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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
x^2 + 3y^2 = 8n + 4
Ink68   0
5 minutes ago
Let $n$ be a positive integer. Let $A$ be the number of pairs of $(x,y)$ satisfying $x^2 + 3y^2 = 8n + 4$ for odd values of $x$. Let $B$ be the number of pairs of $(x,y)$ satisfying $x^2 + 3y^2 = 8n + 4$. Prove that $A = \frac {2}{3} B$.
0 replies
Ink68
5 minutes ago
0 replies
At least k points of S equidistant from P
orl   9
N 26 minutes ago by Twan
Source: IMO 1989/3 , ISL 20, ILL 66
Let $ n$ and $ k$ be positive integers and let $ S$ be a set of $ n$ points in the plane such that

i.) no three points of $ S$ are collinear, and

ii.) for every point $ P$ of $ S$ there are at least $ k$ points of $ S$ equidistant from $ P.$

Prove that:
\[ k < \frac {1}{2} + \sqrt {2 \cdot n}
\]
9 replies
orl
Nov 19, 2005
Twan
26 minutes ago
Gergonne point Harmonic quadrilateral
niwobin   1
N 35 minutes ago by on_gale
Triangle ABC has incircle touching the sides at D, E, F as shown.
AD, BE, CF concurrent at Gergonne point G.
BG and CG cuts the incircle at X and Y, respectively.
AG cuts the incircle at K.
Prove: K, X, D, Y form a harmonic quadrilateral. (KX/KY = DX/DY)
1 reply
niwobin
Yesterday at 8:17 PM
on_gale
35 minutes ago
Find the minimum
sqing   8
N an hour ago by sqing
Source: China Shandong High School Mathematics Competition 2025 Q4
Let $ a,b,c>0,abc>1$. Find the minimum value of $ \frac {abc(a+b+c+8)}{abc-1}. $
8 replies
sqing
Yesterday at 9:12 AM
sqing
an hour ago
Interesting inequalities
sqing   3
N an hour ago by sqing
Source: Own
Let $ a,b >0 $ and $ a^2-ab+b^2\leq 1 $ . Prove that
$$a^4 +b^4+\frac{a }{b +1}+ \frac{b }{a +1} \leq 3$$$$a^3 +b^3+\frac{a^2}{b^2+1}+ \frac{b^2}{a^2+1} \leq 3$$$$a^4 +b^4-\frac{a}{b+1}-\frac{b}{a+1} \leq 1$$$$a^4+b^4 -\frac{a^2}{b^2+1}- \frac{b^2}{a^2+1}\leq 1$$$$a^3+b^3 -\frac{a^3}{b^3+1}- \frac{b^3}{a^3+1}\leq 1$$
3 replies
sqing
May 9, 2025
sqing
an hour ago
Marking vertices in splitted triangle
mathisreal   2
N an hour ago by sopaconk
Source: Mexico
Let $n$ be a positive integer. Consider a figure of a equilateral triangle of side $n$ and splitted in $n^2$ small equilateral triangles of side $1$. One will mark some of the $1+2+\dots+(n+1)$ vertices of the small triangles, such that for every integer $k\geq 1$, there is not any trapezoid(trapezium), whose the sides are $(1,k,1,k+1)$, with all the vertices marked. Furthermore, there are no small triangle(side $1$) have your three vertices marked. Determine the greatest quantity of marked vertices.
2 replies
mathisreal
Feb 7, 2022
sopaconk
an hour ago
distance of a point from incircle equals to a diameter of incircle
parmenides51   5
N an hour ago by Captainscrubz
Source: 2019 Oral Moscow Geometry Olympiad grades 8-9 p1
In the triangle $ABC, I$ is the center of the inscribed circle, point $M$ lies on the side of $BC$, with $\angle BIM = 90^o$. Prove that the distance from point $M$ to line $AB$ is equal to the diameter of the circle inscribed in triangle $ABC$
