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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.

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)!

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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
Diophantine
TheUltimate123   32
N 39 minutes ago by Kempu33334
Source: CJMO 2023/1 (https://aops.com/community/c594864h3031323p27271877)
Find all triples of positive integers \((a,b,p)\) with \(p\) prime and \[a^p+b^p=p!.\]
Proposed by IndoMathXdZ
32 replies
TheUltimate123
Mar 29, 2023
Kempu33334
39 minutes ago
Parallel lines in incircle configuration
GeorgeRP   4
N 40 minutes ago by VicKmath7
Source: Bulgaria IMO TST 2025 P1
Let $I$ be the incenter of triangle $\triangle ABC$. Let $H_A$, $H_B$, and $H_C$ be the orthocenters of triangles $\triangle BCI$, $\triangle ACI$, and $\triangle ABI$, respectively. Prove that the lines through $H_A$, $H_B$, and $H_C$, parallel to $AI$, $BI$, and $CI$, respectively, are concurrent.
4 replies
GeorgeRP
May 14, 2025
VicKmath7
40 minutes ago
Geometry
shactal   1
N an hour ago by ricarlos
Two intersecting circles $C_1$ and $C_2$ have a common tangent that meets $C_1$ in $P$ and $C_2$ in $Q$. The two circles intersect at $M$ and $N$ where $N$ is closer to $PQ$ than $M$ . Line $PN$ meets circle $C_2$ a second time in $R$. Prove that $MQ$ bisects angle $\widehat{PMR}$.
1 reply
shactal
Today at 3:04 PM
ricarlos
an hour ago
Curious inequality
produit   1
N an hour ago by arqady
Positive real numbers x_1, x_2, . . . x_n satisfy x_1 + x_2 + . . . + x_n = 1.
Prove that
1/(1 −√x_1)+1/(1 −√x_2)+ . . . +1/(1 −√x_n)⩾ n + 4.
1 reply
produit
May 10, 2025
arqady
an hour ago
IMO ShortList 2001, combinatorics problem 6
orl   15
N 2 hours ago by awesomeming327.
Source: IMO ShortList 2001, combinatorics problem 6
For a positive integer $n$ define a sequence of zeros and ones to be balanced if it contains $n$ zeros and $n$ ones. Two balanced sequences $a$ and $b$ are neighbors if you can move one of the $2n$ symbols of $a$ to another position to form $b$. For instance, when $n = 4$, the balanced sequences $01101001$ and $00110101$ are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set $S$ of at most $\frac{1}{n+1} \binom{2n}{n}$ balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in $S$.
15 replies
orl
Sep 30, 2004
awesomeming327.
2 hours ago
Number Theory Chain!
JetFire008   64
N 2 hours ago by Primeniyazidayi
I will post a question and someone has to answer it. Then they have to post a question and someone else will answer it and so on. We can only post questions related to Number Theory and each problem should be more difficult than the previous. Let's start!

