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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
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
Partitioning coprime integers to arithmetic sequences
sevket12   2
N 2 minutes ago by E0231
Source: 2025 Turkey EGMO TST P3
For a positive integer $n$, let $S_n$ be the set of positive integers that do not exceed $n$ and are coprime to $n$. Define $f(n)$ as the smallest positive integer that allows $S_n$ to be partitioned into $f(n)$ disjoint subsets, each forming an arithmetic progression.

Prove that there exist infinitely many pairs $(a, b)$ satisfying $a, b > 2025$, $a \mid b$, and $f(a) \nmid f(b)$.
2 replies
sevket12
Feb 8, 2025
E0231
2 minutes ago
Inspired by Bet667
sqing   0
3 minutes ago
Source: Own
Let $ a,b $ be a real numbers such that $a^3+kab+b^3\ge a^4+b^4.$Prove that
$$1-\sqrt{k+1} \leq  a+b\leq 1+\sqrt{k+1} $$Where $ k\geq 0. $
0 replies
1 viewing
sqing
3 minutes ago
0 replies
Prove that lines parallel in triangle
jasperE3   5
N 17 minutes ago by Thapakazi
Source: Mongolian MO 2007 Grade 11 P1
Let $M$ be the midpoint of the side $BC$ of triangle $ABC$. The bisector of the exterior angle of point $A$ intersects the side $BC$ in $D$. Let the circumcircle of triangle $ADM$ intersect the lines $AB$ and $AC$ in $E$ and $F$ respectively. If the midpoint of $EF$ is $N$, prove that $MN\parallel AD$.
5 replies
jasperE3
Apr 8, 2021
Thapakazi
17 minutes ago
JBMO Shortlist 2020 N6
Lukaluce   4
N 34 minutes ago by MR.1
Source: JBMO Shortlist 2020
Are there any positive integers $m$ and $n$ satisfying the equation

$m^3 = 9n^4 + 170n^2 + 289$ ?
4 replies
Lukaluce
Jul 4, 2021
MR.1
34 minutes ago
Nice concyclicity involving triangle, circle center, and midpoints
Kizaruno   0
37 minutes ago
Let triangle ABC be inscribed in a circle with center O. A line d intersects sides AB and AC at points E and D, respectively. Let M, N, and P be the midpoints of segments BD, CE, and DE, respectively. Let Q be the foot of the perpendicular from O to line DE. Prove that the points M, N, P, and Q lie on a circle.

