_of_3

#1 h Sep 1, 2022, 12:37 PM • 1 Y

Y by tiendung2006

Find all functions $f: \mathbb R \to \mathbb R$ such that $f(x+f(x+y))=x+f(f(x)+y)$ holds for all real numbers $x$ and $y$ .

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#2 h Sep 1, 2022, 1:30 PM

Y by

$P(x,y): f(x+f(x+y))=x+f(f(x)+y)$

$P(x,0):f(x+f(x))=x+f(f(x))$
$f(a)=0\Longrightarrow P(a,0):0=a+0\Longrightarrow a=0$

$P(x,-x): f(x)=x+f(f(x)-x)$
$f(x)-x=t\Longrightarrow f(t)=t$

This post has been edited 1 time. Last edited by ratatuy, Sep 1, 2022, 1:31 PM

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#3 h Sep 1, 2022, 1:33 PM • 2 Y

Y by fuzimiao2013, guywholovesmathandphysics

ratatuy wrote:

$P(x,y): f(x+f(x+y))=x+f(f(x)+y)$

$P(x,0):f(x+f(x))=x+f(f(x))$
$f(a)=0\Longrightarrow P(a,0):0=a+0\Longrightarrow a=0$

$P(x,-x): f(x)=x+f(f(x)-x)$
$f(x)-x=t\Longrightarrow f(x)=x$

This is so messed up, well first u have to prove that there exists an $a$ such that $f(a)=0$ now the replace after that doesn't give $a=0$ and finally the fact that $t$ is an identity doesn't mean $f(x)=x$

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hakN

#4 h Sep 1, 2022, 2:00 PM

Y by

Plugging $y=-f(x)$ , we see that $f$ is surjective.

Let $g(x)+x=f(x)$ , so equation rewrites as $g(2x+y+g(x+y))+g(x+y)=g(x)+g(x+y+g(x))$ , and putting $y \to y-x$ , we get $g(x+y+g(y))+g(y)=g(x)+g(y+g(x))$ , call this $P(x,y)$ .

Note that for any $a,b \in \mathbb{R}$ such that $g(a)=g(b)$ , $P(a,y)$ and $P(b,y)$ gives that $g(y+g(y)+a)=g(y+g(y)+b)$ for all $y \in \mathbb{R}$ .

$P(x,0) \implies g(x+g(0))=g(x)+g(g(x))-g(0)$ .

$P(0,x) \implies g(x+g(0))=g(x+g(x))+g(x)-g(0)$ , comparing this with above, we get $g(g(x))=g(x+g(x))$ for all $x \in \mathbb{R}$ .

Now, letting $a=g(x) , b=x+g(x)$ , since $g(a)=g(b)$ , we get $g(y+g(y)+g(x))=g(y+g(y)+x+g(x))$ for all $x,y \in \mathbb{R}$ .

Since $x+g(x)=f(x)$ was surjective, we get $g(z+g(x))=g(z+x+g(x))$ for all $x,z \in \mathbb{R}$ , and putting $z \to z-g(x)$ we get that $g(z)=g(z+x)$ for all $x,z \in \mathbb{R}$ , so $g$ is constant and $f(x)=x+c$ for some constant $c$ . It is easy to see that all such functions work.

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#6 h Sep 3, 2022, 1:17 PM

Y by

The issue was a bit complicated. I hope I have not solved any carelessness in the process
A bit hard problem :wallbash_red:

:wallbash_red:

Let $P(x,y) : f(x+f(x+y))=x+f(f(x)+y)$ .
$P(x,0) : f(x+f(x))=x+f(f(x))$ .
$P(x,f(y)) , P(y,f(x))$ : $f(x+f(x+f(y)))-x=f(y+f(y+f(x)))-y$ . Now put $y=0$ . then $f(x+f(x+f(0)))-x=f(f(f(x)))$ . ( :love:

:love:

)
$P(x,-f(x))$ : $f(x+f(x-f(x)))=x+f(0)$ then $f$ is surjective ! . ( :cool:

:cool:

)
Now we know $\exists \alpha$ : $f(\alpha ) = 0$ . Now by $P(\alpha , 0)$ we get $f(0)=0$ .
Now we know $f(0)=0$ . By ( :love:

:love:

) we get $f(f(x))=f(f(f(x)))$ . ( :blush:

:blush:

)
$P(x,-x)$ : $f(x)=x+f(f(x)-x) \to f(f(x)-x) = x-f(x) \to f(f(f(x)-x)) = f(x-f(x)) \to x+f(f(f(x)-x)) = f(x-f(x))+x$