Question 1
64 replies
JetFire008
Apr 7, 2025
Primeniyazidayi
2 hours ago
equation 2025
mohamed-adam   0
2 hours ago
Source: own
Find all positive integers $a,b$ such that $$a^8-2b^5=2025b$$
0 replies
mohamed-adam
2 hours ago
0 replies
Serbian selection contest for the IMO 2025 - P1
OgnjenTesic   3
N 3 hours ago by MR.1
Source: Serbian selection contest for the IMO 2025
Let \( p \geq 7 \) be a prime number and \( m \in \mathbb{N} \). Prove that
\[\left| p^m - (p - 2)! \right| > p^2.\]Proposed by Miloš Milićev
3 replies
OgnjenTesic
Yesterday at 4:01 PM
MR.1
3 hours ago
Nice "if and only if" function problem
ICE_CNME_4   0
3 hours ago
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
0 replies
ICE_CNME_4
3 hours ago
0 replies
4 variables
Nguyenhuyen_AG   11
N 3 hours ago by arqady
Let $a,\,b,\,c,\,d$ are non-negative real numbers and $0 \leqslant k \leqslant \frac{2}{\sqrt{3}}.$ Prove that
$$a^2+b^2+c^2+d^2+kabcd \geqslant k+4+(k+2)(a+b+c+d-4).$$hide
11 replies
Nguyenhuyen_AG
Dec 21, 2020
arqady
3 hours ago
Hard Number Theory
MuradSafarli   0
3 hours ago
Find all odd natural numbers \( m \) such that \( m^2 - 1 \mid 3^m + 5^m \).
0 replies
MuradSafarli
3 hours ago
0 replies
A and B take stones from a pile
WakeUp   4
N 3 hours ago by reni_wee
Source: CentroAmerican 2003
Two players $A$ and $B$ take turns playing the following game: There is a pile of $2003$ stones. In his first turn, $A$ selects a divisor of $2003$ and removes this number of stones from the pile. $B$ then chooses a divisor of the number of remaining stones, and removes that number of stones from the new pile, and so on. The player who has to remove the last stone loses. Show that one of the two players has a winning strategy and describe the strategy.
4 replies
WakeUp
Nov 29, 2010
reni_wee
3 hours ago
Pisano period
JARP091   3
N 3 hours ago by JARP091
Source: At the time of writing this problem i do not know the source if any
Let $F_n$ denote the $n$th Fibonacci number, defined by the recurrence:
\[
F_0 = 0,\quad F_1 = 1,\quad \text{and} \quad F_{n+2} = F_{n+1} + F_n \quad \text{for } n \geq 0.
\]For a fixed modulus $m \in \mathbb{N}$, define the set
\[
S_m = \left\{ (i, j) \in \mathbb{N}^2 : F_i \cdot F_j \equiv 1 \pmod{m} \right\}.
\]Prove that $|S_m| = \infty$ if and only if $5 \nmid m$.

Note: Its a very beautiful problem
3 replies
JARP091
4 hours ago
JARP091
3 hours ago
FE with conditions on $x,y$
Adywastaken   1
N 4 hours ago by jasperE3
Source: OAO
Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $\forall y>x>0$,
\[
f(x^2+f(y))=f(xf(x))+y
\]
1 reply
Adywastaken
4 hours ago
jasperE3
4 hours ago
(2^n + 1)/n^2 is an integer (IMO 1990 Problem 3)
orl   107
N May 11, 2025 by Rayvhs
Source: IMO 1990, Day 1, Problem 3, IMO ShortList 1990, Problem 23 (ROM 5)
Determine all integers $ n > 1$ such that
\[ \frac {2^n + 1}{n^2}
\]is an integer.
107 replies
orl
Nov 11, 2005
Rayvhs
May 11, 2025
(2^n + 1)/n^2 is an integer (IMO 1990 Problem 3)
G H J
Source: IMO 1990, Day 1, Problem 3, IMO ShortList 1990, Problem 23 (ROM 5)
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BestAOPS
707 posts
#104
Y by
The answer is $n=1,3$. We show that there are no other solutions. Since $n$ must be odd, assume $n \geq 5$.
Let $p$ be the least prime factor of $n$. We have $p > 2$ since $n$ is odd. Furthermore, $2^n + 1 \equiv 0 \implies 4^n \equiv 1 \pmod{p}$.

We claim the order of $4$ mod $p$ is $1$. This is because the order must divide $\gcd{n, p-1}$, which must equal $1$ otherwise $\gcd{n, p-1}$ must have smaller prime factors which also divide $n$, contradicting $p$'s minimality. Thus, $4 \equiv 1 \pmod{p}$, so $p = 3$.

In order for $n^2$ to divide $2^n + 1$, we must have $\nu_3(n^2) = 2\nu_3(n) \leq \nu_3(2^n + 1)$. Since $3 \mid 2 + 1$, using the lifting the exponent lemma yields $\nu_3(2^n + 1) = 1 + \nu_3(n)$. We then conclude that $\nu_3(n) \leq 1$, but since $3$ is the smallest prime factor of $n$, we must have $\nu_3(n) = 1$.

Now, we write $n = 3k$ for some $k$ not divisible by $2$ or $3$. Since $n \geq 5$, we have $k > 1$ and we can let $p$ be the smallest prime factor of $k$. But similarly to before, we see that $2^{6k} \equiv 1 \pmod{p}$ and thus $64 \equiv 1 \pmod{p}$. This means $p \mid 63$. Since $p \geq 5$, we must have $p = 7$.

