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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
Game on 6 by 6 grid
billzhao   26
N 27 minutes ago by Sleepy_Head
Source: USAMO 2004, problem 4
Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing in the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number in that row is colored black. Alice wins if she can then draw a line from the top of the grid to the bottom of the grid that stays in black squares, and Bob wins if she can't. (If two squares share a vertex, Alice can draw a line from one to the other that stays in those two squares.) Find, with proof, a winning strategy for one of the players.
26 replies
+1 w
billzhao
Apr 29, 2004
Sleepy_Head
27 minutes ago
USA GEO 2003
dreammath   21
N an hour ago by lpieleanu
Source: TST USA 2003
Let $ABC$ be a triangle and let $P$ be a point in its interior. Lines $PA$, $PB$, $PC$ intersect sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that
\[ [PAF]+[PBD]+[PCE]=\frac{1}{2}[ABC]  \]
if and only if $P$ lies on at least one of the medians of triangle $ABC$. (Here $[XYZ]$ denotes the area of triangle $XYZ$.)
21 replies
dreammath
Feb 16, 2004
lpieleanu
an hour ago
Balkan Mathematical Olympiad
ABCD1728   0
an hour ago
Can anyone provide the PDF version of the book "Balkan Mathematical Olympiads" by Mircea Becheanu and Bogdan Enescu (published by XYZ press in 2014), thanks!
0 replies
ABCD1728
an hour ago
0 replies
An amazing functional equation over positive reals
ariopro1387   0
an hour ago
Source: Iran Team selection test 2025 - P6
Find all functions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$, such that:
$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$
for all $x, y>0$.
0 replies
1 viewing
ariopro1387
an hour ago
0 replies
Nice "if and only if" function problem
ICE_CNME_4   10
N 2 hours ago by maromex
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!
10 replies
ICE_CNME_4
Friday at 7:23 PM
maromex
2 hours ago
Inequality olympiad algebra
Foxellar   0
2 hours ago
Given that \( a, b, c \) are nonzero real numbers such that
\[
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b},
\]let \( M \) be the maximum value of the expression
\[
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}.
\]Determine the sum of the numerator and denominator of the simplified fraction representing \( M \).
0 replies
Foxellar
2 hours ago
0 replies
Integers on a cube
Rushil   6
N 2 hours ago by SomeonecoolLovesMaths
Source: Indian RMO 2004 Problem 2
Positive integers are written on all the faces of a cube, one on each. At each corner of the cube, the product of the numbers on the faces that meet at the vertex is written. The sum of the numbers written on the corners is 2004. If T denotes the sum of the numbers on all the faces, find the possible values of T.
6 replies
Rushil
Feb 28, 2006
SomeonecoolLovesMaths
2 hours ago
Tangents to circle concurrent on a line
Drytime   9
N 3 hours ago by Autistic_Turk
Source: Romania TST 3 2012, Problem 2
Let $\gamma$ be a circle and $l$ a line in its plane. Let $K$ be a point on $l$, located outside of $\gamma$. Let $KA$ and $KB$ be the tangents from $K$ to $\gamma$, where $A$ and $B$ are distinct points on $\gamma$. Let $P$ and $Q$ be two points on $\gamma$. Lines $PA$ and $PB$ intersect line $l$ in two points $R$ and respectively $S$. Lines $QR$ and $QS$ intersect the second time circle $\gamma$ in points $C$ and $D$. Prove that the tangents from $C$ and $D$ to $\gamma$ are concurrent on line $l$.
9 replies
Drytime
May 11, 2012
Autistic_Turk
3 hours ago
Quadratic
Rushil   8
N 3 hours ago by SomeonecoolLovesMaths
Source: Indian RMO 2004 Problem 3
Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + mx -1 = 0$ where $m$ is an odd integer. Let $\lambda _n = \alpha ^n + \beta ^n , n \geq 0$
Prove that
(A) $\lambda _n$ is an integer
(B) gcd ( $\lambda _n , \lambda_{n+1}$) = 1 .
8 replies
Rushil
Feb 28, 2006
SomeonecoolLovesMaths
3 hours ago
$n^{22}-1$ and $n^{40}-1$
v_Enhance   6
N 3 hours ago by BossLu99
Source: OTIS Mock AIME 2024 #13
Let $S$ denote the sum of all integers $n$ such that $1 \leq n \leq 2024$ and exactly one of $n^{22}-1$ and $n^{40}-1$ is divisible by $2024$. Compute the remainder when $S$ is divided by $1000$.