Now we get $f(x+f(f(f(x)-x))) = f(f(x-f(x))+x)$ . By ( :cool:

:cool:

) and (

:blush:

) we get $f(x+f(f(x)-x))=x$ . ( :w00tb:

:w00tb:

)
$P(0,x+f(f(x)-x)) : f(f(x))=f(x+f(f(x)-x)) = x$ . then we can see $f$ is injective .
Now by ( :blush:

:blush:

) we get $f(x)=x$ . $\blacksquare$

This post has been edited 1 time. Last edited by strong_boy, Sep 3, 2022, 1:18 PM

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#7 h Sep 3, 2022, 1:35 PM • 1 Y

Y by Mango247

strong_boy wrote:

The issue was a bit complicated. I hope I have not solved any carelessness in the process
A bit hard problem :wallbash_red:

:wallbash_red:

Let $P(x,y) : f(x+f(x+y))=x+f(f(x)+y)$ .
$P(x,0) : f(x+f(x))=x+f(f(x))$ .
$P(x,f(y)) , P(y,f(x))$ : $f(x+f(x+f(y)))-x=f(y+f(y+f(x)))-y$ . Now put $y=0$ . then $f(x+f(x+f(0)))-x=f(f(f(x)))$ . ( :love:

:love:

)
$P(x,-f(x))$ : $f(x+f(x-f(x)))=x+f(0)$ then $f$ is surjective ! . ( :cool:

:cool:

)
Now we know $\exists \alpha$ : $f(\alpha ) = 0$ . Now by $P(\alpha , 0)$ we get $f(0)=0$ .
Now we know $f(0)=0$ . By ( :love:

:love:

) we get $f(f(x))=f(f(f(x)))$ . ( :blush:

:blush:

)
$P(x,-x)$ : $f(x)=x+f(f(x)-x) \to f(f(x)-x) = x-f(x) \to f(f(f(x)-x)) = f(x-f(x)) \to x+f(f(f(x)-x)) = f(x-f(x))+x$

Now we get $f(x+f(f(f(x)-x))) = f(f(x-f(x))+x)$ . By ( :cool:

:cool:

) and (

:blush:

) we get $f(x+f(f(x)-x))=x$ . ( :w00tb:

:w00tb:

)
$P(0,x+f(f(x)-x)) : f(f(x))=f(x+f(f(x)-x)) = x$ . then we can see $f$ is injective .
Now by ( :blush:

:blush:

) we get $f(x)=x$ . $\blacksquare$

$P(\alpha,0) \implies \alpha+f(0)=0.$

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Mz_T

#8 h Sep 5, 2022, 7:52 AM • 4 Y

Y by chystudent1-_-, khiemnguyen0620, Gms68bx, Lenzimmi

my solution
Let $P(x,y) : f(x+f(x+y))=x+f(f(x)+y)$ .
$P(x,-f(x)) : f(x+f(x-f(x)))=x+f(0)$ $\Rightarrow$ $f$ is surjective.
Let $f(0)=c$ , because $f$ is subjective then $\exists b \in \mathbb R, \ f(b) = 0$ .
$P(b,0) : b+c=0 \Rightarrow b=-c,\ f(-c)=0$ .
$P(b,y) : f(b+f(y+b))=b+f(y)$ or $f(f(y-c)-c)=f(y)-c$ .
$\Rightarrow f(f(y)-c)=f(y+c)-c \ \forall y \in \mathbb R$ . $(1)$
$P(0,y) : f(f(y))=f(y+f(0))=f(y+c)$ . $(2)$
From $(1)$ and $(2)$ we have $f(f(y)-c)=f(f(y))-c\ \forall y \in \mathbb R$ .
because $f$ is subjective then $f(y-c)=f(y)-c \Rightarrow f(y)+c=f(y+c)\ \forall y \in \mathbb R$ . $(3)$
From $(2)$ and $(3)$ we have $f(f(y))=f(y)+c \Rightarrow f(y)=y+c\ \forall y \in \mathbb R$ .

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ZETA_in_olympiad

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ZETA_in_olympiad

#9 h Sep 5, 2022, 9:09 AM • 1 Y

Y by Mango247

Clearly $f$ is surjective. Define $g(x):=f(x)-x$ and let $z=x+y.$
$\textbf{Claim:}$ The equation $g(x+z+g(z))+g(z)=g(z+g(x))+g(x) \quad \forall x,z \in \mathbb R$ denoted by $E(x,z)$ , is satisfied by constant functions only.

