Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
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
Inspired by Austria 2025
sqing   1
N 3 minutes ago by sqing
Source: Own
Let $ a,b\geq 0 ,a,b\neq 1$ and $  a^2+b^2=1. $ Prove that$$   (a + b ) \left( \frac{a}{(b -1)^2} + \frac{b}{(a - 1)^2} \right) \geq 12+8\sqrt 2$$
1 reply
1 viewing
sqing
8 minutes ago
sqing
3 minutes ago
IMO Genre Predictions
ohiorizzler1434   51
N 12 minutes ago by ethan2011
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
51 replies
+1 w
ohiorizzler1434
May 3, 2025
ethan2011
12 minutes ago
Concurrency from isogonal Mittenpunkt configuration
MarkBcc168   17
N 2 hours ago by Ilikeminecraft
Source: Fake USAMO 2020 P3
Let $\triangle ABC$ be a scalene triangle with circumcenter $O$, incenter $I$, and incircle $\omega$. Let $\omega$ touch the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at points $D$, $E$, and $F$ respectively. Let $T$ be the projection of $D$ to $\overline{EF}$. The line $AT$ intersects the circumcircle of $\triangle ABC$ again at point $X\ne A$. The circumcircles of $\triangle AEX$ and $\triangle AFX$ intersect $\omega$ again at points $P\ne E$ and $Q\ne F$ respectively. Prove that the lines $EQ$, $FP$, and $OI$ are concurrent.

Proposed by MarkBcc168.
17 replies
MarkBcc168
Apr 28, 2020
Ilikeminecraft
2 hours ago
Inequality with a,b,c
GeoMorocco   7
N 2 hours ago by lele0305
Source: Morocco Training
Let $   a,b,c   $ be positive real numbers such that : $   ab+bc+ca=3   $ . Prove that : $$\frac{\sqrt{1+a^2}}{1+ab}+\frac{\sqrt{1+b^2}}{1+bc}+\frac{\sqrt{1+c^2}}{1+ca}\ge \sqrt{\frac{3(a+b+c)}{2}}$$
7 replies
GeoMorocco
Apr 11, 2025
lele0305
2 hours ago
Property of the divisors of k^3 - 2
Scilyse   2
N 3 hours ago by Assassino9931
Source: KoMaL A. 892
Given two integers, $k$ and $d$ such that $d$ divides $k^3 - 2$. Show that there exists integers $a$, $b$, $c$ satisfying $d = a^3 + 2b^3 + 4c^3 - 6abc$.

Proposed by Csongor Beke and László Bence Simon, Cambridge
2 replies
Scilyse
Jan 13, 2025
Assassino9931
3 hours ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   1
N 3 hours ago by sami1618
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
1 reply
BR1F1SZ
4 hours ago
sami1618
3 hours ago
Something nice
KhuongTrang   31
N 3 hours ago by NguyenVanHoa29
Source: own
Problem. Given $a,b,c$ be non-negative real numbers such that $ab+bc+ca=1.$ Prove that

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\sqrt{a+b+c+abc}.$$
31 replies
KhuongTrang
Nov 1, 2023
NguyenVanHoa29
3 hours ago
Nordic 2025 P3
anirbanbz   8
N 4 hours ago by lksb
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
8 replies
anirbanbz
Mar 25, 2025
lksb
4 hours ago
another functional inequality?
Scilyse   32
N 4 hours ago by ihategeo_1969
Source: 2023 ISL A4
Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\]for every $x, y \in \mathbb R_{>0}$.
32 replies
Scilyse
Jul 17, 2024
ihategeo_1969
4 hours ago
Mount Inequality erupts in all directions!
BR1F1SZ   1
N 4 hours ago by sami1618
Source: Austria National MO Part 1 Problem 1
Let $a$, $b$ and $c$ be pairwise distinct nonnegative real numbers. Prove that
\[
(a + b + c) \left( \frac{a}{(b - c)^2} + \frac{b}{(c - a)^2} + \frac{c}{(a - b)^2} \right) > 4.
\](Karl Czakler)
1 reply
BR1F1SZ
4 hours ago
sami1618
4 hours ago
Division involving difference of squares
BR1F1SZ   1
N 4 hours ago by grupyorum
Source: Austria National MO Part 1 Problem 4
Determine all integers $n$ that can be written in the form
\[
n = \frac{a^2 - b^2}{b},
\]where $a$ and $b$ are positive integers.

