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Source: INAMO 2023 P2 (OSN 2023)

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somebodyyouusedtoknow

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somebodyyouusedtoknow

#1 h Aug 29, 2023, 6:48 AM • 2 Y

Y by Amir Hossein, AlexCenteno2007

Determine all functions $f : \mathbb{R} \to \mathbb{R}$ such that the following equation holds for every real $x,y$ :
$f(f(x) + y) = \lfloor x + f(f(y)) \rfloor.$ Note: $\lfloor x \rfloor$ denotes the greatest integer not greater than $x$ .

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#2 h Aug 29, 2023, 7:01 AM • 4 Y

Y by somebodyyouusedtoknow, Furra, Amir Hossein, britishprobe17

INAMO 2023/2 wrote:

Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$f(f(x) + y) = \lfloor x + f(f(y)) \rfloor$ for any $x,y \in \mathbb{R}$ .

Mine! Thought about this problem when I was thinking back about ISL 2022 A6 and proposed this as a cute trivial problem, but somehow got into the contest :play_ball:

:play_ball:

Here's the official solution that I submitted.

We claim that all such solutions for the above functional equation are of the form $\boxed{f(x) = \lfloor x \rfloor + k}$ for all $x$ , for some fixed $k \in \mathbb{Z}$ . To see this, we see that for all $x,y \in \mathbb{R}$ ,
$f(f(x) + y) = \lfloor \lfloor x \rfloor + y \rfloor + 2k = \lfloor x \rfloor + \lfloor y \rfloor + 2k = \lfloor x + \lfloor y \rfloor + 2k \rfloor = \lfloor x + f(f(y)) \rfloor$ We'll now prove that there are no other solutions. Let $P(x,y)$ be the assertion of $x$ and $y$ to the given functional equation.

Claim 01. $\text{Im}(f) \subseteq \mathbb{Z}$ .
Proof. Fix $\alpha \in \mathbb{R}$ . $P(0,\alpha - f(0))$ gives us
$f(\alpha) = f(f(0) + \alpha - f(0)) = \lfloor f(f(\alpha - f(0))) \rfloor \in \mathbb{Z}$
Claim 02. $f(a) = f(b) \implies \lfloor a \rfloor = \lfloor b \rfloor$ .
Proof. As $f(f(0)) \in \mathbb{Z}$ from the first claim, $P(a,0)$ and $P(b,0)$ gives us
$\lfloor a \rfloor + f(f(0)) = f(f(a)) = f(f(b)) = \lfloor b \rfloor + f(f(0)) \implies \lfloor a \rfloor = \lfloor b \rfloor$
To finish this problem, just note that
$f(f(0) + x) \stackrel{P(0,x)}{=} \lfloor f(f(x)) \rfloor \stackrel{\text{Claim 01}}{=} f(f(x)) \stackrel{\text{Claim 02}}{\implies} f(x) \stackrel{\text{Claim 01}}{=} \lfloor f(x) \rfloor = \lfloor x + f(0) \rfloor \stackrel{\text{Claim 01}}{=} \lfloor x \rfloor + f(0)$ as desired.

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Furra

#4 h Aug 29, 2023, 7:19 AM

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A bit more convoluted...

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This post has been edited 2 times. Last edited by Furra, Aug 29, 2023, 7:20 AM

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#5 h Aug 29, 2023, 7:38 AM • 2 Y

Y by Wildabandon, somebodyyouusedtoknow

I think everybody I know solved this problem, but anyway here is a short solution.
Let $P(x,y)$ be the assertion, clearly $P(x,x-f(x))$ gives us $f(x) \in \mathbb{Z} \quad \forall x \in \mathbb{R}$ .
$P(x-f(x),0)$ gives us
$f(x-f(x))=\lfloor x \rfloor-f(x)+f(f(0))$ By substituting it to $P(x,x-f(x))$ , we will get
$f(x)=2\lfloor x\rfloor -f(x)+f(f(0))$ From here, it's already clear what the answer is.

This post has been edited 1 time. Last edited by amogususususus, Aug 29, 2023, 7:39 AM

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#6 h Aug 29, 2023, 8:05 AM

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I was pretty surprised that the FE question reappeared this year, coupled with a bit of seasoning of the floor function, lol.

Let $P(x,y)$ be the assertation of
$f(f(x) + y) = \lfloor x + f(f(y)) \rfloor.$ From $P(x,0)$ , we have $f(f(x)) = \lfloor x +c\rfloor$ where $c=f(f(0))$ i.e.
$f(f(x)+y) = \lfloor x + f(f(y))\rfloor = \lfloor x +\lfloor y+c\rfloor \rfloor = \lfloor x \rfloor + \lfloor y+c\rfloor.$ From $P(x,0)$ we must have $\lfloor x+c\rfloor = \lfloor x\rfloor +\lfloor c\rfloor\implies \lfloor \{x\}+\{c\}\rfloor =0$ for all $x\in \mathbb{R}$ . If $c\not\in \mathbb{Z}$ , take $x=1-\{c\}$ and we have contradiction. Because $c\in \mathbb{Z}$ , we have $f(f(x)+y) = \lfloor x \rfloor + \lfloor y\rfloor +c$ . From $P(0,-f(0))$ we have $f(0) = \lfloor -f(0)\rfloor + c\implies f(0)\in \mathbb{Z}\implies f(0)=\frac{c}{2}=k\in \mathbb{Z}$ . From $P(0,x-f(0))$ we have
$f(x) = \lfloor x-f(0)+c\rfloor = \lfloor x-k+2k\rfloor = \lfloor x \rfloor +k,\;k\in \mathbb{Z}$ and we can check if this solution to the equation.