$\textbf{Proof.}$ $E(x,x)$ yields $g(x+g(x))=g(2x+g(x)).$ Take $u$ such that $g(u)+u=-g(x)-x.$ Comparing $E(x+g(x),u)$ and $E(2x+g(x),u)$ we get $g(x)=g(0).$ $\blacksquare$

This post has been edited 3 times. Last edited by ZETA_in_olympiad, Sep 5, 2022, 9:23 AM
Reason: Typo

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ZETA_in_olympiad

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ZETA_in_olympiad

#10 h Sep 19, 2022, 5:51 PM • 1 Y

Y by MS_asdfgzxcvb

Another solution without shifting.

Let $P(x,y)$ denote the assertion $f(x+f(x+y))=x+f(f(x)+y).$ Clearly $f$ is surjective. Take $f(a)=0$ then $P(a,0)$ gives $a=-f(0).$ By $P(0,x)$ , $f(f(x))=f(x-a).$ Take $x_0$ such that $f(x_0)=x-a$ , then $P(a,x_0-a)$ implies $f(f(x))=f(x)-a$ and so, $f(x')\equiv x'+c$ , which works.

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i3435

#11 h Sep 19, 2022, 6:18 PM • 2 Y

Y by DaRKst0Rm, mathscrazy

I claim that all solutions are of the form $f(x)=x+c$ for some $c$ , these can be shown to work.

Let $P(x,y)$ be the assertion. $P(x,-f(x))$ gives that $f(x+f(x-f(x)))=x+f(0)$ , thus $f$ is surjective. For any $c$ such that $f(c)=0$ , $P(c,-f(c))$ gives that $0=c+f(0)$ , so $f(x)=0$ has exactly $1$ solution. Assume that there are distinct $a$ and $b$ such that $f(a)=f(b)$ . Let $y_1$ be such that $f(f(a)+y_1)=a-f(a)$ , and let $y_2$ be such that $f(f(a)+y_2)=b-f(a)$ . Note that $y_1$ and $y_2$ must be distinct. $P(f(a),y_1)$ and $P(f(a),y_2)$ give that $f(f(f(a))+y_1)=f(f(f(a))+y_2)=0$ , giving a contradiction, so $f$ is injective. $P(0,y)$ gives $f(y)=y+f(0)$ as desired.

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#12 h Sep 25, 2022, 3:06 PM

Y by

chenjiaqi give a good solution ,but it is hard to type ,i am sorry

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#13 h Jan 10, 2023, 7:32 PM • 1 Y

Y by Yunis019

Different solution

Let $P(x,y) : f(x+f(x+y))=x+f(f(x)+y)$

$P(0,x) \implies f(f(x))=f(f(0)+x) \ \ \ (1)$

$P(x,-f(x)) \implies f(x+f(x-f(x)))=x+f(0) \implies \text{ f is surjective }$

So $\exists k$ such that $f(k)=0$

$P(k,0) \implies f(k+f(k))=k+f(f(k)+0) \implies k=-f(0)$

$P(-f(0) , f(0)+y) \implies f(f(y)-f(0))=f(f(0)+y)-f(0) \ \ \ (2)$

Using $(1)$ in $(2)$ and substituting $f(y)$ with $z$ (we can do it since f is surjective) yields

$f(z-f(0))=f(z)-f(0)$

Substituting $z$ with $z+f(0)$ , using $(1)$ and again substituting $f(z)$ with $t$ yields

$f(t)=t+f(0) \implies \boxed{f(x) = x + c} \implies \text { check , works! }$

so we are done :-D

:-D

This post has been edited 1 time. Last edited by Ibrahim_K, Jan 10, 2023, 7:33 PM

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#14 h Jan 10, 2023, 9:38 PM

Y by

Let $g(x)=f(x)-x$ .
$P(x,y-x)\Rightarrow Q(x,y):g(x+y+g(y))+g(y)=g(g(x)+y)+g(x)$
$Q(x,-g(x))\Rightarrow g(x-g(x)+g(-g(x)))+x-g(x)+g(-g(x))=x+g(0)$
so $g(x)+x$ is surjective. Let $g(a)+a=0$ :
$Q(a,a)\Rightarrow a=-g(0)\Rightarrow g(-g(0))=g(0)$
$Q(-g(0),x)-Q(0,x)\Rightarrow g(-g(0)+x+g(x))=g(x+g(x))\Rightarrow g(x)=g(x-g(0))$ since $x+g(x)$ is surjective. Then $g(x+g(0))=g(x)$ as well.
$Q(0,x)\Rightarrow g(x+g(x))=g(0)\Rightarrow g(x)=g(0)$ so $g$ is constant. Then $f$ is linear, testing gives $f(x)=x+c$ .

