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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
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square root problem
kjhgyuio   6
N 2 minutes ago by kjhgyuio
........
6 replies
kjhgyuio
May 3, 2025
kjhgyuio
2 minutes ago
60 posts!(and a question )
kjhgyuio   0
12 minutes ago
Finally 60 posts :D
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kjhgyuio
12 minutes ago
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find angle
TBazar   5
N 16 minutes ago by TBazar
Given $ABC$ triangle with $AC>BC$. We take $M$, $N$ point on AC, AB respectively such that $AM=BC$, $CM=BN$. $BM$, $AN$ lines intersect at point $K$. If $2\angle AKM=\angle ACB$, find $\angle ACB$
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TBazar
Yesterday at 6:57 AM
TBazar
16 minutes ago
2 var inquality
Iveela   19
N 39 minutes ago by sqing
Source: Izho 2025 P1
Let $a, b$ be positive reals such that $a^3 + b^3 = ab + 1$. Prove that \[(a-b)^2 + a + b \geq 2\]
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Iveela
Jan 14, 2025
sqing
39 minutes ago
No more topics!
Simple FE on National Contest
somebodyyouusedtoknow   7
N Apr 15, 2025 by jasperE3
Source: INAMO 2023 P2 (OSN 2023)
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 $x$.
7 replies
somebodyyouusedtoknow
Aug 29, 2023
jasperE3
Apr 15, 2025
Simple FE on National Contest
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Source: INAMO 2023 P2 (OSN 2023)
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somebodyyouusedtoknow
259 posts
#1 • 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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IndoMathXdZ
691 posts
#2 • 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:
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
10 posts
#4
Y by
A bit more convoluted...

Click to reveal hidden text
This post has been edited 2 times. Last edited by Furra, Aug 29, 2023, 7:20 AM
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amogususususus
370 posts
#5 • 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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Wildabandon
507 posts
#6
Y by
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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Seicchi28
252 posts
#7 • 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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StarLex1
816 posts
#8
Y by
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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jasperE3
11310 posts
#10
Y by
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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