Raymond Zhu

6 replies
v_Enhance
Jan 16, 2024
BossLu99
3 hours ago
Parallelogram in the Plane
Taco12   8
N 3 hours ago by lpieleanu
Source: 2023 Canada EGMO TST/2
Parallelogram $ABCD$ is given in the plane. The incircle of triangle $ABC$ has center $I$ and is tangent to diagonal $AC$ at $X$. Let $Y$ be the center of parallelogram $ABCD$. Show that $DX$ and $IY$ are parallel.
8 replies
Taco12
Feb 10, 2023
lpieleanu
3 hours ago
Combinatorial
|nSan|ty   7
N 3 hours ago by SomeonecoolLovesMaths
Source: RMO 2007 problem
How many 6-digit numbers are there such that-:
a)The digits of each number are all from the set $ \{1,2,3,4,5\}$
b)any digit that appears in the number appears at least twice ?
(Example: $ 225252$ is valid while $ 222133$ is not)
[weightage 17/100]
7 replies
|nSan|ty
Oct 10, 2007
SomeonecoolLovesMaths
3 hours ago
pairs (m, n) such that a fractional expression is an integer
cielblue   0
4 hours ago
Find all pairs $(m,\ n)$ of positive integers such that $\frac{m^3-mn+1}{m^2+mn+2}$ is an integer.
0 replies
cielblue
4 hours ago
0 replies
the same prime factors
andria   6
N 4 hours ago by MathLuis
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
6 replies
andria
Sep 6, 2015
MathLuis
4 hours ago
Impossible divisibility
pohoatza   35
N Apr 29, 2025 by cursed_tangent1434
Source: Romanian TST 3 2008, Problem 3
Let $ m,\ n \geq 3$ be positive odd integers. Prove that $ 2^{m}-1$ doesn't divide $ 3^{n}-1$.
35 replies
pohoatza
Jun 7, 2008
cursed_tangent1434
Apr 29, 2025
Impossible divisibility
G H J
Source: Romanian TST 3 2008, Problem 3
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pohoatza
1145 posts
#1 • 6 Y
Y by nguyendangkhoa17112003, Adventure10, megarnie, Mango247, and 2 other users
Let $ m,\ n \geq 3$ be positive odd integers. Prove that $ 2^{m}-1$ doesn't divide $ 3^{n}-1$.
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Rust
5049 posts
#2 • 2 Y
Y by Adventure10, Mango247
$ 3^n-1\not |2^m-1$, because $ 3^n-1$ even, $ 2^m-1$ odd.
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freemind
337 posts
#3 • 18 Y
Y by fish135, gethd, kgo, nguyendangkhoa17112003, Supercali, Williamgolly, The_Giver, karitoshi, Wizard0001, pavel kozlov, Kanep, nguyenvuthanhha, Adventure10, Mango247, and 4 other users
@Rust: divides means "is a divisor of", not "is divisible by".

Nice problem :).

Assume, to the contrary, that there are such odd numbers $ m,n\ge3$ for which $ 2^m-1|3^n-1$.

Let $ p$ be a prime divisor of $ 2^m-1$ of the form $ 4k+3$. Because $ p|3^n-1$, we have that $ d=\text{ord}_3(p)$ is odd. Since $ d|p-1=4k+2$, we have $ d|2k+1$, hence $ 3^{\frac{p-1}2}=\binom{\underline3}p=1$. Then, by the Quadratic Reciprocity Law, we have $ \binom{\underline3}p\cdot\binom{\underline p}3=(-1)^{\frac{3-1}2\cdot\frac{p-1}2}=-1$ hence $ \binom{\underline p}3=-1$, so $ p=3t+2$.

Let now $ p$ be a prime divisor of $ 2^m-1$ of the form $ 4k+1$. A reasoning just as above and $ \binom{\underline3}p\cdot\binom{\underline p}3=(-1)^{\frac{3-1}2\cdot\frac{p-1}2}=1$ leads to $ \binom{\underline p}3=1$, hence $ p=3t+1$.