5 replies
parmenides51
May 21, 2019
Captainscrubz
an hour ago
f(a + b) = f(a) + f(b) + f(c) + f(d) in N-{O}, with 2ab = c^2 + d^2
parmenides51   8
N 5 hours ago by TiagoCavalcante
Source: RMM Shortlist 2016 A1
Determine all functions $f$ from the set of non-negative integers to itself such that $f(a + b) = f(a) + f(b) + f(c) + f(d)$, whenever $a, b, c, d$, are non-negative integers satisfying $2ab = c^2 + d^2$.
8 replies
parmenides51
Jul 4, 2019
TiagoCavalcante
5 hours ago
Functional Inequality Implies Uniform Sign
peace09   33
N 5 hours ago by ezpotd
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 $x\in\mathbb{R}$.
33 replies
peace09
Jul 17, 2024
ezpotd
5 hours ago
Labelling edges of Kn
oVlad   1
N 6 hours ago by TopGbulliedU
Source: Romania Junior TST 2025 Day 2 P3
Let $n\geqslant 3$ be an integer. Ion draws a regular $n$-gon and all its diagonals. On every diagonal and edge, Ion writes a positive integer, such that for any triangle formed with the vertices of the $n$-gon, one of the numbers on its edges is the sum of the two other numbers on its edges. Determine the smallest possible number of distinct values that Ion can write.
1 reply
oVlad
May 6, 2025
TopGbulliedU
6 hours ago
c^a + a = 2^b
Havu   8
N Yesterday at 8:29 PM by MathematicalArceus
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
8 replies
Havu
May 10, 2025
MathematicalArceus
Yesterday at 8:29 PM
Concurrence of lines defined by intersections of circles
Lukaluce   1
N Yesterday at 8:28 PM by sarjinius
Source: 2025 Macedonian Balkan Math Olympiad TST Problem 2
Let $\triangle ABC$ be an acute-angled triangle and $A_1, B_1$, and $C_1$ be the feet of the altitudes from $A, B$, and $C$, respectively. On the rays $AA_1, BB_1$, and $CC_1$, we have points $A_2, B_2$, and $C_2$ respectively, lying outside of $\triangle ABC$, such that
\[\frac{A_1A_2}{AA_1} = \frac{B_1B_2}{BB_1} = \frac{C_1C_2}{CC_1}.\]If the intersections of $B_1C_2$ and $B_2C_1$, $C_1A_2$ and $C_2A_1$, and $A_1B_2$ and $A_2B_1$ are $A', B'$, and $C'$ respectively, prove that $AA', BB'$, and $CC'$ have a common point.
1 reply
Lukaluce
Apr 14, 2025
sarjinius
Yesterday at 8:28 PM
Factorial Divisibility
Aryan-23   45
N Yesterday at 8:28 PM by MathematicalArceus
Source: IMO SL 2022 N2
Find all positive integers $n>2$ such that
$$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$$
45 replies
Aryan-23
Jul 9, 2023
MathematicalArceus
Yesterday at 8:28 PM
Multiple of multinomial coefficient is an integer
orl   14
N Yesterday at 8:23 PM by mickeymouse7133
Source: Romanian Master in Mathematics 2009, Problem 1
For $ a_i \in \mathbb{Z}^ +$, $ i = 1, \ldots, k$, and $ n = \sum^k_{i = 1} a_i$, let $ d = \gcd(a_1, \ldots, a_k)$ denote the greatest common divisor of $ a_1, \ldots, a_k$.
Prove that $ \frac {d} {n} \cdot \frac {n!}{\prod\limits^k_{i = 1} (a_i!)}$ is an integer.