(Walther Janous)
1 reply
BR1F1SZ
4 hours ago
grupyorum
4 hours ago
Erasing the difference of two numbers
BR1F1SZ   0
4 hours ago
Source: Austria National MO Part 1 Problem 3
Consider the following game for a positive integer $n$. Initially, the numbers $1, 2, \ldots, n$ are written on a board. In each move, two numbers are selected such that their difference is also present on the board. This difference is then erased from the board. (For example, if the numbers $3,6,11$ and $17$ are on the board, then $3$ can be erased as $6 - 3=3$, or $6$ as $17 - 11=6$, or $11$ as $17 - 6=11$.)

For which values of $n$ is it possible to end with only one number remaining on the board?

(Michael Reitmeir)
0 replies
BR1F1SZ
4 hours ago
0 replies
4 wise men and 100 hats. 3 must guess their numbers
NO_SQUARES   2
N 5 hours ago by NO_SQUARES
Source: 239 MO 2025 10-11 p5
There are four wise men in a row, each sees only those following him in the row, i.e. the $1$st sees the other three, the $2$nd sees the $3$rd and $4$th, and the $3$rd sees only the $4$th. The devil has $100$ hats, numbered from $1$ to $100$, he puts one hat on each wise man, and hides the extra $96$ hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
2 replies
NO_SQUARES
Yesterday at 5:44 PM
NO_SQUARES
5 hours ago
\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}
NO_SQUARES   2
N 5 hours ago by ektorasmiliotis
Source: 239 MO 2025 8-9 p4
Positive numbers $a$, $b$ and $c$ are such that $a^2+b^2+c^2+abc=4$. Prove that \[\sqrt{2-a}+\sqrt{2-b}+\sqrt{2-c}\geqslant 2+\sqrt{(2-a)(2-b)(2-c)}.\]
2 replies
NO_SQUARES
Yesterday at 5:06 PM
ektorasmiliotis
5 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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
joshualiu315
2534 posts
#55
Y by
The answer is $\boxed{\text{yes}}$.

Let $p_x$ be a prime that divides $2^{3^x}+1$ but not $2^{3^{i}}+1$ for $i<x$. This prime will always exist by Zsigmondy, meaning our desired number is

\[n=3^{2000}p_2p_3\cdots p_{2000}.\]
Seeing that this works is trivial, so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
794 posts
#56
Y by
We note $n_1=3^2$ satisfies the condition. We aim to find a prime $p$ such that $pn_i$ satisfies the condition and $\gcd(p,n_i)=1$, so it suffices to have
\[2^{pn} \equiv -1 \pmod{p} \implies 2^{2n} \equiv 1, 2^n \not\equiv 1 \pmod{p}.\]
This primitive root of 2 modulo $p$ exists by Zsigmondy, as desired. Hence we can induct to find $n_{2000}$ with the desired 2000 prime divisors. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
peppapig_
281 posts
#57
Y by
Overkill?

In general, I claim that there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$.

I prove this by claiming that if there exists a positive integer $n$ with $k$ distinct integer divisors in the form of
\[n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k},\]such that $p_1<p_2\dots<p_k$, then there exists a prime $p_{k+1}>p_k$ such that for all positive integers $e_{k+1}$, it is true that $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the problem condition, where we can make $m$ any positive integer we want.

C1: First, before we begin, I claim that if any prime $q\mid n$, then $nq$ also satisfies problem conditions. This will then prove the part where we can make $m$ whatever we want. This is because
\[n\mid 2^n+1 \iff \nu_q(n)\leq \nu_q(2^n+1),\]and by LTE (which we can use, since $n\mid 2^n+1$ implies that $q\mid 2^n+1$), this gives us that
\[\nu_q(nq)=1+\nu_q(n)\leq 1+\nu_q(2^n+1)=\nu_q(2^{nq}+1),\]which means that $nq$ also satisfies the problem conditions, as desired. Therefore, by induction, we also get that $nq^m$ also satisfies problem conditions.