This tells us that $n^2$ is divisible by $7$. However, $2^n + 1 \equiv 8^k + 1 \equiv 2 \pmod{7}$, so we are done.
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ATGY
2502 posts
#105
Y by
Say $p$ is the smallest prime dividing $n$, so we have $2^n + 1 \equiv 0 \mod{p} \implies 2^{2n} \equiv 1\mod{p}$. Clearly $p \neq 2$. Let $k$ denote the order of $2 \mod{p}$. Observe that $k \mid 2n$, and since $2^{p - 1} \equiv 1 \mod{p}$, we have $k \mid (2n, p - 1) \implies k = 1, 2$. $k = 1$ is impossible, which means $k = 2 \implies 2^2 \equiv 1 \mod{p} \implies p = 3$.

If $q$ is the next smallest prime dividing $2^n + 1$, we have $\text{ord}_q(2) \mid (2n, q - 1) = 3, 6$. $q = 7$ fails upon checking.

So $p = 3$ is the only prime dividing $n$. We need $v_3(2^n + 1) \geq v_3(n^2) = 2v_3(n)$, for $n^2 \mid 2^n + 1$. By LTE, $v_3(2^n + 1) = v_3(3) + v_3(n) \implies 1 + v_3(n) \geq 2v_3(n) \implies v_3(n) \leq 1$.

$v_3(n) = 0 \implies n = 1$, which clearly works. $v_3(n) = 1 \implies n = 3$, which works as well. So, we are done.
This post has been edited 3 times. Last edited by ATGY, Jul 13, 2024, 11:03 AM
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SomeonesPenguin
129 posts
#106
Y by
The problem is equivalent to $2^n\equiv -1\pmod{n^2}$. This implies that $2^{2n}\equiv 1\pmod{n^2}$. Also note that $n=1$ is a solution so suppose $n\neq 1$. Now take the smallest prime factor of $n$ (notice that this can’t be $2$). We have that $ord_p(2)\mid 2n$ and also note that $ord_p(2) \le p-1$ so this order must be equal to $2$. Hence $p$=3.

Now let $n=3^ab$ where $a\ge 1$ and $b$ is an odd integer. We have: $$\nu_3(n^2)\le \nu_3(2^n+1)$$But we clearly have $\nu_3(n^2)=2a$ and from LTE $\nu_3(2^n+1)=a+1$ hence $a$ must be qual to $1$ so we get that $n=3b$.

Plugging this back in, we get that $9k^2\mid 2^{3k}+1$. Now $k=1$ is a solution ($n=3$) so suppose that $k\neq 1$. Let $q$ be the smallest prime factor of $k$ and note that this isn’t $2$ or $3$. We similarly get that $2^{6k}\equiv 1\pmod q$ so $ord_q(2)\mid 6k$ so from minimality this implies $ord_q(2)\mid 6$. Hence $ord_q(2)\in \{3,6\}$. In either case, we get that $q=7$ so $7\mid 2^n+1$ which isn’t possible. Therefore the only solutions are $n=1$ and $n=2$.
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Eka01
204 posts
#109
Y by
Obviously $n$ is odd.
Consider the smallest prime $p$ dividing $n$ then order of $p$ modulo $2$ divides $2n$ but not $n$ and it also divides $p-1$ implying the order must be $3$ giving that the smallest prime must be the smallest prime.
However $LTE$ gives us that $\nu_3(2^n +1)$ =$\nu_3(n) +1$ which is greater than $2\nu_3(n)$ implying $n=3k$ where $k$ is an odd number not divisible by $3$. Plugging this and using similar order arguments, we get that $7$ divides $2^n +1$ which is false implying $k=1$.

Hence $\boxed{n=3}$ is the only solution.
Z K Y
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pie854
243 posts
#110
Y by
The answer is $n=1,3$ which clearly works.