Dan Schwarz, Romania
14 replies
orl
Mar 7, 2009
mickeymouse7133
Yesterday at 8:23 PM
Gcd of N and its coprime pair sum
EeEeRUT   16
N May 9, 2025 by ravengsd
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
16 replies
EeEeRUT
Apr 16, 2025
ravengsd
May 9, 2025
Gcd of N and its coprime pair sum
G H J
G H BBookmark kLocked kLocked NReply
Source: EGMO 2025 P1
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EeEeRUT
82 posts
#1 • 4 Y
Y by dangerousliri, R8kt, MuhammadAmmar, SatisfiedMagma
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
This post has been edited 3 times. Last edited by EeEeRUT, May 11, 2025, 11:49 AM
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MathLuis
1536 posts
#2 • 2 Y
Y by R8kt, HeshTarg
We claim that all $N$ even and power of $3$ work.
To see they do for $N$ even it is trivial as all $c_i$'s are odd and for $N$ power of $3$ just notice the $c_i$'s cycle between being $1,2 \pmod 3$ and thus the sum of consecutive terms is always divisible by $3$.
Now suppose $N$ was odd but not a power of $3$ then notice $c_1=1, c_2=2$ so $3 \mid N$ and thus we can consider $N=3^k \cdot \ell$ for $k \ge 1$ and if $\ell=3m+1$ then notice $3m-1, 3m+2$ are both coprime to $\ell$ and are consecutive coprimes for obvious reasons so we must have $\gcd(N, 6m+1)>1$ however if they did share a prime divisor then from euclid alg it divides $3^k$ and thus it has to be $3$ which is a contradiction, a similar thing can be done for $\ell=3m+2$ thus we are done :cool:.
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ItzsleepyXD
147 posts
#3 • 1 Y
Y by R8kt
Ans is all $N$ even and $N =3^m$ .
note that all even is true (easy to see)
claim 1 if $2 \mid N+1$ then $3 \mid N$
after that is easy (I am lazy to retype the solution again)
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NicoN9
157 posts
#4 • 2 Y
Y by R8kt, shafikbara48593762
Why is that deleted? rewriting!

The answer is all even number, and power of $3$ (and $N\ge 3$.) These $N$ works since if $N$ is even, then all $c_i$ are odd, and if $N$ is power of $3$, then $c_1, c_2, c_3, \dots$ are $1, 2, 1, 2, \dots \pmod 3$, which works.

Also, we see that if $N$ odd, then $c_1=1$ and $c_2=2$, thus $3\mid N$. Assume for a contradiction that $N$ has prime factors $3<p_1<\dots <p_r$ and let $P=p_1p_2\dots p_r$.

$\bullet$ if $P\equiv 1\pmod 3$, then there exists an integer $h$ with $c_h=P-2$, and $c_{h+1}=P+1$. Now,\[
\gcd(N, c_h+c_{h+1})=\gcd(N, 2P-1)=1.
\]contradiction.

$\bullet$ similarly for $P\equiv 2\pmod 3$, we have $c_k=P-1$ and $c_{k+1}=P+2$. $\gcd(N, 2P+1)=1$.

So we are done.
This post has been edited 1 time. Last edited by NicoN9, Apr 16, 2025, 8:05 AM
Reason: 3 divides N
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MuradSafarli
110 posts
#5 • 1 Y
Y by R8kt
Let’s consider two cases depending on whether \( N \) is even or odd:

Case 1: \( N \) is even.
In this case, any number coprime with \( N \) will be odd. Therefore, the sum \( C_i + C_{i+1} \) will be even, which means the GCD of \( N \) and \( C_i + C_{i+1} \) will always be greater than 1. So, the condition is satisfied for all even \( N \).

Now, let’s analyze the odd case.

Let \( C_1 = 1 \), \( C_2 = 2 \). According to the problem condition, \( \gcd(N, 3) > 1 \).
This implies \( N \) must be divisible by 3. Trying some small odd values divisible by 3, we see that \( N = 3, 9, 27 \) satisfy the condition, while \( N = 15, 21 \) do not.