C2: Now, I claim that there exists a prime $p_{k+1}$ such that for all positive integers $e_{k+1}$, $np_k^mp_{k+1}^{e_{k+1}}$ satisfies the conditions. For simplicity, from here on out, I will refer to $p_{k+1}$ as $r$ and $p_k$ as $q$. Note that this implies that
\[r\mid 2^n+1 \iff ord_r(2)\mid 2n,\]and since $ord_r(2)\mid \phi(r)=r-1$, we get that $ord_r(2)\mid \gcd(2n,r-1)$. First, I claim that there always exists an $r$ that divides $2^{q^c}+1$ for some $c$ we can choose such that $r>q$. This is clearly true by Zsigmondy's theorem. I now claim that if we take this $r$, it is true that $rn$ also satisfies problem conditions. This, combined with (C1), will prove our master claim, which states that $nr^e_{k+1}$ satisfies problem conditions for any $e_{k+1}$.

To prove this, first make sure that our previous $n$ is divisible by the $q^c$ we chose. We can do this by making another $n$ that satisfies problem conditions by multiplying it by a power of $q$, which we can do by (C1). Next, by LTE, we have,
\[\nu_r(rn)=1\leq \nu_r(2^{rn}+1)=\nu_r(2^n+1)+1,\]which we can do since we know that $q^c \mid n$ and $r\mid 2^{q^c}+1$, which gives that $r\mid 2^n+1$. This means that $rn$ satisfies problem conditions, since $rn\mid 2^{rn}+1$, which combined with (C1), proves our inductive step claim.

Finally, to complete our induction, note that $3\mid 2^3+1$. Therefore, there exists a positive integer $n$ such that $n\mid 2^n+1$ with exactly $k$ distinct prime divisors for all positive integers $k$, finishing the problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SenorSloth
37 posts
#58
Y by
We claim that the answer is yes. We will induct to show that if there exists such an $n$ with $x$ distinct prime factors, then there also exists such an $n$ with $x+1$ distinct prime factors. By repeating this logic we can find an $n$ with exactly $2000$ distinct prime factors.

Our base case is $n=9$, which works since $9\mid 513$. By Zsigmondy, for any $n>3$, there exists a prime $p$ dividing $2^n+1$ that does not divide any $2^k+1$ for smaller $k$. Since this implies that $2$ has order $2n$ modulo $p$, the prime is at least $2n+1$ and thus cannot divide $n$. Thus we know that $pn\mid 2^n+1$ and consequently $pn \mid 2^{pn}+1$. $pn$ has $1$ more distinct prime factor than $n$, so the induction is complete.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Niku
120 posts
#59
Y by
Do you realise that this post was made 20 years ago.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BestAOPS
707 posts
#60
Y by
Overkill proof while forgetting about Zsigmondy:

Define two sequences as follows: $n_0 = 1$; $p_i$ is (a) the smallest prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$, or (b) if that doesn't exist, the smallest prime factor of $2^{n_i} + 1$; and $n_{i+1} = n_ip_i.$

Notice that each $n_i$ is divisible by all of the previous ones, and that all $n_i$ and $p_i$ are odd.

First, we show that all $n_i$ satisfy $n_i \mid 2^{n_i} + 1$. We proceed by induction. We can see that $n_0 = 1$ works, so assume $n_i$ works (and all the ones before it). We want to prove $n_{i+1} \mid 2^{n_{i+1}} + 1$.

Note that if a prime $p$ divides $n_{i+1}$, then $p = p_j$ for some $j$ satisfying $0 \leq j \leq i$.
This also means that $p_j$ is a factor of $2^{n_j} + 1$.
Then, by LTE, we have
\[ \nu_{p_j}(2^{n_{i+1}} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}\left(\frac{n_{i+1}}{n_j}\right) = \nu_{p_j}(n_{i + 1}) + \nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j). \]However, the strong inductive hypothesis implies $\nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j) \geq 0$, so we have $\nu_{p_j}(2^{n_{i+1}} + 1) \geq \nu_{p_j}(n_{i+1})$.
As this is true for all $p$, the inductive step is complete.

Next, we claim that eventually, the number of distinct prime factors of $n_i$ is always one more than the number of distinct prime factors of $n_{i-1}$. This is equivalent to showing that eventually, there always exists a prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$.

