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Functional Inequality Implies Uniform Sign
peace09   32
N 36 minutes ago by lelouchvigeo
Source: 2023 ISL A2
Let $\mathbb{R}$ be the set of real numbers. Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function such that \[f(x+y)f(x-y)\geqslant f(x)^2-f(y)^2\]for every $x,y\in\mathbb{R}$. Assume that the inequality is strict for some $x_0,y_0\in\mathbb{R}$.

Prove that either $f(x)\geqslant 0$ for every $x\in\mathbb{R}$ or $f(x)\leqslant 0$ for every $x\in\mathbb{R}$.
32 replies
peace09
Jul 17, 2024
lelouchvigeo
36 minutes ago
Calculating sum of the numbers
Sadigly   0
41 minutes ago
Source: Azerbaijan Junior MO 2025 P4
A $3\times3$ square is filled with numbers $1;2;3...;9$.The numbers inside four $2\times2$ squares is calculated,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?

$\text{a)}$ $24,24,25,25$

$\text{b)}$ $20,23,26,29$
0 replies
Sadigly
41 minutes ago
0 replies
JBMO type Combinatorics
Sadigly   0
an hour ago
Source: Azerbaijan Junior MO 2025 P3
Alice and Bob take turns taking balloons from a box containing infinitely many balloons. In the first turn, Alice takes $k_1$ amount of balloons, where $\gcd(30;k_1)\neq1$. Then, on his first turn, Bob takes $k_2$ amount of ballons where $k_1<k_2<2k_1$. After first turn, Alice and Bob alternately takes as many balloons as his/her partner has. Is it possible for Bob to take $k_2$ amount of balloons at first, such that after a finite amount of turns, one of them have a number of balloons that is a multiple of $2025^{2025}$?
0 replies
Sadigly
an hour ago
0 replies
(2^{4n+2}+1)/65 (i) integer, (ii) prime
parmenides51   5
N an hour ago by MR.1
Source: JBMO 2016 Shortlist N3
Find all positive integers $n$ such that the number $A_n =\frac{ 2^{4n+2}+1}{65}$ is
a) an integer,
b) a prime.
5 replies
parmenides51
Oct 14, 2017
MR.1
an hour ago
Divisibility..
Sadigly   0
an hour ago
Source: Azerbaijan Junior MO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
0 replies
Sadigly
an hour ago
0 replies
This question just asks if you can factorise 12 factorial or not
Sadigly   0
an hour ago
Source: Azerbaijan Junior MO 2025 P1
A teacher creates a fraction using numbers from $1$ to $12$ (including $12$). He writes some of the numbers on the numerator, and writes $\times$ (product) between each number. Then he writes the rest of the numbers in the denominator and also writes $\times$ between each number. There is at least one number both in numerator and denominator. The teacher ensures that the fraction is equal to the smallest possible integer possible.

What is this positive integer, which is also the value of the fraction?
0 replies
Sadigly
an hour ago
0 replies
harmonic quadrilateral
Lukariman   1
N an hour ago by Lukariman
Given quadrilateral ABCD inscribed in a circle with center O. CA:CB= DA:DB are satisfied. M is any point and d is a line parallel to MC. Radial projection M transforms A,B,D onto line d into A',B',D'. Prove that B' is the midpoint of A'D'.
1 reply
Lukariman
2 hours ago
Lukariman
an hour ago
Factorising and prime numbers...
Sadigly   4
N an hour ago by Nuran2010
Source: Azerbaijan Senior MO 2025 P4
Prove that for any $p>2$ prime number, there exists only one positive number $n$ that makes the equation $n^2-np$ a perfect square of a positive integer
4 replies
Sadigly
Yesterday at 4:19 PM
Nuran2010
an hour ago
find positive n so that exists prime p with p^n-(p-1)^n$ a power of 3
parmenides51   15
N an hour ago by MR.1
Source: JBMO Shortlist 2017 NT5
Find all positive integers $n$ such that there exists a prime number $p$, such that $p^n-(p-1)^n$ is a power of $3$.