This post has been edited 3 times. Last edited by Wildabandon, Aug 29, 2023, 8:13 AM

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#7 h Aug 29, 2023, 5:46 PM • 1 Y

Y by AlexCenteno2007

Another one...

Denote $P(x,y)$ as the assertion of $x$ and $y$ to the given functional equation on the problem. From $P(x, x - f(x))$ , we obtain $f(x) = \lfloor \text{something} \rfloor$ , so $\text{Im}(f) \subseteq \mathbb{Z}$ .

From $P(f(x), 0)$ we obtain
$f(f(f(x))) = \lfloor f(x) + f(f(0)) \rfloor = f(x) + f(f(0)) \text{ }(\star)$ And therefore, from comparing $P(x, f(y))$ and $P(y, f(x))$ we obtain:
$\begin{align*} \lfloor x + f(f(f(y))) \rfloor = f(f(x) + f(y)) = \lfloor y + f(f(f(x))) \rfloor &\stackrel{\text{Im}(f) \subseteq \mathbb{Z}}{\implies} \lfloor x \rfloor + f(f(f(y))) = \lfloor y \rfloor + f(f(f(x))) \\ &\stackrel{(\star)}{\implies} \lfloor x \rfloor + f(y) + f(f(0)) = \lfloor y \rfloor + f(x) + f(f(0)) \\ &\implies f(x) - \lfloor x \rfloor = K, \end{align*}$ for some constant $K$ . Since $f(x)$ and $\lfloor x \rfloor$ are both integers, then $K$ must be an integer as well, so the solutions are of the form $\boxed{f(x)=\lfloor x \rfloor + K}$ , where $K \in \mathbb{Z}$ . It should be easy to check this into the original equation.

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#8 h Sep 1, 2023, 3:29 PM

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Let $P(x,y)$ be the assertion

$P(x,y-f(x)) = f(y) = \lfloor{x+f(f(y-f(x))}\rfloor$ for each real y the function is integer

$P(x-f(f(y)),y) = f(f(x-f(f(y)))+y) = \lfloor{x}\rfloor$ this implies f surjective over integer

so we can pick u integer so that $f(u) = 0$ proof

le $x = 0$ and $y \in \mathbb{Z}$ then we are done

$P(x,u) = f(f(x)+u) = u+f(0)$
$P(u,u) = 0 = u+f(0)$
$u = -f(0)$

$P(u,y) = f(y) = u+f(f(y))$
$P(-f(0),y) = f(y)+f(0) = +f(f(y))$ (1)

Rewrite the main equation

$P(x,y) = f(f(x)+y) = \lfloor{x}\rfloor+f(y)+f(0)$
$P(x,0) = f(f(x)) = \lfloor{x}\rfloor+2f(0) = f(x)+f(0)$ (From equation (one) we get )

$f(x) = \lfloor{x}\rfloor +f(0)$

This post has been edited 1 time. Last edited by StarLex1, Sep 1, 2023, 3:29 PM
Reason: edit

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#10 h Apr 15, 2025, 2:27 AM

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somebodyyouusedtoknow wrote:

Determine all functions $f : \mathbb{R} \to \mathbb{R}$ such that the following equation holds for every real $x,y$ :
$f(f(x) + y) = \lfloor x + f(f(y)) \rfloor.$ Note: $\lfloor x \rfloor$ denotes the greatest integer not greater than $x$ .

Let $c=f(f(0))$ , and $P(x,y)$ the assertion $f(f(x)+y)=\lfloor x+f(f(y))\rfloor$ .
$P(y,x-f(y))\Rightarrow f(x)=\lfloor y+f(f(x-f(y)))\rfloor\Rightarrow f(x)\in\mathbb Z$ for all $x\in\mathbb R$
$P(x,0)\Rightarrow f(f(x))=\lfloor x\rfloor+c$ (since $c\in\mathbb Z$ )
Then:
$f(\lfloor x\rfloor+f(f(0)))=f(f(f(x)))=\lfloor f(x)\rfloor+c=f(x)+c,$ so if $g:\mathbb Z\to\mathbb Z$ is defined by $g(n)=f(n+c)-c$ for all $n\in\mathbb Z$ we have $f(x)=g(\lfloor x\rfloor)$ for all $x\in\mathbb R$ . Then $P(x,y)$ rewrites as (using $f(f(x))=\lfloor x\rfloor+c$ as well):
$g(g(n)+m)=n+m+c$ for all $n,m\in\mathbb Z$ . Now $g$ is injective, and swapping $n,m$ we get $g(n)+m=g(m)+n$ , so $g(n)=n+g(0)$ . Then we can write $\boxed{f(x)=\lfloor x\rfloor+a}$ for some constant $a\in\mathbb Z$ (where $a=g(0)$ ) and all $x\in\mathbb R$ , all such functions work.

This post has been edited 1 time. Last edited by jasperE3, Apr 15, 2025, 2:28 AM

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