Suppose $n>1$ and let $p_0$ be the smallest prime divisor of $n=p_0^{e_0} p_1^{e_1} \cdots p_k^{e_k}$ with $e_i\geq 0$ for $i>0$. Let $a=\text{ord}_{p_0^2}(2)$. Since $p_0^2\mid 2^n+1$, it follows that $a\nmid n$ and $a\mid 2n$. So $a=2p_0^{f_0}p_1^{f_1} \cdots p_k^{f_k}$ where $f_i\leq e_i$ for all $0\leq i\leq k$. But $a\mid p_0(p_0-1)$ and thus $a\leq p_0(p_0-1)$. From this it follows that $a\in \{2, 2p_0, 2p_i\}$ for some $1\leq i\leq k$. If $a=2$ then $p_0^2\mid 2^2-1=3$ which is absurd. If $a=2p_i$ then from $2p_i\mid p_0(p_0-1)$ it follows that $p_i\mid p_0-1$ which is absurd as well.

Thus $a=2p_0$ and $2^{2p_0}\equiv 1\pmod{p_0^2}$. From here it's easy to get that $p_0=3$. So $n=3m$ (by LTE $2v_3(n)\leq v_3(2^n+1)=v_3(2+1)+v_3(n)$ and thus $v_3(n)\leq 1$ and so $3\nmid m$) and $m^2\mid 8^m+1$. We have $m=p_1^{e_1} p_2^{e_2}\cdots p_k^{e_k}$ and let's assume $m>1$ and that $p_1$ is the smallest prime divisor. Let $b=\text{ord}_{p_1^2}(8)$ and as before we get $b\in \{2,2p_1\}$ ($b$ cannot be $2p_i$ because of the same reason as before). If $b=2$ then $p_1^2\mid 8^2-1=3^2\cdot 7$ but this is impossible. So $b=2p_1$ and $p_1^2\mid 8^{2p_1}-1$. From this we get $p_1\mid 8^2-1=3^2\cdot 7$ and so $p_1=7$. Note that $7^2\mid 42799\cdot 7^2=8^7-1$, so $b\leq 7$ but this is a contradiction since $b=2\cdot 7=14$. From this contradiction it follows that $m=1$ and we get the desired solution set.
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EVKV
71 posts
#111
Y by
For min prime p|n
p≠3
2^2n = 1 mod p
Ord2 mod p | 2n

ord 2 mod p = 2,,2n,n,d,2d
Here d|n d>1
So it can only be 2,2d

As ord2 mod p ≤ p-1
d≥p
2d≥ 2p> p-1
Thus ord 2 mod p ≠ 2d for any d>1
Thus ord 2 mod p = 2

4= 1 mod p
So p = 3

Thus 3| n

Let q be second smallest prime
Ord 2 mod q = 2,2d,6,2n,n,3d
Where d|n
If d>3
3d>q-1
So d≤3
Now 3 cases
d= 3 X
d=2 X
d= 1 works
So ord 2 mod q = 3

Same reasoning with 2d gives
ord 2 mod q = 6

Implies q=7

Which is not possible

Hence no such q exists
Thus only 3|n
So
2vp(n) ≤ vp(3) + vp(n)
n = 3
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AshAuktober
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#112
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We claim the only solution is $n = 3$, which clearly works.

Note that $2^{2n} \equiv 1 \pmod{p}$, where $p$ is the smallest prime factor of $n$. But $2^{p-1} \equiv 1 \pmod p$, and thus $2^2 = 2^{\gcd(2n, p-1)} \equiv 1 \pmod{p} \implies \boxed{p = 3}.$

Now note that we have from LTE and $\nu_p$ stuff that

$$2\nu_p(n) \le \nu_p(2^n + 1) = \nu_p(3) + \nu_p(n) = 1 + \nu_p(n) \implies \nu_p(n) = 3.$$Now let $n = 2m$ where $\gcd(m, 3) = 1$.
Then we have $m^2 \mid 8^m + 1 \implies 8^{2m} \equiv 1 \pmod{q}$ where $q$ is the smallest prime divisor of $m$.
But $8^{q-1} \equiv 1 \pmod{q} \implies 8^2 \equiv 1 \pmod{q} \implies q \in \{3, 7\}$.
$q = 3$ is impossible because then $\nu_p(n) \ge 2$, and $q = 7$ is impossible because $8^m + 1 \equiv 2 \pmod{7}$, and so $7$ can never divide the numerator. Thus in fact no such prime exists, and $m = 1 \implies \boxed{n = 3}$.
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ItsBesi
146 posts
#116
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Solution:

$\frac{2^n+1}{n^2}$ is an integer $\implies n^2 \mid 2^n+1$

Let $p$ be the smallest prime divisor of $n$ (such a prime exists because $n > 1$).