This post has been edited 1 time. Last edited by jasperE3, Jan 11, 2023, 5:29 PM

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#15 h Jan 11, 2023, 2:00 AM

Y by

jasperE3 wrote:

Let $g(x)=f(x)-x$ .
$P(x,y-x)\Rightarrow Q(x,y):g(x+y+g(y))+g(y)=g(g(x)+y)+g(x)$
$Q(x,-g(x))\Rightarrow g(x-g(x)+g(-g(x)))+x-g(x)+g(-g(x))=x+g(0)$
so $g(x)+x$ is surjective. Let $g(a)+a=0$ :
$Q(a,a)\Rightarrow a=-g(0)\Rightarrow g(-g(0))=g(0)$
$Q(-g(0),x)-Q(0,x)\Rightarrow g(-g(0)+x+g(x))=g(x+g(x))\Rightarrow g(x)=g(x-g(0))$ since $x+g(x)$ is surjective. Then $g(x+g(0))=g(x)$ as well.
$Q(0,x)\Rightarrow g(x+g(x))=g(0)\Rightarrow g(x)=g(0)$ so $g$ is constant. Then $f$ is linear, testing gives $f(x)=x$ .

i think that f(x)=x+c is the solution.

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#16 h Mar 8, 2023, 4:22 PM

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EDIT: I made a mistake.

This post has been edited 2 times. Last edited by Parsia--, Mar 8, 2023, 5:00 PM

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Jjesus

#17 h Mar 8, 2023, 4:37 PM

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Parsia-- wrote:

Setting $y = -f(x)$ we find that $f$ is surjective. $P(0,y) \rightarrow f(f(y)) = f(f(0)+y)) \rightarrow f(y) = y+c$ which works.

$f$ is inyective?

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F10tothepowerof34

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F10tothepowerof34

#18 h May 4, 2023, 5:46 PM

Y by

Let the assertion $P(x,y)=f(x+f(x+y))=x+f(f(x)+y)$
$P(0,x)$ yields: $f(f(x))=f(f(0)+x)$ (1)
Since $f$ is surjective, $\exists \alpha, \text{such that}, f(\alpha)=0$
$P(\alpha,0)$ yields: $f(\alpha)=\alpha +f(0)\Longrightarrow \alpha=-f(0)$
$P(\alpha,x)$ yields: $f(f(x-f(0))-f(0))=f(x)-f(0)$ , let $Q(x)$ denote $P(\alpha,x)$
$Q(y+f(0))$ yields: $f(f(y)-f(0))=f(y+f(0))-f(0)$
$\therefore f(f(y)-f(0)=f(f(y))-f(0)$ from (1)
Furthermore let $f(y)=z\Longrightarrow f(z-f(0))=f(z)-f(0)$ , thus let $z=k+f(0)$ . This implies that: $f(k)=k+f(0)\Longrightarrow f(x)=x+c$ $\blacksquare$ .
And we are done! :coool:

:coool:

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Medira0103080201

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Medira0103080201

#19 h Feb 25, 2024, 7:53 AM

Y by

f(x + f(x+y)) = x + f(f(x) + y)(1)
(1) x --> 0 => f(f(y)) = f(y + f(0))(2)
(1) y--> -f(x) => f : surjective
There is exist some c such that f(c) = 0
(1) x--> c, y --> 0 => f(c) = c + f(0) => c = -f(0)
(1) y --> -x+c => f(x) = x + f(f(x) - x +c ) => f(x) - x = f(f(x) - x +c)
Let a = f(x) - x + c => f(a) = a+f(0)
(1) x--> a, y --> y - a => f(a + f(y)) = a + f(f(a) + y - a) => f(a + f(y)) = a + f(y + f(0)) => f(a + f(y)) = a + f(f(y))
By (2) we can replace f(y) by z => f(z + a) = a + f(z)(3)
(1) y --> y - x + c => f(x + f(y + c)) = x + f(f(x) + y - x +c) and a = f(x) - x +c => f(x + f(y+c)) = x + f(y) + f(x) - x + c =>
f(x + f(y - f(0))) = f(x) + f(y) + c, replace y by y + f(0) => f(x + f(y)) = f(x) + f(y + f(0)) + c => f(x + f(y)) = f(x) + f(y) + c
By (2) we can replace f(y) by y => f(x + y) = f(x) + f(y) + c then g(x) = f(x) - f(0) => g(x+y) = g(x) + g(y)
(1) => g(x + g(x + y) + f(0)) + f(0) = x + f(0) + g(g(x) + y + f(0)) => g(x + g(x + y)+ f(0)) - g(g(x) + y + f(0)) = x =>
g(x - y + g(y)) = x => put y = a => g(x) = x then f(x) = x+f(0) - is answer