Suppose, for some $i$, that every prime factor of $2^{n_i} + 1$ is also a factor of $n_i$.
Then, let $p$ be a prime factor of $n_i$, and pick the minimal $j$ such that $p_j = p$. This minimality implies that $\nu_{p_j}(n_j) = 0$.
We have
\[ \nu_{p_j}(2^{n_i} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i) - \nu_{p_j}(n_j) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i). \]We can raise $p_j$ to the power of both sides to get
\[ {p_j}^{\nu_{p_j}(2^{n_i} + 1)} = {p_j}^{\nu_{p_j}(2^{n_j} + 1)} {p_j}^{\nu_{p_j}(n_i)}. \]Doing this for every prime factor $p$ of $n_i$ (notice that $j$ is now a one-to-one function of $p$) and multiplying the resulting equations, we get
\[ 2^{n_i} + 1 = n_i \prod_p p^{\nu_p(2^{n_{j(p)}} + 1)}. \]For all $p$, we have $p^{\nu_p(2^{n_{j(p)}} + 1)} \leq 2^{n_{j(p)}} + 1$. Thus,
\[ 2^{n_i} + 1 \leq n_i \prod_p (2^{n_{j(p)}} + 1). \]Since $j$ is one-to-one, every $j(p)$ is unique and in the set $\{0, 1, \ldots, i-1\}$. Therefore, we have the inequality
\[ 2^{n_i} + 1 \leq n_i \prod_{j=0}^{i-1} (2^{n_j} + 1). \]Taking the log base $2$ of both sides, we have
\[ n_i < \log _2 (2^{n_i} + 1) \leq \log _2 n_i + \sum _{j=0}^{i-1} \log_2(2^{n_j} + 1) < \log_2 n_i + \sum _{j=0}^{i-1} (n_j + 1) = \log_2 n_i + i + \sum_{j=0}^{i-1}n_j. \]Now, in order to achieve a bound on $n_i$, we notice that $p_i \geq 3$ for all $i$, so therefore, $n_i \geq 3^i$. It is then easy to see that $\sum_{j=0}^{i-1}n_j \leq \frac12 n_i$ for all $i \geq 1$. Then,
\[ n_i < \log_2 n_i + i + \frac12 n_i \leq \log_2 n_i + \log_3 n_i + \frac12 n_i \iff n_i < 2(\log _2 n_i + \log _3 n_i). \]This inequality obviously cannot be satisfied as $n_i$ grows large. Thus, eventually, we must have that there exists a prime factor of $2^{n_i} + 1$ that is not a factor of $n_i$. Hence, there must eventually exist an $n$ in our sequence with exactly 2000 distinct prime factors.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
LQFNB
12 posts
#61
Y by
是否存在一个恰有2000个素因子的正整$n$$n \mid 2^n + 1$?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
LQFNB
12 posts
#62
Y by
是否存在一个恰有2000个不同素因子的正整$n$$n \mid 2^n + 1$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
eibc
600 posts
#63
Y by
The answer is yes. We will construct a positive integer $n = p_1^2 p_2p_3 \cdots p_{2000}$ where $p_1 < p_2 < \cdots < p_{2000}$ are distinct primes such that $n \mid 2^n + 1$.

First, we set $p_1 = 3$. For $i \ge 2$, let $q_i = 3^2 \cdot p_2 \cdots p_3 \cdots p_{i - 1}$. Then, we shall construct $p_i$ recursively by picking a primitive divisor of $2^{2q_i} - 1$, which exists by Zsigmondy's theorem.

Now, we show that $n \mid 2^n + 1$. It suffices to show that for $1 \le i \le 2000$, we have $\nu_{p_i}(2^n + 1) \ge \nu_{p_i}(n)$. For $i = 1$, we note that by LTE,
$$\nu_3(2^n + 1) = \nu_3(2 + 1) + \nu_3(n) = 1 + \nu_3(n) > 1.$$For $i > 1$, we note that by construction, $\text{ord}_{p_i}(2) = 2q_i$. Thus, $2^{q_i} \equiv -1 \pmod {p_i}$, so by LTE, we have
$$\nu_{p_i}(2^n + 1) = \nu_{p_i}((2^{q_i})^{n/q_i} + 1) = \nu_{p_i}(2^{q_i} + 1) + \nu_{p_i}(n/q_i) \ge 1 + \nu_{p_i}(n) > \nu_{p_i}(n),$$so we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Natrium
50 posts
#64
Y by
Lemma. If $a\mid 2^a+1$ and $b\mid 2^b+1$, where $a=3^\alpha a'$, $b=3^\beta b'$, $3 \nmid a', b'$ and $a'$ and $b'$ are coprime, then $ab\mid 2^{ab}+1$.
Proof. Obviously, $a$ and $b$ are odd.
$$2^{ab}\equiv (-1)^b\equiv -1 \pmod{a}$$$$2^{ab}\equiv (-1)^a\equiv -1 \pmod{b}$$By LTE, $v_3(2^{ab}+1)=v_3(ab)+1=\alpha + \beta + 1$. Finally, as $a', b', 3^{\alpha+\beta}\mid 2^{ab}+1$ and all three numbers are pairwise coprime, $ab=a'b'3^{\alpha+\beta}\mid 2^{ab}+1$. $\square$