So let $ M$ be the multiset of prime divisors $ p$ of $ 2^m-1$ of the form $ 4k+3$, containing each prime with multiplicity equal to its exponent in the prime factorization of $ 2^m-1$. Because $ 2^m-1\equiv 3\pmod 4$, $ |M|$ is odd. But $ M$ contains precisely all prime divisors $ p$ of the form $ 3t+2$ of $ 2^m-1$. Then considering $ \text{mod}$ $ 3$, we have $ 2^m-1\equiv 2^{|M|}=2\pmod 3$, Contradiction.
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Erken
1363 posts
#4 • 1 Y
Y by Adventure10
Suppose that $ 2^{m} - 1$ divides $ 3^{n} - 1$.As we know $ n$ is odd,let $ n = 2k - 1$,where $ k\geq 2$.
It is easy to understand that $ 2^{m} - 1\equiv - 5$ mod $ 12$.
Since $ 2^{m} - 1$ is not divisible by $ 3$,we conclude that all prime divisors of $ 2^{m} - 1$ are congruent to either $ + 5, - 5$,or $ + 1, - 1$ mod $ 12$,but as we know $ 2^{m} - 1\equiv - 5$,
it follows that, there exist $ p\equiv 5$ or $ p\equiv - 5$ mod $ 12$ and $ p$ divides $ 2^{m} - 1$,thus it divides $ 3^{n} - 1$,too,but then $ p$ divides $ 3(3^{n} - 1)$ as well,hence $ 3^{2k} = 3^{n + 1}\equiv 3$ mod $ p$,so $ 3$ is a quadratic residue mod $ p$,but this contradicts to the quadratic reciprocity theorem,because $ p\equiv 5$ or $ - 5$ mod $ 12$.
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TTsphn
1313 posts
#5 • 3 Y
Y by Adventure10, Mango247, Want-to-study-in-NTU-MATH
Other solution (same free mind but have a bit difference)
Call p is a odd prime divisor of $ 3^n - 1$
$ \Rightarrow p|3^{\frac {n + 1}{2}} - 3$
$ \Rightarrow \frac (\frac {3}{p}) = 1$
$ \Rightarrow p\equiv 1 (\mod 12)$ or $ p\equiv - 1 (\mod 12)$
So if $ k|3^n - 1$ then $ k\equiv 1 (\mod 12)$ or $ k\equiv - 1(\mod 12)$
But $ 2^n - 1\equiv 7 (\mod 12)$ ,it give contradiction .
More than $ rad(2^m - 1) \not |3^m - 1$
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kim_ina_88
9 posts
#6 • 1 Y
Y by Adventure10
freemind wrote:
Let $ p$ be a prime divisor of $ 2^m - 1$ of the form $ 4k + 3$. Because $ p|3^n - 1$, we have that $ d = \text{ord}_3(p)$ is odd. Since $ d|p - 1 = 4k + 2$, we have $ d|2k + 1$, hence $ 3^{\frac {p - 1}2} = \binom{\underline3}p = 1$.
how you can say that if dl 4k+2 then dl 2k+1 for example p=7 7l 3^6-1 but 7l 3^3-1
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freemind
337 posts
#7 • 2 Y
Y by Adventure10, Mango247
$ d$ is odd because $ d|n$ and $ n$ is odd.
kim_ina_88 wrote:
freemind wrote:
Let $ p$ be a prime divisor of $ 2^m - 1$ of the form $ 4k + 3$. Because $ p|3^n - 1$, we have that $ d = \text{ord}_3(p)$ is odd. Since $ d|p - 1 = 4k + 2$, we have $ d|2k + 1$, hence $ 3^{\frac {p - 1}2} = \binom{\underline3}p = 1$.
how you can say that if dl 4k+2 then dl 2k+1 for example p=7 7l 3^6-1 but 7l 3^3-1
Z K Y
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AndreiAndronache
88 posts
#8 • 2 Y
Y by pavel kozlov, Adventure10
Very simple, if we use Jacobi's symbol:
We note $3^{\dfrac{n-1}{2}}=a\Rightarrow 3a^2\equiv 1(mod\,\; 2^m-1)\Rightarrow (3a)^2\equiv 3(mod\,\; 2^m-1)\Rightarrow (\dfrac{3}{2^m-1})=1$. By use quadratic reciprocity, $(\dfrac{2^m-1}{3})*(-1)^{\dfrac{3-1}{2}*\dfrac{2^m-1-1}{2}}=1\Rightarrow (\dfrac{2^m-1}{3})=-1$, obviously false, because $2^m\equiv 2(mod\,\; 3)$.
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anantmudgal09
1980 posts
#10 • 1 Y
Y by Adventure10
After noticing that there exists a prime $p$ such that $p \mid 2^m-1$ and that $p \equiv 5,7 (mod 12)$ we conclude by quadratic reciprocity law.
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niyu
830 posts
#12 • 2 Y
Y by Adventure10, Mango247
I will solve the following alternative formulation of this problem:
Alternative Formulation wrote:
Let $a$ and $b$ be positive integers such that $2^a - 1$ divides $3^b - 1$. Prove that either $a = 1$ or $b$ is even.

Solution: Suppose $a > 1$ and $b$ is odd. Let $p \mid 2^a - 1$ be a prime. Then,
\begin{align*}
		p &\mid 3^b - 1 \\
		p &\mid 3^{b + 1} - 3.
	\end{align*}Hence $\left(\frac{3}{p}\right) = 1$. If $p \equiv 1 \pmod{4}$, we have by Quadratic Reciprocity that
\begin{align*}
		\left(\frac{p}{3}\right) &= \left(\frac{3}{p}\right) \\
		&= 1,
	\end{align*}so $p \equiv 1 \pmod{12}$. If $p \equiv -1 \pmod{4}$, we have by Quadratic Reciprocity that
\begin{align*}
		\left(\frac{p}{3}\right) &= -\left(\frac{3}{p}\right) \\
		&= -1,
	\end{align*}so $p \equiv -1 \pmod{12}$. Hence, $p$ must be of the form $12k \pm 1$. Since $a > 1$, we have
\begin{align*}
		2^a - 1 &\equiv -1 \pmod{4}.
	\end{align*}Also, note that
\begin{align*}
		2^a - 1 \equiv 0, 1 \pmod{3}.
	\end{align*}Hence
\begin{align*}
		2^a - 1 \equiv 3, 7 \pmod{12}.
	\end{align*}Hence there exists a prime $q \mid 2^a - 1$ that is not of the form $12k \pm 1$, contradiction. Thus, either $a = 1$ or $b$ is even, as desired. $\Box$
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shankarmath
544 posts
#13 • 2 Y
Y by Adventure10, Mango247
QR
This post has been edited 1 time. Last edited by shankarmath, Mar 27, 2019, 3:39 PM
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yayups
1614 posts
#14 • 6 Y
Y by pad, AlastorMoody, skrublord420, Kobayashi, Adventure10, Mango247
Suppose $p$ is some prime factor of $3^n-1$. Then, we have that
\[\left(3^{\frac{n+1}{2}}\right)^2\equiv 3\pmod{p},\]so $3$ is a quadratic residue mod $p$. Using quadratic reciprocity, this tells us that $p\equiv\pm 1\pmod{12}$.

