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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
Surjective number theoretic functional equation
snap7822   3
N 7 minutes ago by internationalnick123456
Source: 2025 Taiwan TST Round 3 Independent Study 2-N
Let $f:\mathbb{N} \rightarrow \mathbb{N}$ be a function satisfying the following conditions:
[list=i]
[*] For all $m, n \in \mathbb{N}$, if $m > n$ and $f(m) > f(n)$, then $f(m-n) = f(n)$;
[*] $f$ is surjective.
[/list]
Find the maximum possible value of $f(2025)$.

Proposed by snap7822
3 replies
snap7822
May 1, 2025
internationalnick123456
7 minutes ago
nice inequality with condition
junior2001   3
N 9 minutes ago by produit
Let $ a,b,c \in R $ such that $ a+b+c=ab+bc+ca $.Prove that $ a+b+c+1 \ge 4abc $.
3 replies
+1 w
junior2001
Apr 24, 2015
produit
9 minutes ago
FE with devisibility
fadhool   0
11 minutes ago
if when i solve an fe that is defined in the set of positive integer i found m|f(m) can i set f(m) =km such that k is not constant and of course it depends on m but after some work i find k=c st c is constant is this correct
0 replies
fadhool
11 minutes ago
0 replies
Many Equal Sides
mathisreal   3
N 13 minutes ago by QueenArwen
Source: Brazil EGMO TST 2023 #1
Let $ABC$ be a triangle with $BA=BC$ and $\angle ABC=90^{\circ}$. Let $D$ and $E$ be the midpoints of $CA$ and $BA$ respectively. The point $F$ is inside of $\triangle ABC$ such that $\triangle DEF$ is equilateral. Let $X=BF\cap AC$ and $Y=AF\cap DB$. Prove that $DX=YD$.
3 replies
mathisreal
Nov 10, 2022
QueenArwen
13 minutes ago
LOTS of recurrence!
SatisfiedMagma   4
N 16 minutes ago by Reacheddreams
Source: Indian Statistical Institute Entrance UGB 2023/5
There is a rectangular plot of size $1 \times n$. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size $1 \times 1$, the blue tiles are of size $1 \times 1$ and the black tiles are of size $1 \times 2$. Let $t_n$ denote the number of ways this can be done. For example, clearly $t_1 = 2$ because we can have either a red or a blue tile. Also $t_2 = 5$ since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.

[list=a]
[*]Prove that $t_{2n+1} = t_n(t_{n-1} + t_{n+1})$ for all $n > 1$.

[*]Prove that $t_n = \sum_{d \ge 0} \binom{n-d}{d}2^{n-2d}$ for all $n >0$.
[/list]
Here,
\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]for integers $m,r$.
4 replies
SatisfiedMagma
May 14, 2023
Reacheddreams
16 minutes ago
combi/nt
blug   1
N 18 minutes ago by blug
Prove that every positive integer $n$ can be written in the form
$$n=a_1+a_2+...+a_k,$$where $a_m=2^i3^j$ for some non-negative $i, j$ such that
$$a_x\nmid a_y$$for every $x, y\leq k$.
1 reply
blug
Yesterday at 3:37 PM
blug
18 minutes ago
Inequality, inequality, inequality...
Assassino9931   9
N 25 minutes ago by ZeroHero
Source: Al-Khwarizmi Junior International Olympiad 2025 P6
Let $a,b,c$ be real numbers such that \[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]Find the smallest possible value of $a^2 + b^2 + c^2$.

Binh Luan and Nhan Xet, Vietnam
9 replies
Assassino9931
Today at 9:38 AM
ZeroHero
25 minutes ago
Vectors in a tilted square
mathwizard888   20
N 28 minutes ago by HamstPan38825
Source: 2016 IMO Shortlist A3
Find all positive integers $n$ such that the following statement holds: Suppose real numbers $a_1$, $a_2$, $\dots$, $a_n$, $b_1$, $b_2$, $\dots$, $b_n$ satisfy $|a_k|+|b_k|=1$ for all $k=1,\dots,n$. Then there exists $\varepsilon_1$, $\varepsilon_2$, $\dots$, $\varepsilon_n$, each of which is either $-1$ or $1$, such that
\[ \left| \sum_{i=1}^n \varepsilon_i a_i \right| + \left| \sum_{i=1}^n \varepsilon_i b_i \right| \le 1. \]
20 replies
mathwizard888
Jul 19, 2017
HamstPan38825
28 minutes ago
Combi Geo
Adywastaken   0
38 minutes ago
Source: NMTC 2024/8
A regular polygon with $100$ vertices is given. To each vertex, a natural number from the set $\{1,2,\dots,49\}$ is assigned. Prove that there are $4$ vertices $A, B, C, D$ such that if the numbers $a, b, c, d$ are assigned to them respectively, then $a+b=c+d$ and $ABCD$ is a parallelogram.
0 replies
Adywastaken
38 minutes ago
0 replies
Japan MO Finals 2023
parkjungmin   1
N 42 minutes ago by EvansGressfield
It's hard. Help me
1 reply
parkjungmin
Yesterday at 2:35 PM
EvansGressfield
42 minutes ago
Non-homogenous Inequality
Adywastaken   0
an hour ago
Source: NMTC 2024/7
$a, b, c\in \mathbb{R}$ such that $ab+bc+ca=3abc$. Show that $a^2b+b^2c+c^2a \ge 2(a+b+c)-3$. When will equality hold?
0 replies
Adywastaken
an hour ago
0 replies
Calculus
youochange   1
N an hour ago by youochange
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
1 reply
youochange
2 hours ago
youochange
an hour ago
Classic Diophantine
Adywastaken   0
an hour ago
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
0 replies
Adywastaken
an hour ago
0 replies
Indian Geo
Adywastaken   0
an hour ago
Source: NMTC 2024/5
$\triangle ABC$ has $\angle A$ obtuse. Let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, $C$ respectively. Let $A_1$, $B_1$, $C_1$ be arbitrary points on $BC$, $CA$, $AB$ respectively. The circles with diameter $AA_1$, $BB_1$, $CC_1$ are drawn. Show that the lengths of the tangents from the orthocentre of $ABC$ to these circles are equal.
0 replies
Adywastaken
an hour ago
0 replies
$n$ 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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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.
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shendrew7
795 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. $\blacksquare$
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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.
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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.
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Niku
120 posts
#59
Y by
Do you realise that this post was made 20 years ago.
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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.
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LQFNB
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#61
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是否存在一个恰有2000个素因子的正整$n$$n \mid 2^n + 1$?
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LQFNB
12 posts
#62
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是否存在一个恰有2000个不同素因子的正整$n$$n \mid 2^n + 1$
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eibc
600 posts
#63
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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.
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Natrium
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#64
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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.
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smileapple
1010 posts
#65
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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 $\fbox{\text{yes}}$. $\blacksquare$
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amapstob
19 posts
#66
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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. $\blacksquare$
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cursed_tangent1434
625 posts
#67
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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.
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ray66
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#68
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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.
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Ilikeminecraft
626 posts
#69
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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.
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