0 replies
Kizaruno
37 minutes ago
0 replies
non-perfect square is non-quadratic residue mod some p
SpecialBeing2017   3
N an hour ago by ilovemath0402
If $n$ is not a perfect square, then there exists an odd prime $p$ s.t. $n$ is a quadratic non-residue mod $p$.
3 replies
SpecialBeing2017
Apr 14, 2023
ilovemath0402
an hour ago
Circles tangent at orthocenter
Achillys   62
N an hour ago by Rayvhs
Source: APMO 2018 P1
Let $H$ be the orthocenter of the triangle $ABC$. Let $M$ and $N$ be the midpoints of the sides $AB$ and $AC$, respectively. Assume that $H$ lies inside the quadrilateral $BMNC$ and that the circumcircles of triangles $BMH$ and $CNH$ are tangent to each other. The line through $H$ parallel to $BC$ intersects the circumcircles of the triangles $BMH$ and $CNH$ in the points $K$ and $L$, respectively. Let $F$ be the intersection point of $MK$ and $NL$ and let $J$ be the incenter of triangle $MHN$. Prove that $F J = F A$.
62 replies
Achillys
Jun 24, 2018
Rayvhs
an hour ago
Unsymmetric FE
Lahmacuncu   1
N an hour ago by ja.
Source: Own
Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ that satisfies $f(x^2+xy+y)+f(x^2y)+f(xy^2)=2f(xy)+f(x)+f(y)$ for all real $(x,y)$
1 reply
Lahmacuncu
2 hours ago
ja.
an hour ago
find angle
TBazar   3
N an hour 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 $\angle ACB$
3 replies
TBazar
Today at 6:57 AM
TBazar
an hour ago
Polynomial and Irrational Coefficients
Iamsohappy   1
N 2 hours ago by tom-nowy
Find all natural numbers $n$ such that there exists a monic polynomial of degree $n$ and with irrational coefficients ( excepts its leading term ) and such that it has $n$ distinct irrational roots.
1 reply
Iamsohappy
Jul 5, 2019
tom-nowy
2 hours ago
Divisibility on 101 integers
BR1F1SZ   5
N 3 hours ago by Grasshopper-
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
5 replies
BR1F1SZ
Aug 9, 2024
Grasshopper-
3 hours ago
Nice number theory problem
ItsBesi   9
N 3 hours ago by Jupiterballs
Source:  Kosovo Math Olympiad 2025, Grade 8, Problem 3
Let $m$ and $n$ be natural numbers such that $m^3-n^3$ is a prime number. What is the remainder of the number $m^3-n^3$ when divided by $6$?
9 replies
ItsBesi
Nov 17, 2024
Jupiterballs
3 hours ago
My Unsolved Problem
ZeltaQN2008   1
N 3 hours ago by wh0nix
Let \(f:[0,+\infty)\to\mathbb{R}\) be a function which is differentiable on \([0,+\infty)\) and satisfies
\[
\lim_{x\to+\infty}\bigl(f'(x)/e^x\bigr)=0.
\]Prove that
\[
\lim_{x\to+\infty}\bigl(f(x)/e^x\bigr)0.
\]
1 reply
ZeltaQN2008
4 hours ago
wh0nix
3 hours ago
Find the value
sqing   8
N 3 hours ago by xytunghoanh
Source: Own
Let $a,b,c$ be distinct real numbers such that $ \frac{a^2}{(a-b)^2}+ \frac{b^2}{(b-c)^2}+ \frac{c^2}{(c-a)^2} =1. $ Find the value of $\frac{a}{a-b}+ \frac{b}{b-c}+ \frac{c}{c-a}.$
Let $a,b,c$ be distinct real numbers such that $\frac{a^2}{(b-c)^2}+ \frac{b^2}{(c-a)^2}+ \frac{c^2}{(a-b)^2}=2. $ Find the value of $\frac{a}{b-c}+ \frac{b}{c-a}+ \frac{c}{a-b}.$
Let $a,b,c$ be distinct real numbers such that $\frac{(a+b)^2}{(a-b)^2}+ \frac{(b+c)^2}{(b-c)^2}+ \frac{(c+a)^2}{(c-a)^2}=2. $ Find the value of $\frac{a+b}{a-b}+\frac{b+c}{b-c}+ \frac{c+a}{c-a}.$
8 replies
sqing
Mar 17, 2025
xytunghoanh
3 hours ago
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
Valentin Vornicu   66
N Apr 25, 2025 by Ilikeminecraft
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
66 replies
Valentin Vornicu
Oct 24, 2005
Ilikeminecraft
Apr 25, 2025
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
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Valentin Vornicu
7301 posts
#1 • 16 Y
Y by Amir Hossein, Davi-8191, MarkBcc168, Understandingmathematics, itslumi, Adventure10, megarnie, RedFlame2112, Leo890, clevereagle, Mango247, cookie130, Dansman2838, PikaPika999, and 2 other users
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
This post has been edited 1 time. Last edited by Amir Hossein, Mar 21, 2016, 7:33 PM
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grobber
7849 posts
#2 • 15 Y
Y by viperstrike, laegolas, Wizard_32, to_chicken, Adventure10, DCMaths, Mango247, cookie130, MS_asdfgzxcvb, PikaPika999, and 5 other users
For any $k\ge 1$ there is such an $n$ with exactly $k$ prime factors.

For $k=1,\ n=3^t$ works for every $t\ge 1$. Take $t$ s.t. for $n_1=3^t,\ 2^{n_1}+1$ has a prime factor $p_2$ larger than $3$. Now take $n_2=n_1p_2$. Then $n_2|2^{n_1}+1|2^{n_2}+1$, and $2^{p_2}+1$ has a prime factor $p_3\not|n_2$. This is because $(2^{n_1}+1,2^{p_2}+1)=3,\ p_2\not|2^{p_2}+1$, and $2^{p_2}+1$ cannot be a power of $3$, since $p_2>3$ (I'm using the well known and easy to prove fact that the only positive integer solution $(a,b),a>1$ to $3^a=2^b+1$ is $(a,b)=(2,3)$). Then take $n_3=n_2p_3$. Just like above, we deduce that $2^{p_3}+1$ has a prime factor $p_4$ which is coprime to $n_3$, and take $n_4=n_3p_4$, and so on. $n_k$ will have exactly $k$ prime factors and will satisfy $n_k|2^{n_k}+1$.
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pluricomplex
390 posts
#3 • 3 Y
Y by Adventure10, Mango247, cookie130
Valentin Vornicu wrote:
Does there exist a positive integer $n$ such that $n$ has exactly 2000 prime divisors and $n$ divides $2^n + 1$?