Therefore, all the prime factors of $2^m-1$ are $\pm 1\pmod{12}$, so $2^m-1\equiv\pm 1\pmod{12}$, so in particular, $2^m\equiv 2\pmod{4}$. Thus, $m=1$, which is a contradiction since $m\ge 3$, so we're done.
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amuthup
779 posts
#15 • 1 Y
Y by Want-to-study-in-NTU-MATH
Suppose $b$ is odd and $2^a-1\mid 3^b-1$ for some $a.$ Consider some prime $p$ dividing $2^a-1.$ We have $$3^b-1\equiv 0\pmod{p}$$$$\implies 3^{b+1}\equiv 3\pmod{p}$$$$\implies \left(\frac{3}{p}\right)=1.$$If $p\equiv 1\pmod{4},$ then $$p\equiv \left(\frac{p}{3}\right)\stackrel{QR}{=}\left(\frac{3}{p}\right)=1\pmod{3}.$$If $p\equiv 3\pmod{4},$ then $$p\equiv \left(\frac{p}{3}\right)\stackrel{QR}{=}-\left(\frac{3}{p}\right)=-1\pmod{3}.$$Therefore, $p\equiv\pm 1\pmod{12}.$ Since our choice of $p$ was arbitrary, we must have $$2^a-1\equiv\pm 1\pmod{12}.$$This is true if and only if $a=1,$ so we are done.
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GeronimoStilton
1521 posts
#16 • 3 Y
Y by Mango247, Mango247, Mango247
We do the following stronger problem.
Stronger Statement wrote:
Let $a$ and $b$ be positive integers such that $2^a-1\mid 3^b-1$. Prove that either $a=1$ or $b$ is even.
Suppose otherwise. As $2^a-1 > 1$, $2^a-1=N$ has some prime divisors $p$.

Claim: For $p\mid N$, we have that $p\equiv \pm 1\pmod{12}$.

Solution: Let $o_p(3)$ be the order of $b$ modulo $p$. As $p\mid N\mid 3^b-1$, we have $o_p(3)\mid b$. Since $b$ is odd, we must have $o_p(3)\mid \frac{p-1}{2}$, since we can disregard powers of $2$ that divide $p-1$. This implies that $3$ is a quadratic residue modulo $p$, so $p\equiv \pm 1\pmod{12}$ as desired. $\fbox{}$

Now, note that this implies $N\equiv \pm 1\pmod{12}$. As $N\equiv -1\pmod{4}$, we get that $N\equiv -1\pmod{12}$. In particular, $2^a-1=N\equiv -1\pmod{3}$. But this would imply $3\mid 2^a$, absurd.
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algebra_star1234
2467 posts
#17
Y by
Since $b$ is odd, we have $3^{b+1} \equiv 3 \pmod{2^{a}-1}$ is a quadratic residue. By quadratic reciprocity,
\[ \left( \frac{3}{2^{a}-1}\right)=\left(\frac{2^{a}-1}{3}\right) (-1)^{2^{a-1}-1} = 1 \cdot -1 = -1 .\]for $a > 1$. Therefore $b$ cannot be odd.
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fireflame241
8 posts
#18
Y by
algebra_star1234 wrote:
Since $b$ is odd, we have $3^{b+1} \equiv 3 \pmod{2^{a}-1}$ is a quadratic residue. By quadratic reciprocity,
\[ \left( \frac{3}{2^{a}-1}\right)=\left(\frac{2^{a}-1}{3}\right) (-1)^{2^{a-1}-1} = 1 \cdot -1 = -1 .\]for $a > 1$. Therefore $b$ cannot be odd.

This is close, but it doesn't work because it's missing the $(-1)^\frac{3-1}{2}=-1$ term, so
\[
  \left(\frac{3}{2^a-1}\right) = 1
\]
It would work if $3$ was changed in the original problem to a value that is 1 mod 4, such as $5$.
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Quantum_fluctuations
1282 posts
#19
Y by
Rust wrote:
$ 3^n-1\not |2^m-1$, because $ 3^n-1$ even, $ 2^m-1$ odd.

Odd number can divide an even number but even number can never divide an odd number.
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VulcanForge
626 posts
#20 • 2 Y
Y by pavel kozlov, Mathandski
We have that $3^{b+1} \equiv 3 \pmod{2^a-1}$ is a quadratic residue; by quadratic reciprocity we have $$\left( \frac{3}{2^a-1} \right) \left( \frac{2^a-1}{3} \right) = (-1)^{(2^a-2)(2)/4} = -1$$so $2^a-1$ is a quadratic nonresidue $\pmod{3}$. This is impossible.
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jj_ca888
2726 posts
#21 • 1 Y
Y by Mathandski
The case where $a = 1$ is trivial. Otherwise, FTSoC assume $a > 1$ and $b$ is odd. Also ote that $a$ must also be odd, else $3 \mid 2^a - 1$.