So we explore two subcases:
Case 2.1: \( N = 3^a \)

In this form, every number congruent to 1 or 2 mod 3 is coprime with \( N \).
Let \( C_j = 3t + 1 \), then \( C_{j+1} = 3t + 2 \), so the sum \( C_j + C_{j+1} = 3t + 1 + 3t + 2 = 6t + 3 \), which is divisible by 3.
Similarly, for \( C_j = 3t + 2 \), we have \( C_{j+1} = 3t + 4 \), so \( C_j + C_{j+1} = 6t + 6 \), again divisible by 3.
Thus, any \( N = 3^a \), where \( a \) is a positive integer, satisfies the condition.

Case 2.2: \( N = 3^a \cdot k \), where \( \gcd(k, 3) = 1 \).
Then \( k \equiv 1 \) or \( 2 \mod 3 \).

Case 2.2.1: Let \( k \equiv 2 \mod 3 \).
Since \( N \) is odd, \( k \equiv 5 \mod 6 \), so write \( k = 6t + 5 \).
There exists an integer \( h\) such that \( C_h = 6f + 4 \).
Here, \( \gcd(6f + 4, 3) = 1 \), and \( \gcd(6f + 5, 6f + 4) = 1 \).
Then \( C_{h+1} = 6f + 7 \), and clearly \( \gcd(6f + 7, 6f + 5) = 1 \), \( \gcd(6f + 7, 3) = 1 \).
Now the sum \( C_h + C_{h+1} = 12f + 11 \).
Then,
\[
\gcd(12f + 11, N) = \gcd(12f + 11, 3^a \cdot (6f + 5)) = \gcd(12f + 11, 6f + 5)
\]\[
= \gcd(6f + 6, 6f + 5) = 1 \quad \text{(Contradiction!)}
\]Case 2.2.2:\( k \equiv 1 \mod 3 \), i.e., \( k = 6f + 1 \).
Then take \( C_h = 6f - 1 \), and \( C_{h+1} = 6f + 2 \).
Now:
\[
\gcd(N, C_h + C_{h+1}) = \gcd(12f + 1, N) = \gcd(12f + 1, 6f + 1) = \gcd(6f, 6f + 1) = 1 \quad \text{(Contradiction!)}
\]Final Answer:
1) \( N = 2k \), for any natural number \( k > 1 \)
2) \( N = 3^a \), for any natural number \( a \)
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SimplisticFormulas
118 posts
#6 • 1 Y
Y by R8kt
overcomplication be like
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Safal
170 posts
#7 • 1 Y
Y by R8kt
My Solution
This post has been edited 6 times. Last edited by Safal, Apr 17, 2025, 7:08 PM
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de-Kirschbaum
201 posts
#8 • 1 Y
Y by R8kt
We claim that the solutions are $N=2^kt, 3^k$.

First note that if $2 \mid N$ then we only have odd numbers left, and odd plus odd must be even so every $N=2^kt$ works. If $2 \nmid N$ then we know that $1=c_1, 2=c_2$ so $1+2=3 \mid N$. If $N=3^k$ then $c_1, c_2, \ldots , c_m \equiv 1, 2, \ldots \mod{3}$ and clearly $3 \mid c_i+c_{i+1}$ for all $i$ so $N=3^k$ also works.

If $N=3^kt$ where $2,3 \nmid t \neq 1$ then we know that there is some $h \in \{1,2\}$ such that $ht \equiv 2 \mod{3}$. Then, $ht$ won't be in $\{c_1,\ldots,c_m\}$ because it isn't coprime with $t$, $ht+1$ is coprime with $t$ but will be $0 \mod{3}$ so it won't be in $\{c_1,\ldots,c_m\}$ either. $ht+2$ will be $1 \mod {3}$ and $(ht+2,t)=(2,t)=1$ so $ht+2$ is coprime with $N$. Similarly $ht-1$ is coprime with $t$ and is $1 \mod{3}$ so $ht-1$ is coprime with $N$ and $ht-1, ht+2$ are actually consecutive coprime numbers, so we can take $ht-1+ht+2 =2ht+1$. This is clearly coprime with $t$ and it is $2 \mod{3}$ so it is coprime with $N$. Thus no $3^kt$ works if $2, 3 \nmid t$. So the only solutions are $N=2^kt, 3^k$.