Note. A power of $3$ is a number of the form $3^a$ where $a$ is a positive integer.
15 replies
parmenides51
Jul 25, 2018
MR.1
an hour ago
Cute Inequality
EthanWYX2009   1
N an hour ago by lbh_qys
Let $a_1,\ldots ,a_n\in\mathbb R\backslash\{0\},$ determine the minimum and maximum value of
\[\frac{\sum_{i,j=1}^n|a_i+a_j|}{\sum_{i=1}^n|a_i|}.\]
1 reply
1 viewing
EthanWYX2009
Today at 2:01 AM
lbh_qys
an hour ago
Two permutations
Nima Ahmadi Pour   12
N Apr 23, 2025 by Zhaom
Source: Iran prepration exam
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
12 replies
Nima Ahmadi Pour
Apr 24, 2006
Zhaom
Apr 23, 2025
Two permutations
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Source: Iran prepration exam
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Nima Ahmadi Pour
160 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Suppose that $ a_1$, $ a_2$, $ \ldots$, $ a_n$ are integers such that $ n\mid a_1 + a_2 + \ldots + a_n$.
Prove that there exist two permutations $ \left(b_1,b_2,\ldots,b_n\right)$ and $ \left(c_1,c_2,\ldots,c_n\right)$ of $ \left(1,2,\ldots,n\right)$ such that for each integer $ i$ with $ 1\leq i\leq n$, we have
\[ n\mid a_i - b_i - c_i
\]

Proposed by Ricky Liu & Zuming Feng, USA
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ZetaX
7579 posts
#2 • 3 Y
Y by Adventure10, Mango247, and 1 other user
This problem and the following generalisation appeared 1979 in Ars Combinatoria (thanks to Darij who found it):

Let $ (G, + )$ be a finite abelian group of order $ n$.
Let also $ a_1,a_2,...,a_{n - 1} \in G$ be arbitrary.
Then there exist pairwise distinct $ b_1,b_2,...,b_{n - 1} \in G$ and pairwise distinct $ c_1,c_2,...,c_{n - 1} \in G$ such that $ a_k = b_k + c_k$ for $ k = 1,2,...,n - 1$.

[Moderator edit: The Ars Combinatoria paper is:
F. Salzborn, G. Szekeres, A problem in Combinatorial Group Theory, Ars Combinatoria 7 (1979), pp. 3-5.]
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epitomy01
240 posts
#3 • 3 Y
Y by Dan37kosothangnao, Adventure10, Mango247
so could someone post a proof of either the problem or its generalization?
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spdf
136 posts
#4 • 1 Y
Y by Adventure10
you can find the proof in file shortlist 2005 which has been posted by orl
The main idea is given two sequence $a_{1}...a_{n}$ and $b_{1}...b_{n}$ s.t $\sum a_{i}\equiv 0(mod n)$ and $\sum b_{i}\equiv 0(mod n)$ and there are exactly two i;j s.t $a_{i}\neq\ b_{i}(modn)$ and $a_{j}\neq\ b_{j}(modn)$.Then if we know two permutation good for the sequence (a_1...a_n) then we can build two permuttionm good for (b_1...b_n)
i will come back with detail if you need
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ZetaX
7579 posts
#5 • 1 Y
Y by Adventure10
Well, I will post the solution from Ars Combinatoria if a re-find that two sheets of paper...
It's a bit different from the ISL one.
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keira_khtn
485 posts
#6 • 2 Y
Y by Adventure10, Mango247
I think you didnt keep promise,Zetax :lol: Please post it here and now!
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bodom
123 posts
#7 • 2 Y
Y by Adventure10, Mango247
to spdf: that was also my idea when i first saw the problem but i can't find a good way to contruct those 2 permutations for $ b_j$.you said you can post details.please do so :)
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ZetaX
7579 posts
#8 • 2 Y
Y by Adventure10, Mango247
Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).
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arandomperson123
430 posts
#9 • 2 Y
Y by Adventure10, Mango247
ZetaX wrote:
Sorry for not repsonding (I merely forgot...). But I just saw that problem again: In a slightly different manner (but being equivalent to this one here) it is solved in "The Mathematics of Juggling", called the "Converse of the Average Theorem".