Select a prime $p$ dividing $2^a - 1$; note that $2^{a+1} \equiv 2 \pmod p$ and $3^{b+1} \equiv 3 \pmod p$. Hence,\[\left(\frac{2}{p}\right) = \left(\frac{3}{p}\right) = 1\]must be true. From $\left(\tfrac{2}{p}\right) = 1$, we get that $\tfrac18(p^2 - 1)$ is even which yields $p \equiv \pm 1 \pmod 8$.
  • If $p \equiv 1 \pmod 8$, then\[1 = \left(\frac{3}{p}\right) = \left(\frac{p}{3}\right)\]and clearly $p \neq 3$ since $a$ is odd, so $p \equiv 1 \pmod 3$. Thus $p \equiv 1 \pmod {24}$.
  • If $p \equiv -1 \pmod 8$, then\[1 = \left(\frac{3}{p}\right) = -\left(\frac{p}{3}\right)\]so $p \equiv -1 \pmod 3$. Thus $p \equiv -1 \pmod {24}$.
Since all primes dividing $2^a - 1$ are $\pm 1$ modulo $24$, the number $2^a - 1$ itself must be $\pm 1$ mod $24$. This reduces to either $24 \mid 2^a$ or $24 \mid 2^{a} - 2$ which are both impossible, so we have our desired contradiction. $\blacksquare$
This post has been edited 2 times. Last edited by jj_ca888, Dec 31, 2020, 6:43 PM
Reason: formatting
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pad
1671 posts
#22
Y by
Suppose $2^m-1\mid 3^n-1$. Suppose $p\mid 2^m-1$, so $p\mid 3^n-1$. Hence
\[ \left(3^{\frac{n+1}{2}}\right)^2 \equiv 3^{n+1} \equiv 3 \pmod{p} \implies \left(\frac{3}{p}\right)=1. \]Hence $p\equiv \pm1 \pmod{12}$. Hence all prime factors of $2^m-1$ are $\pm 1 \pmod{12}$, so $2^m-1 \equiv \pm 1\pmod{12}$. So $2^m\equiv 2 \pmod{12}$, which is a contradiction by mod 4.
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bora_olmez
277 posts
#23
Y by
Cool.

We will use the generalization of the Law of Quadratic Reciprocity for the Jacobi Symbol.
Assume FTSOC that $2^m-1 \mid 3^{2k+1}-1$, then notice that $3$ is a quadratic residue $\pmod{2^m-1}$. Moreover that $m$ must be odd as otherwise $3 \mid 2^m-1$ which is not possible and consequently, $$2^m-1 \equiv 1 \pmod{3}$$We have that , $$1 = \left(\frac{2^m-1}{3}\right)= \left(\frac{3}{2^m-1}\right) \cdot \left(\frac{2^m-1}{3}\right) = (-1)^{\frac{2(m-1)}{4}} = -1$$as $m$ is odd which is a contradiction. $\blacksquare$
This post has been edited 1 time. Last edited by bora_olmez, Aug 21, 2021, 10:07 PM
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sriraamster
1492 posts
#24
Y by
$a=1$ is trivial so assume $a>1.$

Consider some $p \mid 2^{a}-1.$ Then, $2^{a+1} \equiv 2 \pmod{p},$ meaning \[ \left( \frac{2}{p} \right) = (-1)^{1/8 (p^2-1)} =1 \iff p \equiv \pm 1 \pmod{8} \]as $a+1$ is even, meaning $2$ is a QR mod $p.$ Similarly, $p \mid 3^{b}-1$ and furthermore \[ \left( \frac{3}{p} \right) \left( \frac{p}{3} \right) = (-1)^{1/2(p-1)}. \]If $p \equiv 2 \pmod{3},$ then $ \left( \frac{p}{3} \right) = -1,$ which also gives $p \equiv -1 \pmod{3}.$ Combined, these give $p \equiv -1 \pmod{24}.$ The other case gives $p \equiv 1 \pmod{24},$ so $p \equiv \pm 1 \pmod{24}.$

Therefore, if $p \mid 2^{a}-1$ we also have $p \equiv \pm 1 \pmod{24}.$ Combined this implies that $2^{a} -1 \equiv \pm 1 \pmod{24}$ which are both contradictions due to divisibility reasons. Thus there are no solutions.
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trying_to_solve_br
191 posts
#25
Y by
Can someone check this sol?

