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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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[*]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
Iranians playing with cards module a prime number.
Ryan-asadi   0
6 minutes ago
Source: Iranian Team Selection Test - P2
Let $p$ be an arbitrary prime number.we have a deck of cards which a number is written on the back of each of them. such that for every $i \in \{1,…,p-1\}$ we have at most one card with number $i$ written on its back and we also have exactly one card with number zero on its back.we want to design a game in which we need to decide for every two cards $X,Y$ witch one wins from the other one using the following rules.

$(I)$: If $x$ wins from $y$ and also $y$ wins from $z$ , then $x$ wins from $z$.
$(II)$: If $x$ doesn’t fail $y$ and $z$ doesn’t fail $t$ , then in case of existing of both cards $y+t$ and $x+z$ module $p$, card $x+z$ also doesn’t fail $y+t$.

What is the maximum number of cards which designing such game is possible?

0 replies
Ryan-asadi
6 minutes ago
0 replies
Number theory
Foxellar   0
7 minutes ago
It is known that for all positive integers $k$,
\[
1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k + 1)(2k + 1)}{6}
\]Find the smallest positive integer $k$ such that $1^2 + 2^2 + 3^2 + \ldots + k^2$ is divisible by 200.
0 replies
Foxellar
7 minutes ago
0 replies
Combinatorics
P162008   4
N 34 minutes ago by cazanova19921
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
4 replies
P162008
5 hours ago
cazanova19921
34 minutes ago
An analytic sequence
Ryan-asadi   0
37 minutes ago
Source: Iran Team Selection Test - P1
Let $a_n$ be a sequence of real positive numbers such that for all $n>2025$ :
$$a_n = \max_{1 \le i \le 2025}{a_{n-i}}-\min_{1 \le i \le 2025}{a_{n-i}} $$Prove that for all large enough natural $n$ we have that $a_n < \frac{1}{1404}$.
0 replies
Ryan-asadi
37 minutes ago
0 replies
Aime type Geo
ehuseyinyigit   4
N 39 minutes ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
4 replies
ehuseyinyigit
Monday at 9:04 PM
ehuseyinyigit
39 minutes ago
n variables with n-gon sides
mihaig   1
N 43 minutes ago by mihaig
Source: Own
Let $n\geq3$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$\frac{24}{(n-1)(n-2)}\cdot\sum_{1\leq i<j<k\leq n}{a_ia_ja_k}\geq3\sum_{i=1}^{n}{a_i}+n.$$
1 reply
mihaig
Apr 25, 2025
mihaig
43 minutes ago
Geometry
gggzul   5
N an hour ago by nabodorbuco2
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
5 replies
gggzul
Yesterday at 8:22 AM
nabodorbuco2
an hour ago
Inspired by lgx57
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b>0, a^4+ab+b^4=60 $. Prove that
$$  2\sqrt{15} \leq a^2+ab+b^2 \leq \frac{3(\sqrt{481}-1)}{4}$$$$\frac{\sqrt{481}-1}{4}\leq a^2-ab+b^2  \leq 2\sqrt{15} $$Let $ a,b>0, a^4-ab+b^4=60 $. Prove that
$$ 2\sqrt{15} \leq a^2+ab+b^2 \leq \frac{3(\sqrt{481}+1)}{4}$$$$ 5<a^2-ab+b^2 \leq 2\sqrt{15} $$
1 reply
sqing
an hour ago
sqing
an hour ago
Geometry
Lukariman   4
N an hour ago by lbh_qys
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
4 replies
Lukariman
Yesterday at 12:43 PM
lbh_qys
an hour ago
Need help with barycentric
Sadigly   0
2 hours ago
Hi,is there a good handout/book that explains barycentric,other than EGMO?
0 replies
Sadigly
2 hours ago
0 replies
Find min and max
lgx57   0
3 hours ago
Source: Own
$x_1,x_2, \cdots ,x_n\ge 0$,$\displaystyle\sum_{i=1}^n x_i=m$. $k_1,k_2,\cdots,k_n >0$. Find min and max of
$$\sum_{i=1}^n(k_i\prod_{j=1}^i x_j)$$
0 replies
lgx57
3 hours ago
0 replies
Find min
lgx57   0
3 hours ago
Source: Own
$a,b>0$, $a^4+ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
$a,b>0$, $a^4-ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
0 replies
lgx57
3 hours ago
0 replies
III Lusophon Mathematical Olympiad 2013 - Problem 5
DavidAndrade   2
N 4 hours ago by KTYC
Find all the numbers of $5$ non-zero digits such that deleting consecutively the digit of the left, in each step, we obtain a divisor of the previous number.
2 replies
DavidAndrade
Aug 12, 2013
KTYC
4 hours ago
Maximum number of terms in the sequence
orl   11
N 4 hours ago by navier3072
Source: IMO LongList, Vietnam 1, IMO 1977, Day 1, Problem 2
In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.
11 replies
orl
Nov 12, 2005
navier3072
4 hours ago
(2^n + 1)/n^2 is an integer (IMO 1990 Problem 3)
orl   106
N Apr 26, 2025 by Ilikeminecraft
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.
106 replies
orl
Nov 11, 2005
Ilikeminecraft
Apr 26, 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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ddot1
24719 posts
#103
Y by
Oh, duh - sorry, I didn't read it carefully enough.
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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
128 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.
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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
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#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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