(It's impossible for $ht+2 \leq 2t+2$ to be out of range because $2t+2 < 3t \leq N \implies 2<t$, since $t=1,2$ are both impossible this construction is always safe)
This post has been edited 1 time. Last edited by de-Kirschbaum, Apr 17, 2025, 8:09 PM
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MathematicalArceus
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#9 • 1 Y
Y by R8kt
Answers: The only possible solutions are $N=3^k$ and $N=2k \quad \forall k$.

$N=2k$ Note that obviously this works, because $c_i \in \{2k+1\}$ and sum of any odd number is always even and hence $c_i+c_{i+1}=2m \implies \gcd(2m,2k)>1$, the condition, we required.

$N=3^k$ Note that taking the set $\{c_i\}$ and writing the set under $\pmod{3}$ gives us: $\{1,-1,1,-1,\dots -1\}$ so for any $i$, $c_i+c_{i+1} \equiv 0 \pmod{3}$, granting our condition to be true.

No other sols exist: We only consider for odd $N$, because all even $N$ works. Note that if any odd $N$ works, this means $3 \mid N$ because for any odd $N, \text{ } c_1+c_2=1+2=3 \implies \gcd(N,3)>1 \implies 3 \mid N$. Now, let $N=3^kp_1^{a_1}p_2^{a_2}\dots p_n^{a_n}$. We consider $p_1p_2\dots p_n \pmod{3}$.

$p_1p_2\dots p_n \equiv 1 \pmod{3}$ Note this implies that $p_1p_2\dots p_n -2 \equiv 2 \pmod{3}$ and $p_1p_2\dots p_n+1 \equiv 2 \pmod{3}$. Adding we get $2p_1p_2\dots p_n-1 \equiv 1 \pmod{3}$, which is obviously coprime to $N$. Note that $c_i = p_1p_2\dots p_n-2, c_{i+1} = p_1p_2\dots p_n+1$ and hence we get $c_i+c_{i+1}$, such that its gcd with $N$ is 1.

Similarly $p_1p_2\dots p_n \equiv 2 \pmod{3}$ follow the same argument and hence, we are done!

Remark
This post has been edited 2 times. Last edited by MathematicalArceus, Apr 17, 2025, 7:41 PM
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Jupiterballs
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#10 • 2 Y
Y by R8kt, GeoKing
Just for the sake of not typing twice, here is a pdf solution :gleam:
Attachments:
EGMO 2025 Solution.pdf (32kb)
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kamatadu
480 posts
#11 • 2 Y
Y by R8kt, SilverBlaze_SY
Solved with SilverBlaze_SY.

We prove that $N=2k$ for $k\ge 2$ and $N=3^k$ for $k\ge 1$ are the only solutions.

First we show that these work.

For $N=2k$, note that all the integers coprime to $N$ are odd and summing two of them would give an even result. So the $\gcd$ is at least $2$ and we are done.

For $N=3^{k}$, note that the integers coprime to it alternate between $1$ and $2\pmod 3$. This means that sum of two consecutive integers among them is going to be divisible by $3$ which makes the $\gcd$ at least $3$ and we are done.

Now we show that the other cases are not possible. FTSOC assume that some other integer which is not even or a power of $3$ works.

Note that clearly $\gcd(N,1)=1$ and $\gcd(N,2)=1$ as $N$ is odd. Since we must have $\gcd(N,3)\not=1$, we get $3\mid N$.

Represent $N=3^{\alpha_1}\cdot q$ where $\alpha_1\ge 1$ and $\gcd(q,3) = 1$, $q\not=1$.

If $q\equiv 1\pmod 3$, then note that $\gcd(N,q-2)=\gcd(N,q+1)=1$. Also, $\gcd(N,(q-2)+(q+1))=\gcd(N, 2q-1) = 1$ which gives a contradiction.