Obviously there exists a prime $p$ congruent to 3 mod 4 dividing $2^m-1$. Now, notice that as $m,n$ are odd, $m=2x+1$ and $n=2y+1$ thus $2^m\equiv 1 (p) \implies 2^{2x}.2\equiv 1 (p) \implies (\frac{1/2}{p})=1$, thus we have by legendre symbol properties that $(\frac{2.1/4}{p})=1=(\frac{1/4}{p}).(\frac{2}{p})=1$ which implies $2$ is a quadratic residue mod $p$, similarly for $3$. Respectively, this implies that $p \equiv \{1,-1\} (8), p \equiv \{1,-1\} (12)$, but as $p\equiv 3 (4)$ this implies $p\equiv -1 (mod 8)$ and $p \equiv -1 (mod 12)$. Quadratic recyprocity gives: $(\frac{3}{p}).(\frac{p}{3})=(-1)^{p-1/2}.(-1)^{1}=1$ and thus $p\equiv 1 (3)$ or $p=3$, which both contradict the fact that $p\equiv -1 (mod 12)$
This post has been edited 1 time. Last edited by trying_to_solve_br, Oct 4, 2021, 3:30 PM
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nguyenvuthanhha
482 posts
#27
Y by
yayups wrote:
Suppose $p$ is some prime factor of $3^n-1$. Then, we have that
\[\left(3^{\frac{n+1}{2}}\right)^2\equiv 3\pmod{p},\]so $3$ is a quadratic residue mod $p$. Using quadratic reciprocity, this tells us that $p\equiv\pm 1\pmod{12}$.

Therefore, all the prime factors of $2^m-1$ are $\pm 1\pmod{12}$, so $2^m-1\equiv\pm 1\pmod{12}$, so in particular, $2^m\equiv 2\pmod{4}$. Thus, $m=1$, which is a contradiction since $m\ge 3$, so we're done.

very good and correct solution :)
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AwesomeYRY
579 posts
#28
Y by
If $a=1$, clearly $2^a-1=1\mid 3^b-1$. Otherwise, if $a$ is even then $3\mid 2^a-1$, so $3\mid 2^a-1\mid 3^b-1$, a contradiction. Thus, we will only consider odd $a$.

Claim 1:For $q\geq 5$, 3 is a quadratic residue mod $q$ if and only if $q\equiv 1,-1\pmod{12}$.
Proof:If $q\equiv 1\pmod{4}$, then $q$ must be a quadratic residue mod 3, so $q\equiv 1\pmod{3}$ too, so $q\equiv 1$. Otherwise, if $q\equiv 3\pmod{4}$, then $q\equiv 2\pmod{3}$ by quadratic recirprocity, so $q\equiv -1$.$\square$


Claim 2:For odd $a$, 3 is not a quadratic residue mod $2^a-1$
Proof: Note that $2,3 \nmid 2^a-1$, and $2^a-1 \equiv 3 \pmod{4}$ and $2^a-1\equiv 1 \pmod{3}$, so $2^a-1\equiv 7\pmod{12}$. Thus, there exists some $q\not\equiv -1,1$ such that $q\mid 2^a-1$. Since by the previous claim, $a$ is not a quadratic residue mod $q$, $a$ is not a quadratic residue mod $2^a-1$.$\square$

Thus, if $b$ is odd, then $3^b=1$ is also not a quadratic residue mod $2^a-1$ a clear contradiction.
This post has been edited 2 times. Last edited by AwesomeYRY, Nov 1, 2021, 3:40 AM
Reason: aops duplicate
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IAmTheHazard
5003 posts
#30 • 1 Y
Y by centslordm
Ok how broken is quadratic reciprocity

We make use of the fact that if a prime $p$ divides $a^k-1$ for $k$ odd, then $a$ is a quadratic residue $\pmod{p}$, since $\mathrm{ord}_p(a)=\tfrac{p-1}{d}$ for some $d \mid k$.
Assume otherwise, and let $p$ be a prime dividing $2^m-1$. Then from the above fact we obtain $(\tfrac{2}{p})=1 \implies p \equiv 1,7 \pmod{8}$. Now, prime factorize
$$2^m-1=p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}.$$Since $2^m-1 \equiv 7 \pmod{8}$, there are an odd number of indices $i$ such that $e_i$ is odd and $p_i \equiv 7 \pmod{8}$call these indices special. Note that the non-special indices $i$ either satisfy $p_i \equiv 1 \pmod{8}$ or $p_i \equiv 7 \pmod{8}$ and $e_i$ is even.
Now, by assumption, we must have $p_i \mid 3^n-1 \implies (\tfrac{3}{p_i})=1$. Consider the following cases:
  • $p_i \equiv 1 \pmod{8}$. Then by quadratic reciprocity
    $$\left(\frac{3}{p_i}\right)\left(\frac{p_i}{3}\right)=1 \implies \left(\frac{p_i}{3}\right)=1 \implies p_i \equiv 1 \pmod{3}$$
  • $p_i \equiv 7 \pmod{8}$. Then by quadratic reciprocity
    $$\left(\frac{3}{p_i}\right)\left(\frac{p_i}{3}\right)=1 \implies \left(\frac{p_i}{3}\right)=-1 \implies p_i \equiv 2 \pmod{3}$$
Take both sides of $2^m-1=p_1^{e_1}\ldots p_k^{e_k}$ modulo $3$. The LHS is clearly $1 \pmod{3}$. However, since the only $i$ with $p_i^{e_i} \equiv 2 \pmod{3}$ are the special $i$, of which there are an odd number, and the rest are $1 \pmod{3}$either because $p_i \equiv 1 \pmod{3}$ or $e_i$ is even—it follows that the RHS is $2 \pmod{3}$, which is a contradiction. Thus no such $(m,n)$ exist, as desired. $\blacksquare$
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HamstPan38825
8868 posts
#31
Y by
Suppose that $b$ is odd. If $a$ is even, then for mod $3$ yields a contradiction.