We construct a sequence of primes $p_2, p_3, \dots, p_{2000}$ in the following manner. Let $p_2=19$ (so that $p_2\mid 2^{3^2}+1$). Inductively assume we have constructed $p_2, p_3, \dots, p_{i}$, for some $i\ge 2$, such that:
$\bullet$ all $p_j$ with $2\le j\le i$ are distinct,
$\bullet$ $p_j>3$ and $p_j\mid2^{3^{i}}+1$ for each $2\le j\le i$.
As $2^{3^{i+1}}+1=(2^{3^i}+1)(4^{3^i}-2^{3^i}+1)$ and the greatest common divisor of the terms in the parenthesis is $3$, we conclude that $4^{3^i}-2^{3^i}+1$ has all its prime divisors distinct from $p_j$ for $2\le j\le i$. As $9\mid 2^{3^i}+1$ and $4^{3^i}-2^{3^i}+1>3$, it has a prime factor $p_{i+1}>3$, so the inductive claim holds for $i+1$ as well.

Having constructed $p_2, p_3, \dots, p_{2000}$, let $n_i=3^i p_i$ for each $2\le i\le 2000$. By LTE, $v_3(2^{n_i}+1)=i+1$. By Fermat's Little Theorem, $2^{n_i}\equiv 2^{3^i p_i}\equiv 2^{3^i}\equiv -1\pmod{p_i}$. Therefore, $n_i = 3^i p_i\mid 2^{n_i} + 1$ for each $i$. Let $n=n_2n_3\dots n_{2000}$. By repeated application of the lemma, $n\mid 2^n + 1$, and by construction, $n$ has $2000$ distinct prime divisors.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
smileapple
1010 posts
#65
Y by
Write $x\sim y$ if $p\mid x$ iff $p\mid y$.

Let $n$ be some odd integer such that $n>3$, $n\mid 2^n+1$, and $n\nsim 2^n+1$. We claim that there exists a prime $p\nmid n$ for which $np\mid 2^{np+1}$ and $np\nsim 2^{np}+1$. Indeed, take some $p\nmid n$ and $p\mid 2^n+1$. Note that $2^{np}+1\equiv 2^n+1\equiv0\pmod p$, and since $p$ is odd, we also have $n\mid 2^{np}+1$. Hence $np\mid 2^{np}+1$. Furthermore, since $n>3$, there exists some prime $q$ satisfying $q\nmid 2^n+1$ and $q\mid 2^{np}+1$ by Zsigmondy.

Applying the above claim $1999$ times on $n=9$ implies that the answer is $\fbox{\text{yes}}$. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
amapstob
19 posts
#66
Y by
The answer is yes.
Lemma. For all primes $p>3$, $2^p+1$ has a prime divisor greater than $p$.
Proof. $2^p\equiv -1\pmod{q}\implies 2^{\gcd(q-1,2p)}\equiv 1\pmod{q}$. If $\gcd(q-1,2p)=2$, then $q=3$, so we have to rule out $2^p+1$ being a power of three, which only happens with $p=3$ by Mihailescu's theorem. Since $p>3$, there exists $q\neq 3$ dividing $2^p+1$. Then $p\mid q-1\implies p<q$, as desired. $\blacksquare$

Claim. If $n\mid 2^{n}+1$ and $n$ is odd and has a prime divisor greater than $3$, there exists a prime with $p\mid 2^n+1$ and $p\nmid n$ such that $np\mid 2^{np}+1$.
Proof. Let $q$ be the greatest prime divisor of $n$. Then $2^q+1\mid 2^n+1$. But $2^q+1$ has a prime divisor $p$ greater than $q$ by the above lemma. So $p\nmid n$ and $p\mid 2^n+1$. Then since $p\mid 2^n+1$ and $n\mid 2^n+1$ and $n,p$ are coprime, $np\mid 2^n+1$. But $2^n+1\mid 2^{np}+1$, so we're done. $\blacksquare$

Now observe that $3^2\cdot 19 \mid 2^{3^2\cdot 19}+1$, so applying the lemma above $1998$ times finishes. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cursed_tangent1434
618 posts
#67
Y by
We claim that the answer is yes. In fact we can show the more general statement that for any positive integer $d$ there exists some positive integer $n$ for which $n$ has exactly $d$ distinct prime divisors and $n \mid 2^n+1$. To do this, we employ induction.