Main ideas:
You show that this property (being a sum of two permutations) is invariant under the operations $ a_{i,j,d}$ that add $ d$ to $ a_i$ and subtract $ d$ from $ a_j$.
For this, you need to do it algorithmically (but describing it is a bit hard without that graphics...).

that is what I tried to do, but I can not prove that we can do it for the general case... can someone please help?
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ThE-dArK-lOrD
4071 posts
#10 • 16 Y
Y by Lam.DL.01, Mosquitall, nmd27082001, Arc_archer, MathbugAOPS, iceillusion, Aryan-23, magicarrow, k12byda5h, gabrupro, Mop2018, Adventure10, Mango247, bhan2025, CyclicISLscelesTrapezoid, winniep008hfi
Since it's almost twelve years without complete solution, here's the official solution:

Suppose there exists permutations $\sigma$ and $\tau$ of $[n]$ for some sequence $\{ a_i\}_{i\in [n]}$ so that $a_i\equiv_n \sigma (i)+\tau (i)$ for all $i\in [n]$.
Given a sequence $\{ b_i\}_{i\in [n]}$ with sum divisible by $n$ that differ, in modulo $n$, from $\{ a_i\}_{i\in [n]}$ in only two positions, say $i_1$ and $i_2$.
We want to construct permutations $\sigma'$ and $\tau'$ of $[n]$ so that $b_i\equiv_n \sigma' (i) +\tau' (i)$ for all $i\in [n]$.
Recall that $b_i\equiv a_i\pmod{n}$ for all $i\in [n]$ that $i\neq i_1,i_2$.
Construct a sequence $i_1,i_2,i_3,...$ by, for each integer $k\geq 2$, define $i_{k+1}\in [n]$ to be the unique integer satisfy $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$.
Let (clearly exists) $p<q$ are the indices that $i_p=i_q$ with minimal $p$, and then minimal $q$.

If $p>2$. This means $i_j\not\in \{ i_1,i_2\} \implies \sigma (i_j) +\tau (i_j) \equiv_n b_{i_j}$ for all $j\in \{ p,p+1,...,q\}$.
Summing the equation $\sigma (i_{k-1})+\tau (i_{k+1})\equiv_n b_{i_k}$ for $k\in \{ p,p+1,...,q-1\}$ gives us
$$\sum_{j=p-1}^{q-2}{\sigma (i_j) } +\sum_{j=p+1}^{q}{\tau (i_j)} \equiv_n\sum_{j=p}^{q-1}{b_{i_j}} \implies \sigma (i_{p-1}) +\sigma (i_p) +\tau (i_{q-1}) +\tau (i_q) \equiv_nb_{i_p}+b_{i_{q-1}}.$$Plugging $i_p=i_q$ and use $\sigma (i_p) +\tau (i_p)\equiv_n b_{i_p}$ gives us $\sigma (i_{p-1}) +\tau (i_{q-1})\equiv_n b_{i_{q-1}} \equiv_n \sigma (i_{q-1})+\tau (i_{q-1})$.
Hence, $\sigma (i_{p-1}) \equiv_n \sigma (i_{q-1})\implies i_{p-1}=i_{q-1}$, contradiction to the definition of $p,q$.