Thus assume $a$ is odd, Let $p \mid 2^a - 1$; thus $2^a \equiv 1 \pmod p$ for $a$ odd implies $\left(\frac 2p\right) = 1$. Similarly, $\left(\frac 3p\right) = 1$.

On the other hand, the first equality holds for $p \equiv \pm 1 \pmod 8$, and the other equality holds for $$(-1)^{(p-1)/2}\left(\frac p3\right) = 1 \iff p \equiv \pm 1 \pmod {12}.$$So $p \equiv \pm 1 \pmod {24}$ and $2^a - 1 \equiv \pm 1 \pmod {24}$, contradiction.
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kamatadu
480 posts
#32 • 1 Y
Y by HoripodoKrishno
Here is the phrasing I received the problem in.
Romania TST 2008 modified wrote:
Let $a > 1$ and $b > 1$ be positive integers such that $2^a-1\mid 3^b-1$. Prove that $b$ is even.

The following below is my solution.

Firstly, if $a$ is even, then we get that $3 \mid 2^a-1\mid 3^b-1$ which gives a contradiction. So $a$ must be odd. Furthermore, FTSOC assume that $b$ is odd too. So we now use $a=2m+1$ and $b=2n+1$, where $m$ and $n$ are +ve integers.

Now pick any prime $p$ such that $p\mid 2^{2m+1}-1$. Clearly $p\not=3$. This gives us that $2^{2m+1}\equiv 1\pmod{p}\implies 2^{2(m+1)}\equiv 2\pmod{p}$, which means that $2$ is a quadratic residue $\pmod{p}$. Similarly, we get that $p\mid 2^{2m+1}\mid 3^{2n+1}$ which further gives that $3$ is a quadratic residue too. Now we proceed using Legendre's notation.

Firstly, from the fact that $\left(\dfrac{2}{p}\right)=1$, we get that $p\equiv \left\{+1,-1\right\}\pmod{8}$. Also, from $\left(\dfrac{3}{p}\right)=1$, we get that $\left(\dfrac{p}{3}\right)=(-1)^{\dfrac{3-1}{2}\cdot\dfrac{p-1}{2}}\cdot\left(\dfrac{3}{p}\right)=(-1)^{\dfrac{p-1}{2}}$.

Now if $p\equiv 1\pmod{8}$, then we get that $\left(\dfrac{p}{3}\right)=-1$ which further gives us that $p\equiv 1\pmod{3}$. Similarly if $p\equiv -1\pmod{8}$, then we get that $p\equiv -1\pmod{3}$. Now finally combining these two using C.R.T., we get that $p\equiv\left\{+1,-1\right\}\pmod{24}$.

Now as $2^{2m+1}-1$ is just a product of a bunch of such primes, we get that $2^{2m+1}-1\equiv \left\{+1, -1\right\}\pmod{24}$. Both of the cases give simple modular contradictions and we are done. :stretcher:
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Pyramix
419 posts
#33 • 1 Y
Y by Mathandski
Suppose there exist $a,b$ such that $b$ is odd and $a>1$ and $2^a-1\mid 3^b-1$.

Note that $3\nmid 3^b-1$, which means $3\nmid 2^a-1$, and hence $a$ is odd. Let $p$ be an odd prime such that $p\mid 2^a-1\mid 3^b-1$. Since $a,b$ are odd, $\left(\frac 2p\right)=\left(\frac 3p\right)=1$.

Claim: $p\equiv\pm1\pmod{24}$.
Proof. Since $\left(\frac 2p\right)=1$, we have $p\equiv\pm1\pmod{8}$.
Case 1. $p\equiv 1\pmod{8}$. Then, by Quadratic Reciprocity Law, we have $\left(\frac p3\right)=\left(\frac 3p\right)\left(\frac p3\right)=1$, which means $p\equiv 1\pmod{3}$. Hence, $p\equiv1\pmod{3}$. Combining, we get $p\equiv 1\pmod {24}$.
Case 2. $p\equiv -1\pmod{8}$. Similar to above, we get $p\equiv -1\pmod{24}$.