If $q\equiv 2\pmod 3$, then note that $\gcd(N, q-1) = \gcd(N, q+2)=1$. Also, $\gcd(N,(q-1)+(q+2))=\gcd(N, 2q+1)=1$ which gives a contradiction too.

Therefore such a choice of $N$ is not possible and we are done.
This post has been edited 1 time. Last edited by kamatadu, Apr 18, 2025, 2:22 PM
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Mathgloggers
88 posts
#12
Y by
WE can use a type of "infinite ascent" here to show that only $3^k$ and all even numbers is the only solution:
FTSOC,consider
$N=3^K.p_1^{a_1}...p_i{a_i}$,but there must exist some $x>N$ for which :$gcd(3x+1,N) \neq 1\implies gcd(3x-1,N)=1=gcd(3x+2,N)$,hence we must have :
The prime dividing $3x-1+3x+2=6x+1$ in the prime list of our $N$ ,but here we see that $gcd(6x+1,N)=1$ hence we have to include another prime $q$ which is not in our list hence we can make our list infinitely big hence no numbers like this exists.

Why $(6x+1,N) =1$ as explained by Luis above also that if we continue on applying the EDA,we would reach somewhere : $6x+1|3^k.(3x+1)$ which is not not possible.
This post has been edited 2 times. Last edited by Mathgloggers, Apr 19, 2025, 9:32 AM
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John_Mgr
70 posts
#13
Y by
Claim: The only solution are N is even number and $N=3^k$ $\forall$$k\ge1$
Proof:
For N is even:
$c_i\equiv 1,2(mod3)$ as $gcd(N,c_i)=1$
i.e if $c_i\equiv 1(mod3)$ then, $c_{i+1}\equiv 2 (mod3)$ or vice versa $\implies gcd(N,c_i + c_{i+1})>1$
For $N=3^k$, similar to above
For $N=3^k\cdot t$ and $(3,t)=1$ $t>1$
case 1: $t\equiv1(mod3)$
$3\mid t-1$ so $3\nmid (t-2, t+1) $ $\implies gcd(3^k\cdot t, (t-2)+(t+1))=gcd(3^k\cdot t, 2t-1)=1$, we are done..
case 2:$t\equiv 2(mod3)$
$3\mid t-2$ and $3\nmid (t-1, t+2)$ $\implies gcd(3^k\cdot t,(t-1)+(t+2))=gcd(3^k\cdot t, 2t+1)=1$, we are done...
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EVKV
71 posts
#14 • 1 Y
Y by Nobitasolvesproblems1979
CASE 1 N is even

Claim: All even N works
Proof: As all $c_{i}$ are even the sum of 2 consecutive co-prime 2 divides the gcd thus its not 1

CASE 2 N is odd

Claim: 3|N
Proof: 1,2 are consecutive co primes and so gcd(N,3)$\neq$1 so 3|N

Claim: Only prime dividing N is 3
Proof: Assume the contrary lets assume $q_{1}, \cdots ,q_{n}$ are all the primes except 3 which divide N

Claim if $\prod q_{i}$ $\equiv 1$ mod 3 then there exists an r such that the product divides 3r+1 and $ 3r+1 \leq N$

Proof: clearly 3r+1= $\prod q_{i}$ $\leq N$ satisfies this

Claim if $\prod q_{i}$ $\equiv 2$ mod 3 then there exists an r such that the product divides 3r+1 and $ 3r+1 \leq N$

Proof: clearly 3r+1= $2\prod q_{i}$ $\leq N$ satisfies this

Now 3(r-1)+2 and 3r +2 are consecutive co primes
3(r-1)+2 + 3r +2 $\equiv -2+1$ $\equiv 1$ mod $q_{i}$ for all i $\leq n$
3(r-1)+2 + 3r +2 $\equiv 2+2$ $\equiv 1$ mod $3$
Thus gcd(N,3(r-1)+2 + 3r +2)=1
Contradiction

Claim: N=$3^a$ for $a \in N$

Proof: Clearly the consecutive co primes are $1+2 \equiv 0$ mod 3 thus gcd(N,$c_{i} + c_{i+1}$) $\neq1$



REMARK:
A nice problem tho I changed my solution making it more rigorous crt was def grt motivation

I rate it 5 mohs

SOLVED ON 21/4/25
ALSO 50TH POST YAY
This post has been edited 2 times. Last edited by EVKV, Apr 22, 2025, 5:42 PM
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zRevenant
14 posts
#15
Y by
(Livesolved on YouTube: Art of Olympiad Mathematics)

Answer: $n$ is even or a power of $3$.