Note that $p \neq 2$ because RHS$=2^n+1 \equiv 1 \pmod 2$

So $p \mid n^2 \mid 2^n+1 \implies p  \mid 2^n+1 \iff 2^n+1 \equiv 0 \pmod p \implies 2^n \equiv -1 \pmod p \implies$ $$\boxed{2^{2n} \equiv 1 \pmod p}$$
Also by Fermat's Little Theorem we have that : $$\boxed{2^{p-1} \equiv 1 \pmod p} (\because p \neq 2 \implies gcd(2,p)=1)$$
Hence $$2^{\gcd(2n,p-1)} \equiv 1 \pmod p$$
Note that since $p$ is the smallest prime divisor of $n$ we get that $\gcd(n,p-1)=1$ so $\gcd(2n,p-1)=1 \vee 2$

Note that since $2n \equiv 0 \pmod 2$ and $p-1 \equiv 0 \pmod 2$ we get:

$\gcd(2n,p-1)=2 \implies 2^2 \equiv 1 \pmod p \implies p \mid 3 \implies p=3 (\because p$-prime)

Hence $3=p \mid n \implies n=3k \implies 9k^2 \mid 8^k+1$

Now suppose that $k >1$ so simmilarly as before let $q$ be the smallest prime divisor of $k$ hence we find that $$8^{\gcd(2k,q-1)} \equiv 1 \pmod q $$
$\gcd(2k,q-1)=2 \implies 8^2 \equiv 1 \pmod q \implies q \mid 63=7 \cdot 3^2 \implies q=3 \vee q=7$ however if $q=7$ then:
$7=q \mid 9k^2 \mid 8^k+1 \implies 7 \mid 8^k+1$ but RHS$=8^k+1 \equiv 1+1=2 \pmod 7$ so $7 \nmid RHS \rightarrow \leftarrow$

So $3=q \mid k \implies k=3 \ell \implies$ $$81 \ell^2 \mid 8^{3 \ell}+1$$
Now by taking $\nu_3$ we get:

$4+2 \cdot \nu_3(\ell)=\nu_3(81 \ell^2) \leq \nu_3(8^{3 \ell}+1) \stackrel{LTE}{=} \nu_3(8+1)+\nu_3(3 \cdot \ell)=3+\nu_3(\ell) \implies 4+2 \cdot \nu_3(\ell) \leq 3+\nu_3(\ell) \implies$ $$\nu_3(\ell) \leq -1 \rightarrow \leftarrow$$
Hence there doesn't exist a prime $q$ such that $q \mid k$ hence $k=1 \implies n=3$ which obviously works $\blacksquare$
This post has been edited 2 times. Last edited by ItsBesi, Jan 22, 2025, 2:21 PM
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mathfortaleza23
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#117
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what is r ?
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smileapple
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#118
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Observe that $n$ must be odd, and suppose that $n>1$. Letting $p$ be the minimal prime divisor of $n$, we have $2^n\equiv-1\pmod p$ and thus $4^n\equiv1\pmod p$. We also have $4^{p-1}\equiv1\pmod p$. By minimality $\gcd(n,p-1)=1$, so it follows that $p=3$.

Let $k=\nu_3(n)$ and $m=\frac{n}{3^k}$. Then $\nu_3(2^n+1)=\nu_3(2^{3^km}-(-1)^{3^km})=v_3(3^km)+v_3(3)=k+1$ from exponent lifting, giving $2k=\nu_3(n^2)\le\nu_3(2^n+1)=k+1$. Thus $k=1$.

Hence, write $n=3m$ for $3\nmid m$, so that $n^2\mid 2^n+1$ occurs if and only if $m^2\mid 8^m+1$. Let $q$ be the smallest prime factor of $m$. Then similarly $64\equiv64^{\gcd(m,q-1)}\equiv1\pmod q$. Hence $q\in\{3,7\}$, but $q\neq 3$ as $3\nmid m$. and $q\neq 7$ as $8^m\equiv-1\pmod q$. Thus $q$ cannot exist; in other words, we have $m=1$.