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trying_to_solve_br

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trying_to_solve_br

#20 h Mar 8, 2024, 7:30 PM

Y by

Hardish. Good FE spam.
Set $y=-f(x)$ to get surjectivity.

Then, take $y=0$ to get $f(f(x))+x=f(x+f(x))$ (#).

Then, rewrite the initial condition to $f(x+f(y))=x+f(f(x)+y-x)$ (1), by substituting $y$ by $y-x$ . Put now $y=0$ here to get $f(f(x))=f(f(x)-x)+x$ (*).

Now, let for any $x$ , $f(x)-x=t$ . Let $a_t$ be the number (by surjectivity) that makes $f(a_t)=t$ .

Putting $y=a_t$ on (1), we get that $f(f(x))=x+f(t+a_t)$ . But by (*) we may substitute $f(f(x))$ in the last expression to get $f(t)=f(t+a_t)$ .

But by (#), putting $x=a_t$ , we get that $f(t+a_t)=a_t+f(f(a_t))=a_t+f(t)$ . This implies, with the equality above, $a_t=0$ and thus $f(x)=x+f(0)$ .

The motivation behind this short solution is just trying everything and seeing how much $f(x)-x$ appears when you plug nice stuff.

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#21 h Nov 2, 2024, 5:32 PM

Y by

Answer is $f(x)=x+c$ for any real constant $c$ . Note that plugging $y=-f(x)$ yields the surjectivity of $f$ . Let $f(x)-x=g(x)$ .
$g(x+y)+2x+y+g(g(x+y)+2x+y)=x+y+x+g(x)+g(g(x)+x+y)$ Pick $x,y-x$ to get
$g(y)+g(g(y)+x+y)=g(x)+g(g(x)+y)$ $x=y$ gives $g(g(x)+2x)=g(g(x)+x)$ or $g(f(x)+x)=g(f(x))$ . Comparing $P(f(x),y)$ with $P(f(x)+x,y)$ implies
$g(f(y)+f(x))=g(f(y)+f(x)+x)$ Since $f$ is surjective, we can replace $y-f(x)$ with $f(y)$ . So $g(y)=g(y+x)$ which yields $g$ is constant. If $g\equiv c,$ then $f(x)=x+c$ as desired. $\blacksquare$

This post has been edited 2 times. Last edited by bin_sherlo, Nov 4, 2024, 4:18 PM

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#22 h Nov 2, 2024, 7:00 PM • 1 Y

Y by Tony_stark0094

solution

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#23 h Mar 13, 2025, 6:49 AM

Y by

f is surjective
$f(-f(0))=0$
$ff(y)=f(y+f(0))=f(y-c)\ where \ f(0)=-c$
x=c and y=y-c
$f(c+f(y))=c+f(y-c)=c+f(fy)$
$let\ \ f(y)=u$ where u can be any real
we get
$f(c+u)=c+f(u)$
u=u-c
$f(u)=c+f(u-c)=c+ffu==> f(y)=y-c$

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#24 h Mar 13, 2025, 7:04 AM

Y by

InterLoop wrote:

solution

how do you get $f(x)=f(x+c)$

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#25 h Mar 29, 2025, 4:06 PM

Y by

$P(x,y): f(x+f(x+y))=x+f(f(x)+y)$
$P(x, -f(x)): f(x+f(x-f(x)))=x+f(0) \implies f$ is surjective.
$\exists$ $t: f(t)=0$ then $P(t,0)$ gives $f(0)=0$
$P(x, -x): f(x)-x=f(f(x)-x)$ . And we get $f(x)=x$

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#26 h Mar 29, 2025, 4:36 PM

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John_Mgr wrote:

$P(x, -x): f(x)-x=f(f(x)-x)$ . And we get $f(x)=x$

How do you deduce $f(x)=x$ $\forall x$ ?
Have you proved somewhere that $f(x)-x$ is surjective ? (which would be very surprising

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