Proof. If $n$ is even, then all $c_i$ are all odd, meaning that $c_i + c_{i+1}$ is even, which is never coprime to $n$ so it works. If $n=3^k$, then all the $c_i$'s are just numbers that are not divisible by $3$ meaning that the residues go $1, 2, 1, 2, ... $ mod $3$. Hence this works as well.

Now, let's suppose $n=3^k \cdot a$. Then we look at $a+1$ and $a-1$. At least one of them is not divisible by $3$, and by looking at the sums of adjacent ones we are done - either we get $a+(a+2)=2a+2$ which is coprime to $n$ if $3 | a+1$ or we get $a$+$a-2$ which works as well.
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Tuvshuu
11 posts
#16 • 1 Y
Y by Arcturuss
We will easy check all even $N$ works.
Now prove $N$ is odd. Then $c_1=1$ and $c_2=2$. Then $3 \mid N$.
Let $N = 3^k \times m$ and $m > 1$.
If $m \equiv 1\pmod{3}$ then $(m - 2, N) = 1$ and $(m + 1, N) = 1$. Now $(2m - 1, N) \ne 1$. But $(2m - 1, N) = (2m - 1, 3^k \times m) = (2m - 1, m) = (-1, m) = 1$ (Because $(2m - 1, 3) = 1$)
$m \equiv 2\pmod{3}$ same. Then $m = 1$ or $N = 3^k$.
Answer is $N$ is even or $N = 3^k$.
This post has been edited 1 time. Last edited by Tuvshuu, Apr 24, 2025, 2:02 PM
Reason: how i know
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ravengsd
17 posts
#17
Y by
Case 1: $N$ is even. Then all $c_i$ 's are odd, so $c_i+c_{i+1}$ is even, $\forall 1\leq i\leq N-1$, so $\gcd(N, c_i+c_{i+1})>1$, so all even $N$ work.
Case 2: $N$ is odd. Clearly, $N-1=c_{\varphi(N)}$ and since $N$ is odd, $N-2=c_{\varphi(N)-1}$, implying $\gcd(N,2N-3)=\gcd(N,3)>1$ so $N$ is a multiple of $3$. Write $N=3^a\cdot b$, with $(b,2)=(b,3)=1$. FTSoC suppose $b>1$.
Case 2.1: $b\equiv 1(\mod 3)$
Then, if $d=\gcd(3^a\cdot b, b-2)$, we have: $d\mid 3^a\cdot b$ and $d\mid 3^a\cdot b-2\cdot 3^a$, implying $d\mid 2\cdot 3^a$. However, the second condition implies $(d,6)=1$, so there exists $i<\varphi(N)$ such that $c_i=b-2$. Since $b-1\equiv 0(\mod 3)$, $c_{i+1}\geq b+1$. However, if there is some $d>1$ such that $d\mid 3^a\cdot b$ and $d\mid b+1$, then $d\mid 3^a$ which is impossible. So $c_{i+1}=b+1$, implying $\gcd(3^a\cdot b, 2b-1)>1$, which is clearly a contradiction.
Case 2.2: $b\equiv 2(\mod 3)$. Then we may do the same thing but for $b-1$ and $b+2$, which will also yield a contradiction.
So $b=1$ is the only possible case, which can be easily checked to work, because $c_i$s alternate $(\mod 3)$, so we're done $\blacksquare$
This post has been edited 2 times. Last edited by ravengsd, May 14, 2025, 10:58 AM
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