Note that this means $2^a-1\equiv\pm1\pmod{24}$, either of which is impossible. $\blacksquare$
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OronSH
1748 posts
#34 • 1 Y
Y by Mathandski
If $3^n \equiv 1\pmod{2^m-1}$ then $3$ has odd order w.r.t. all primes dividing $2^m-1,$ so it is a QR mod all primes dividing $2^m-1.$ Then quadratic reciprocity gives $\left(\frac3{2^m-1}\right)=-\left(\frac{2^m-1}3\right)=-\left(\frac13\right)=-1,$ thus there is some $p\mid 2^m-1$ for which $\left(\frac3p\right)=-1,$ contradiction
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Markas
150 posts
#35
Y by
Let b be odd for the sake of contradiction. Now let a be even $\Rightarrow$ $3 \mid 2^a - 1$ $\Rightarrow$ $3 \mid 3^b - 1$, which is impossible $\Rightarrow$ a is odd. We have that a and b are odd. Let $p \mid 2^a - 1$ $\Rightarrow$ $p \mid 3^b - 1$ $\Rightarrow$ $2^a \equiv 1 \pmod p$ and $3^b \equiv 1 \pmod p$ and now considering a and b are odd we get that $2^{a+1} \equiv 2 \pmod p$ and $3^{b+1} \equiv 3 \pmod p$ $\Rightarrow$ $\left(\frac 2p\right) = \left(\frac 3p\right) = 1$. Since $\left(\frac 2p\right) = (-1)^{\frac{p^2-1}{8}} = 1$ we have that $p \equiv  \pm 1 \pmod 8$. Also $\left(\frac 3p \right)\left(\frac p3 \right) = (-1)^{\frac{(p-1)(3-1)}{4}} = (-1)^{\frac{p-1}{2}}$. If $p \equiv 1 \pmod 8$ we have that $\left(\frac p3 \right) = 1$ $\Rightarrow$ $p \equiv 1 \pmod 3$ $\Rightarrow$ $p \equiv 1 \pmod {24}$. If $p \equiv -1 \pmod 8$ we have that $\left(\frac p3 \right) = -1$ $\Rightarrow$ $p \equiv -1 \pmod 3$ $\Rightarrow$ $p \equiv -1 \pmod {24}$ $\Rightarrow$ in conclusion $p \equiv \pm 1 \pmod {24}$. Since each different prime divisor of $2^a - 1$ is $\equiv \pm 1 \pmod {24}$ it follows that $2^a - 1 \equiv \pm 1 \pmod {24}$. If $2^a - 1 \equiv -1 \pmod {24}$ we have that $24 \mid 2^a$, which is obviously impossible $\pmod 3$. If $2^a - 1 \equiv 1 \pmod {24}$ we have that $24 \mid 2^a - 2$ $\Rightarrow$ $8 \mid 2^a - 2$, which is impossible for $a > 3$ $\Rightarrow$ we get the desired contradiction.
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onyqz
195 posts
#36 • 1 Y
Y by Mathandski
posting for storage
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maxal
629 posts
#37
Y by
My generalization published as Problem 3883 in Crux Mathematicorum 39:9 (2013):

Let $a,b,c,d$ be positive integers such that $a+b$ and $ad+bc$ are odd. Prove that if $2^a - 3^b>1$, then $2^a - 3^b$ does not divide $2^c + 3^d$.
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Mathandski
773 posts
#38
Y by
$            $
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zuat.e
66 posts
#39
Y by
Assume otherwise, that is $2^a-1\mid3^b -1$ with $b$ even and $a>1$, hence $3^b\equiv1\pmod{2^a-1}$, therefore $3^{b+1}\equiv3\pmod{2^a-1}$, so $3$ is a quadratic residue $\pmod{2^a-1}$, consequently: $\prod_{i=1}^k(\frac{3}{p_i})=(\frac{3}{2^a-1})\equiv-(\frac{2^a-1}{3})\equiv-(\frac{1}{3})\equiv-1$, hence $3$ isn't a $QR$ for some $p_i\mid k$, hence $3$ isn't a $QR$ $\pmod{2^a-1}$
This post has been edited 1 time. Last edited by zuat.e, Jan 26, 2025, 5:23 PM
Reason: My solution was wrong
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cursed_tangent1434
640 posts
#40
Y by
We consider $a>1$ and $b$ odd in what follows. Clearly $a$ must be odd since if $a$ is even, $3 \mid 2^a-1 \mid 3^b-1$ which is a clear contradiction for all positive integers $b$. Note that for each prime $p \mid 2^a-1$,
\begin{align*}
    3^b & \equiv 1 \pmod{p}\\
    3 & \equiv \left (\frac{1}{3^{{\frac{b-1}{2}}}}\right)^2 \pmod{p}
\end{align*}which implies that $3$ is a quadratic residue $\pmod{p}$ for each prime divisor $p$ of $2^a-1$. Further, the Law of Quadratic Reciprocity states that
\[\left(\frac{p}{3}\right)=\left(\frac{3}{p}\right)\left(\frac{p}{3}\right) = (-1)^{\frac{p-1}{2}}\]since $3$ is a quadratic residue $\pmod{p}$. Now if $p \equiv 1 \pmod{4}$ this implies $p$ is a quadratic residue $\pmod{3}$ so $p \equiv 1 \pmod{3}$. And similarly if $p \equiv 3 \pmod{4}$ this implies $p$ is a non-quadratic residue $\pmod{p}$ so $p \equiv 2 \pmod{3}$. Hence, any prime divisor $p$ of $2^a-1$ must be $\pm 1\pmod{12}$.

However, if $a$ is even, $2^a-1 \equiv -1 \pmod{4}$ since $a>1$ and $2^a-1 \equiv 1 \pmod{3}$. Hence, we must have $2^a-1 \equiv 7 \pmod{12}$. But if each prime divisor of $2^a-1$ is $\pm 1 \pmod{12}$, $2^a-1$ must also be $\pm 1 \pmod{12}$ which is a clear contradiction. Hence it is impossible to have $a>1$ and $b$ being odd simultaneously which finishes the problem.
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