First note that $3 \mid 2^3+1$. Now, say there exists a positive integer $n_r = 3^{r}\cdot p_1p_2 \dots p_r$ for which $n_r \mid 2^{n_r}+1$. Consider
\[2^{3n_r}+1=2^{3^{r+1}\cdot p_1p_2\dots p_{r}} +1\]Now, by Zsigmondy's Theorem there exists a prime $p_{r+1} \mid 2^{3^{r+1}\cdot p_1p_2\dots p_{r}} +1$ but $p_{r+1} \nmid 2^{3^r \dot p_1p_2\dots p_r}$. Thus, this prime factor $p_{r+1} \not \in \{p_1,p_2,\dots , p_r\}$. Further,
\[3n_r \mid 2^{n_r}+1 \mid 2^{3n_r\cdot p_{r+1}}+1\]since by Lifting the Exponent Lemma, $\nu_3(2^{n_r}+1)= \nu_3(3)+\nu_3(n_r) = \nu_3(3n_r)$. Finally,
\[2^{3n_r\cdot p_{r+1}}+1 \equiv 2^{3n_r}+1 \equiv 0 \pmod{p_{r+1}}\]by construction. Thus, $3n_r\cdot p_{r+1} \mid 2^{3n_r\cdot p_{r+1}}+1$ so we can let $n_{r+1}=3n_r\cdot p_{r+1}$ which completes the induction and proves the result.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ray66
35 posts
#68
Y by
We will prove the result by induction.

First take the base case $n_1=9$ so that $9$ divides $513$. Now consider the number $n_2=n_1p_2$ where $p_2$ is a unique prime number dividing $2^{n_1}+1$. We know that such a $p$ exists by Zsigmondy. Therefore $2^{n_2}+1$ is also divisible by $p_2$, so we finish the induction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ilikeminecraft
615 posts
#69
Y by
I claim that there \boxed{\text{exists}} a number that has $n$ distinct prime numbers that satisfies our conditions.

Let $k_n = 9 \cdot \prod\limits_{i = 1}^{n - 1} p_i$ be a construction for $n$. I will prove this with induction.

Clearly, $k_1 = 9$ is a construction for $n = 1.$

Now, assume that $n = l$ has a valid construction. Let $p_{l}$ be a prime dividing $2^{k_l} + 1$ such that $(p_l, k_l).$ I will prove the existence of such a $p_l:$

Notice that $k$ is not even. We have that $\nu_3(2^{k_l} + 1) = \nu_3(k_l) + \nu_3(3) = 1 + 2 = 3,$ and $$\nu_{p_i}(2^{k_l} + 1) = \nu_{p_i}\left(\left(2^{k_{i + 1}}\right)^{\frac{k_l}{k_{i + 1}}} + 1\right) = \nu_{p_i}(2^{k_{i + 1}} + 1) + \nu_{p_i}\left(\prod_{j = i + 1}^{l - 1}p_j\right) = 1$$Thus, since $3k_{l} < 2^{k_l} + 1,$ there must exist a non-3 value greater than 1 that divides $\frac{2^{k_l} + 1}{k_l}.$ By picking a $p_i$ that divides that, we can gaurantee $p_i$ exists and is relatively prime to all other primes in $k_l.$

Finally, I claim that $k_{l + 1} = k_lp_{l}$ is a valid construction for $n = l + 1.$ We have that $k_l \mid 2^{k_l} + 1 \mid 2^{k_l p_l} + 1,$ and $p_l \mid 2^{k_lp_l} + 1\implies k_{l + 1} = k_lp_l \mid 2^{k_l p_l} + 1 = 2^{k_{l + 1}} + 1.$

Thus, we are done.
Z K Y
N Quick Reply
G
H
=
a