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Source: FKMO 2025 P2

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#1 h Mar 29, 2025, 10:16 AM

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Let $\mathbb{R}$ be the set of real numbers. Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ that satisfy the following condition. Here, $f^{100}(x)$ is the function obtained by composing $f(x)$ $100$ times, that is, $(\underbrace{f \circ f \circ \cdots \circ f}_{100 \ \text{times}})(x).$

(Condition) For all $x, y \in \mathbb{R}$ , $f(x + f^{100}(y)) = x + y \ \ \ \text{or} \ \ \ f(f^{100}(x) + y) = x + y$

This post has been edited 1 time. Last edited by pokmui9909, Mar 29, 2025, 10:25 AM

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#2 h Mar 29, 2025, 10:54 AM • 4 Y

Y by Acorn-SJ, seoneo, Lufin, k2i_forever

$100$ has no meaning here, so replace it with a given integer $m$ . Let $P(x,y)$ be the assertion
$f(x+f^{m}(y)) = x + y \quad \text{or} \quad f(f^{m}(x) + y) = x + y$ From $P(0,0)$ , we can easily observe that $f^{m+1}(0) = 0$ .

Claim. $f$ is bijective.
Proof. From $P(x,x)$ , we have that $f( x + f^m(x)) = 2x$ for all reals $x$ . This implies that $f$ is surjective. Now, assume that $f(a) = f(b)$ for some $a, b \in \mathbb{R}$ . Since $f^m(a) = f^m(b)$ , we have that $2a = f(a + f^m(b))$ and $2b = f(b + f^m(a))$ , implying with $P(a,b)$ that $a = b$ . Hence, $f$ is injective, so we can conclude that $f$ is bijective.

Claim. $f(0) = 0$ .
Proof. Let $f(0) = c$ be a nonzero real. With induction, we will prove that $f^k(0) = kc$ for any $k \in \mathbb{Z}_{\ge 0}$ . The inductive hypothesis holds when $k = 1$ . Assume that $f^{\ell}(0) = \ell c$ for any $\ell < k$ , for $k \ge 2$ . From $P(f^{k-1}(0), f(0))$ , we have that
$f(f^{k-1}(0) + f^{m+1}(0)) = f^{k-1}(0) + f(0) \quad \text{or} \quad f(f^{m + k - 1}(0) + f(0)) = f^{k-1}(0) + f(0)$ Since $f^{k-1}(0) = (k-1)c$ and $f^{m + 1 + k-2}(0) = f^{k-2}(0) = (k-2)c$ , and $f^{m+1}(0) = 0$ , we have that $f^k(0) = kc$ should hold. Therefore, from $f^{m+1}(0) = 0$ , we can obtain that $c = 0$ , or, $f(0) = 0$ .

From $P(x, 0)$ , we have that $f^{m+1}(x) = x$ or $f(x) = x$ must hold. In both cases, we have that $f^{m+1}(x) = x$ . Hence, if a function $g : \mathbb{R} \to \mathbb{R}$ is defined as $x \mapsto f^m(x)$ , we have that $g$ is the inverse of $f$ , and so bijective. Also, $g^{m+1}(x) = x$ always holds, and $g(0) = 0$ . Now, we can simplify $P(x,y)$ as follows.
$g(x+y) = g(x) + y \quad \text{or} \quad g(x+y) = x + g(y)$

Claim. $g(x) = x$ for any $x \in \mathbb{R}$ .
Proof. Assume the contrary, so that there is a real number $a$ such that $g(a) \neq a$ . When $g^{k-1}(a) = g^k(a)$ for some $k$ , from injectivity, we can deduce that $g(a) = a$ , so $g^{k-1}(a) \neq g^k(a)$ for all positive integer $k$ . From $g(0) = 0$ , we have that $g(g^{k-1}(a) - g^k(a)) \neq 0$ also holds for all positive integer $k$ . From $P(g^{k-1}(a) - g^k(a), g^k(a))$ , we have that
$g^k(a) = g^k(a) + g( g^{k-1}(a) - g^k(a) ) \quad \text{or} \quad g^k(a) = g^{k+1}(a) + g^{k-1}(a) - g^k(a)$ The first one cannot hold from the reasons mentioned above. Therefore, for any positive integer $k$ , we have that $2g^k(a) = g^{k+1}(a) + g^{k-1}(a)$ . This implies that the sequence(orbit)
$\mathcal{O}: \quad a, g(a), g(g(a)), \cdots, g^i(a), \cdots$ is an arithmetic progression. Hence, when $g(a) \neq a$ , $g^k(a) \neq a$ for any positive integer $k$ , so $g^{m+1}(a) \neq a$ , a contradiction. Therefore, $g(x) = x$ must hold for any $x \in \mathbb{R}$ .

Since $g$ was the inverse of $f$ , we can conclude that $f(x) = x$ for all reals $x$ , and the identity function clearly satisfies the problem condition.

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#4 h Mar 31, 2025, 2:04 AM

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Another proof for the last claim
We have $x+f^{-1}(y)=f^{-1}(x+y)$ or $y+f^{-1}(x)=f^{-1}(x+y)$
if there exist different real numbers $a$ and $b$ in which $f(a)=b$ ,
$P(b, b); b+a=f^{-1}(2b)$ , so $f(a+b)=2b$
I don't remember how I got this, but $f(2a)=a+b$ holds.
Let $a_{1}\xrightarrow[]{f}a_{2}\xrightarrow[]{f}a_{3}\cdots a_{101}\xrightarrow[]{f}a_{1}$
$2a_{1}\xrightarrow[]{f}a_{1}+a_{2}\xrightarrow[]{f}2a_{2}\cdots 2a_{101}\xrightarrow[]{f}a_{101}+a_{1}\xrightarrow[]{f}2a_{1}$
some calculation gives $a_{i+2}-a_{i}=c$ where c is a constant.
So that c must be 0, contradiction.

This post has been edited 1 time. Last edited by k2i_forever, Mar 31, 2025, 2:08 AM

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#5 h Apr 1, 2025, 6:06 AM

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GOTCHA!
$P(-y, y); f(-y+f^{100}(y))=0$ or $f(y+f^{100}(-y))=0$ , so by the claim that $f$ is bijective,
$f^{-1}(y)=y$ or $f^{-1}(-y)=-y$ , so $f(y)=y$ or $f(-y)=-y$ for all $y$
Hence $P(a+b, -a); f(a+b+f^{-1}(-a))= b$ or $f(f^{-1}(a+b)-a)=b$
Since $f(a)\neq a$ , $f(-a)=-a$ , and $f^{-1}(-a)=-a$
So the first condition is equivalent to $f(b)=b$ , so by $f$ bijective, $a=b$ , contradiction
The second condition is equivalent to $f^{-1}(a+b)-a=a$ by bijective, $f^{-1}(a+b)=2a$ , $f(2a)=a+b$
Hence we earn $f(2a)=a+b$

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#6 h Apr 1, 2025, 6:07 AM

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Kinda straightforward, huh?

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