Our solution set for $n$ is thus $\boxed{\{1,3\}}$. $\blacksquare$
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John_Mgr
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#119
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smileapple wrote:
Observe that $n$ must be odd, and suppose that $n>1$. Letting $p$ be the minimal prime divisor of $n$, we have $2^n\equiv-1\pmod p$ and thus $4^n\equiv1\pmod p$. We also have $4^{p-1}\equiv1\pmod p$. By minimality $\gcd(n,p-1)=1$, so it follows that $p=3$.

Let $k=\nu_3(n)$ and $m=\frac{n}{3^k}$. Then $\nu_3(2^n+1)=\nu_3(2^{3^km}-(-1)^{3^km})=v_3(3^km)+v_3(3)=k+1$ from exponent lifting, giving $2k=\nu_3(n^2)\le\nu_3(2^n+1)=k+1$. Thus $k=1$.

Hence, write $n=3m$ for $3\nmid m$, so that $n^2\mid 2^n+1$ occurs if and only if $m^2\mid 8^m+1$. Let $q$ be the smallest prime factor of $m$. Then similarly $64\equiv64^{\gcd(m,q-1)}\equiv1\pmod q$. Hence $q\in\{3,7\}$, but $q\neq 3$ as $3\nmid m$. and $q\neq 7$ as $8^m\equiv-1\pmod q$. Thus $q$ cannot exist; in other words, we have $m=1$.

Our solution set for $n$ is thus $\boxed{\{1,3\}}$. $\blacksquare$

$n>1$
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John_Mgr
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#120
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I too posted it.
Attachments:
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cursed_tangent1434
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#121
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We claim that the only positive integers $n$ which satisfy the given condition are $n=1$ and $n=3$. It is clear that these solutions work, so we shall now show that they are the only ones.

Since $n=1$ clearly works, we consider $n>1$ in what follows. Further, the left hand side is clearly odd for all $n \ge 1$ so we must have $n$ being odd. We first show the following.

Claim : For any positive integer $n$ which satisfies the given divisibility, $3$ must be the smallest prime divisor of $n$.

Proof : Let $q$ denote the smallest prime divisor of $n$. Since $q \mid n^2 \mid 2^n+1$ it follows that $2^{2n} \equiv 1 \pmod{q}$. Thus, $\text{ord}_q(2) \mid 2n$. Also, $\text{ord}_q(2) \mid q-1$ so if there exists an odd prime $r \mid \text{ord}_q(2)$ we have $r \mid 2n$ so $r \mid n$. But, $r \mid q-1$ so $r \le q-1 <q$ which contradicts the minimality of $q$. Thus, $\text{ord}_q(2)$ is a perfect power of two. Now, since $\nu_2(2n)=1$ as $n$ is odd we must have $\text{ord}_q(2) =2$. Thus,
\[0 \equiv 2^n+1 \equiv 2 +1 \equiv 3 \pmod{q}\]which holds if and only if $q=3$ as desired.

Now note,
\[2\nu_3(n)=\nu_3(n^2) \le \nu_3(2^n+1) = \nu_3(2+1)+\nu_3(n)=\nu_3(n)+1\]which implies that $\nu_3(n)=1$. Thus, if $n$ is a power of $3$ it must be $3$ itself, which works. Next, we consider $n>3$ and hence, there exists a second smallest prime divisor $s>3$ of $n$. As before note that if $\text{ord}_s(2)$ is not a power of two, $n$ must have a smaller prime divisor than $s$. Thus, $\text{ord}_s(2)$ can only have factors of $2$ and $3$ in it's prime factorization. However, as noted before, $\nu_2(2n) =1$ and $\nu_3(2n)=1$ so the only possibilities are $\text{ord}_s(2)=2$ and $\text{ord}_s(2)=6$. The former implies that $s=3$ which is a contradiction while the second implies $2^6 \equiv 1 \pmod{s}$ which requires $s \mid 63$. Thus we must have $s=7$. But it is not hard to check that $2^n+1 \not \equiv 0 \pmod{7}$ over all positive integers $n$ so this is a contradiction and we are done.
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Ilikeminecraft
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#123
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I claim that the only solution is $n = 1, 3.$ Let $p$ be the smallest prime dividing $n.$ If $p$ doesn't exist, we get that $n = 1$, which is a valid solution. If $p$ exists, I claim that $p = 3.$