You can find my paper for a general problem of this problem in This file
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Philip_Leszczynski
327 posts
#4 • 3 Y
Y by Adventure10, Mango247, cookie130
Let $N=2^n+1$. We will assume for the sake of contradiction that $n|N$.

$2^n+1 \equiv 0$ (mod $n$) $\Rightarrow 2^n \equiv -1$ (mod $n$). So 2 does not divide $n$, and so $n$ is odd.

Select an arbitrary prime factor of $n$ and call it $p$. Let's represent $n$ in the form $p^am$, where $m$ is not divisible by $p$.

Note that $p$ and $m$ are both odd since $n$ is odd. By repeated applications of Fermat's Little Theorem:

$N = 2^n+1 = 2^{p^am} + 1 = (2^{p^{a-1}m})^p + 1 \equiv 2^{p^{a-1}m} + 1$ (mod $p$)

Continuing in this manner, and inducting on k from 1 to $a$,

$2^{p^{a-k}m}+1 \equiv (2^{p^{a-k-1}m})^p + 1$ (mod $p$) $\equiv 2^{p^{a-k-1}m} + 1$ (mod $p$)

So we have $N \equiv 2^m+1$ (mod $p$)

Since $p$ is relatively prime to $m$, $N \equiv 1+1$ (mod $p$) $\equiv 2$ (mod $p$)

Since $p$ is odd, $N$ is not divisible by $p$. Hence $N$ is not divisible by $n$. So we have a contradiction, and our original assumption was false, and therefore $N$ is still not divisible by $n$.
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Philip_Leszczynski
327 posts
#5 • 3 Y
Y by Adventure10, Mango247, cookie130
Hmmm... I made a mistake here somewhere but I do not see it.
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Arne
3660 posts
#6 • 4 Y
Y by The_fandangos, Adventure10, Mango247, cookie130
Yes, since there are lots of integers $n$ such that $n$ divides $2^n + 1$, and the statement is true!
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Johann Peter Dirichlet
376 posts
#7 • 3 Y
Y by Adventure10, Mango247, cookie130
There exists a pretty beautiful generalization:

"
Let $s, a, b$ positive integers, such that $GCD(a,b) = 1$ and $a+b$ is not a 2-power.
Show that there exists infinitely many $n \in N$ such that

--- $n=p_1^{e_1} \cdot p_2^{e_2} \cdot p_3^{e_3} \cdots p_s^{e_s} \cdot$ is the canonical factoring of $n$.

--- $n|(a^n+b^n)$
"
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The QuattoMaster 6000
1184 posts
#8 • 3 Y
Y by Adventure10, Mango247, cookie130
Valentin Vornicu wrote:
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
Here is a solution that I don't think has been mentioned yet:
Solution
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Arquimedes
8 posts
#9 • 3 Y
Y by Adventure10, Mango247, cookie130
For any i,
2^3^i+1
is divisible by
3^i
(the proof is easy with euler`s theorem+induction and maybe with primitive roots (2 is primitive root modulo 3^i for any i)). Hence, for i=1999,
3^1999
has 2000 divisors and it satisfies the asked in the problem.

it is correct??

please , answer me.

bye

sorry for my english.
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L-b
11 posts
#10 • 3 Y
Y by Adventure10, Mango247, cookie130
Well, the point is to find a number which has exactly $ 2000$ prime divisiors, whereas $ 3^{1999}$ has only one ($ 3$).

But it is a very nice thought to look at powers of $ 3$, when the problem considers powers of $ 2$ (vide grobber's solution, which I do not completely understand yet, but it seems very nice and simple)
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Binomial-theorem
3982 posts
#11 • 5 Y
Y by JasperL, Anar24, Adventure10, Mango247, cookie130
Solution overkilling with Zsigmondy's
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v_Enhance
6877 posts
#12 • 18 Y
Y by Binomial-theorem, WL0410, e_plus_pi, Omeredip, Not_real_name, Takeya.O, TheHerculean11, ZHEKSHEN, Quidditch, HamstPan38825, Msn05, samrocksnature, joseph02, aidan0626, Adventure10, Mango247, MarioLuigi8972, cookie130
Answer: Yes.

We say that $n$ is Korean if $n \mid 2^n+1$. First, observe that $n=9$ is Korean. Now, the problem is solved upon the following claim:

Claim: If $n > 3$ is Korean, there exists a prime $p$ not dividing $n$ such that $np$ is Korean too.