So, we've $p\in \{ 1,2\}$. Let $p'=3-p$. Define the desired permutations $\sigma'$ and $\tau'$ as follows:
$$\sigma' (i_l)=\begin{cases} 
\sigma (i_{l-1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\sigma (i_{q-1}), & \text{ if } l=1
\end{cases} ,\tau' (i_l)= \begin{cases} 
\tau (i_{l+1}), & \text{ if } l\in \{ 2,3,...,q-1\} \\
\tau (i_{p'}), & \text{ if } l=1
\end{cases}  $$and $\sigma' (i) =\sigma (i),\tau' (i)=\tau (i)$ for the rest $i\in [n]$ that $i\not\in \{ i_1,i_2,...,i_{q-1}\}$.
Note that the reason we choose $\tau (i_{p'})$ is just to not use $\tau (i_p)=\tau (i_{(q-1)+1})$ more than one time.
This construction gives us $\sigma' (i)+\tau' (i)\equiv_n b_i$ for all $i\in [n]$ except when $i=i_1$.
But since both $\sigma'$ and $\tau'$ are permutations of $[n]$, we've $\sum_{i\in [n]}{(\sigma' (i)+\tau' (i))} \equiv_n 2\times \frac{n(n-1)}{2}\equiv_n 0\equiv_n \sum_{i\in [n]}{b_i}$.
This guarantee that $\sigma' (i) +\tau' (i)\equiv_n b_i$ when $i=i_1$ too. This prove the validity of permutations we constructed.
This post has been edited 3 times. Last edited by ThE-dArK-lOrD, Jan 16, 2018, 3:07 PM
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mathleticguyyy
3217 posts
#11
Y by
The case with $n$ prime is also resolved in this paper
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Bataw
43 posts
#13
Y by
any other solutions ?
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Zhaom
5124 posts
#16
Y by
:star_struck:

Plot $\left(b_k,c_k\right)$ for $k=1,2,\cdots,n$ on the coordinate plane with coordinates modulo $n$. We want to show that it is possible to choose the $n$ points plotted to have $\left\{b_1+c_1,b_2+c_2,\cdots,b_n+c_n\right\}$ be any multiset of $n$ elements $\pmod{n}$ such that sum of the elements is $0\pmod{n}$. Note that the $n$ points plotted can be any $n$ points not in the same row or column.

Instead, we will prove that we can choose $n-1$ points not in the same row or column such that the multiset $S$ of $x+y$ for all of the points $(x,y)$ can be any multiset of $n-1$ elements $\pmod{n}$. Then, we can choose the unique point not in the same row or column as the $n-1$ points to obtain $n$ points for the original statement, as the sum of the coordinates of all $n$ points is necessarily $2\cdot\left(0+1+\cdots+(n-1)\right)\equiv(n-1)n\equiv0\pmod{n}$.

We will construct the $n-1$ points in the following way.

We start with $n-1$ arbitrary points not in the same row or column. We will change the $n-1$ points in such a way that we can replace any element of $S$ by any residue $\pmod{n}$. Suppose we change the element in $S$ corresponding to a point $P$. Then, we will shift $P$ horizontally until the sum of the coordinates of $P$ is the desired value $\pmod{n}$. Now, we might have two points in the same column. We will repeatedly perform the following operation.

Note that there are exactly $n-1$ distinct rows taken up by the $n-1$ points. If the last point moved was $P$ and it is in the same column as $Q$, we will move $Q$ on the line through $Q$ with slope $-1$ until $Q$ occupies the previously empty row.

It suffices that this cannot last forever.

Claim. This operation is a reflection over a fixed line.

Proof. Suppose that this operation moves $P$ to a point $P'$, then on the next step it moves $Q$ to a point $Q'$. We see that $\overline{PP'}$ and $\overline{QQ'}$ both have slope $-1$. Now, note that $P'$ and $Q$ must be in the same row. Furthermore, after the operation moves $P$ to $P'$, the row containing $P$ must be empty, meaning that $Q$ must be moved to that row. This implies that $P$ and $Q'$ are in the same row. Therefore, we see that $PP'QQ'$ is a cyclic isosceles trapezoid and in particular the perpendicular bisectors of $\overline{PP'}$ and $\overline{QQ'}$ are the same, proving the claim.

Now, assume for the sake of contradiction that the operation lasts forever. We will label the $n-1$ points to distinguish them. Now, there are finitely many states of the $n-1$ points, so eventually the operation must loop through a cycle of states. In this cycle, start at an arbitrary state. Now, since at least $1$ point must be moved and then return back to its original position, consider the point $P$ which returns to its original position the earliest. When $P$ was moved, the next point $Q$ which was moved must have occupied the original row of $P$. However, if $P$ was moved back to its original position, then the original row of $P$ must be vacant, so $Q$ must have been moved. However, this implies that $Q$ was moved back to its original position before $P$ by the claim, a contradiction, so we are done.
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