We can write that $2^{2n}\equiv1\pmod p,$ while $2^{p - 1} \equiv 1\pmod p.$ Thus, $\operatorname{ord}_p(2) \mid \gcd(2n, p - 1).$ However, since $p$ is the smallest prime dividing $n,$ we have that $\operatorname{ord}_p(2) \mid 2.$ Clearly, the order can't be 1. Thus, the order is 2. Hence, $2^{2}\equiv1\pmod p\implies p = 3.$

Now, let $n = 3k_0.$ Let $p$ be the smallest prime dividing $k_0.$ If $p$ exists, I claim that $3\mid k_0$. If $p$ doesn't exist, we have that $k_0 = 1,$ and thus $n = \boxed{3},$ which is indeed a solution.

We have that $2^{3k_0} \equiv1\pmod{3p},$ while $2^{2(p - 1)}\equiv1\pmod{3p}$ by Euler's theorem. Thus, $\operatorname{ord}_{3p}(2) \mid \gcd(3k_0, 2(p - 1)) = 2, \text{ or }6.$ If it is 1, this clearly doesn't work. If it is $2,$ we have that $p = 1,$ which means that $k_0 = 1.$ If the order is $3,$ this is clearly impossible because we can't have $-1$. If the order is $6,$ then $2^{6} \equiv 1\pmod p \implies p = 3.$ Thus, we have that $3\mid k_0.$

Thus, let $n = 3k_0 = 9k_1.$ Thus, we have that $$\nu_3(2^{9k_1} + 1) = \nu_3(\left(2^{3}\right)^{3k_1} + 1) = 2 + \nu_3(3k_1) = 3 + \nu_3(k_1)$$However, we have that $\nu_3(2^{9k_1} + 1) \geq4,$ and hence $3\mid k_1.$ By infinite descent, we are done.
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Rayvhs
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#124
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I'm not sure if this is fully correct, but here's my approach.
Observe that $n=3$ is a solution and suppose that $n>3$.
Now note that $n$ is odd.
1st case) $3\nmid n$
Let $q$ be a prime divisor of $n$, $q \ne 2, 3$. Then:
\[q \mid n \quad \text{and} \quad n^2 \mid 2^n + 1 \Rightarrow q \mid 2^n + 1\]Applying LTE:
\[v_q(2^n + 1) = v_q(2 + 1) + v_q(n) = v_q(3) + v_q(n) = v_q(n)\]On the other hand:
\[v_q(n^2) = 2v_q(n)\]Since $n^2 \mid 2^n + 1$, we must have:
\[v_q(n^2) \le v_q(2^n + 1)
\Rightarrow 2v_q(n) \le v_q(n)
\Rightarrow v_q(n) \le 0\]But $q \mid n \Rightarrow v_q(n) \ge 1$, a contradiction.

2nd case) Suppose $n = 3^k \cdot t$, where $k \ge 1$ and $\gcd(t, 6) = 1$.

Let $p$ be an odd prime divisor of $t$. Since $p \mid n$, and $n^2 \mid 2^n + 1$, $p \mid 2^n + 1$.

Applying LTE gives
$v_p(2^n + 1) = v_p(2 + 1) + v_p(n) = v_p(3) + v_p(n) = v_p(n)$

But $v_p(n^2) = 2v_p(n)$. So:
$v_p(2^n + 1) = v_p(n) \ ge  2v_p(n) = v_p(n^2)$
Contradiction.

3rd case) Now check when $n =3^k$ satisfies the condition.
Apply LTE again:

$v_3(2^{3^k} + 1) = v_3(2 + 1) + v_3(3^k) = 1 + k$

So:
$1 + k \ge 2k \Rightarrow 1 \ge k \Rightarrow k \le 1$
This post has been edited 4 times. Last edited by Rayvhs, May 11, 2025, 4:24 PM
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