Proof. I claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$, which exists by Zsigmondy theorem. Obviously $p \neq 2$. Then:
  • Since $p \nmid 2^{\varphi(n)}-1$ it follows then that $p \nmid n$.
  • Moreover, $p \mid 2^n+1$ since $p \nmid 2^n-1$;
Hence $np \mid 2^{np} + 1$ by Chinese Theorem, since $\gcd(n,p) = 1$. $\blacksquare$

EDIT: The version of the proof I posted four years ago was incorrect. This one should work.
This post has been edited 1 time. Last edited by v_Enhance, May 3, 2017, 1:08 AM
Reason: Wanlin Li pointed out a mistake
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lfetahu
134 posts
#13 • 6 Y
Y by Takeya.O, gghx, Adventure10, Mango247, cookie130, and 1 other user
I feel like it isn't interesting to make any new remark on this problem, but anyway I'm posting my approach too.

If we show that for any positive integer k, there exists a positive integer n with exactly k distinct prime divisors such that n | 2^n + 1, then we are done, since the problem asks us to examine a special case, more exactly k = 2000. Furthermore, we can even show that we can find these n's divisible by a power of 3, which will help us on our proof.

We use induction on k. k = 1, we can choose n(1) = 3, which clearly satisfies the conditions. Assume that k >= 1, and there exists n(k) = 3^t * m, where gcd(3, m) = 1, and m has exactly (k - 1) distinct prime divisors. So, we have n(k) | 2^(n(k)) + 1.
Before generating n(k+1) from n(k), let us look at the number 3n(k), which clearly has k distinct prime divisors. 2^(3n(k)) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1). Since we must have n(k) always odd because of the fact that n(k) = 1 (mod 2), we deduce that 3 | (2^(2n(k)) - 2^(n(k)) + 1), so we have that 3n(k) | 2^(3n(k)) + 1. It is enough to find a prime p, such that p | 2^(3n(k)) + 1 and p doesn't divide (2^(n(k)) + 1), which could guarantee us that p doesn't divide n(k) and consequently, we could generate n(k + 1) = 3n(k)*p, which could clearly work by observing that 2^(3n(k)*p) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1)*A. But, since we can pick up this prime p by Zsigmondy, we are done.

Note that in case of not using Zsigmondy, we can observe that gcd(a^2 - a + 1, a + 1) = gcd(3, a + 1) = 1 if a is not 2 mod 3 and 3 if a = 3k + 2. But if a = 3k + 2, then a^2 - a + 1 is divisible by 3 but not by 9, so we could pick up any prime p that divides (a^2 - a + 1) / 3.
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sayantanchakraborty
505 posts
#14 • 3 Y
Y by Adventure10, Mango247, cookie130
Let $n=\prod_{i=1}^{2000}{p_i}$ and wlog let $p_1<p_2<\dots<p_{2000}$.I dscribe a process on how to construct such an $n$.By the problem we have $2^{2n} \equiv 1\pmod{p_1} \Rightarrow ord_{p_1}{2} |2n$ and so by the minimality of $p_1$ we get that $ord_{p_1}=1$ or $ord_{p_1}=2$.In the first case we get $p_1|1$ which is absurd while in the second case we get $p_1|2^2-1=3 \Rightarrow \boxed{p_1=3}$.Similarly it is easy to note that $ord_{p_2}{2}|2n$ and so by the choice of $p_2$ we get $ord_{p_2}{2}|2*3=6$.Clearly there exists such a prime such that $ord_{p_2}{2}=6$(By Zsigmondy!!)In general we have $2^{2n} \equiv 1\pmod{p_k} \Rightarrow ord_{p_k}2|p_1p_2\dots p_{k-1}$.Clearly there exists a prime such that $ord_{p_k}2=p_1p_2\dots p_{k-1}$(Again Zsigmondy!!).So we are done!!!(In fact by this procedure we may fix any number of primes).
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junioragd
314 posts
#15 • 3 Y
Y by Adventure10, Mango247, cookie130
Since $N=3^k$ all satisfay the condition.Now,it is enough to prove that numbers of the form $N=2{}^3{}^k+1$ have infinitely many primes dividing them,but this is easy to prove,since we have for $n<m$ $2{}^3{}^n+1$ divides $2{}^3{}^m+1$ so suppose opposite.Now,we just need to prove that $a+1$ and $a^3+1$ can't have identical sets of primes for $a>2$,and this is true because $GCD(a+1,a^2-a+1)$ is at most $3$ and $a^2-a+1>3$ for $a>2$ so we are done.
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