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

G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

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

Prealgebra 2 Self-Paced

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

Introduction to Algebra A Self-Paced

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

Introduction to Counting & Probability Self-Paced

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

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

Introduction to Algebra B Self-Paced

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

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

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

Intermediate: Grades 8-12

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

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

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

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

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

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

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

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

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

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

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

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

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

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

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


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

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

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

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

Physics

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

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

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
Balkan Mathematical Olympiad
ABCD1728   0
13 minutes 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
13 minutes ago
0 replies
A sharp one with 3 var
mihaig   4
N 36 minutes ago by arqady
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
4 replies
mihaig
May 13, 2025
arqady
36 minutes ago
Find all p(x) such that p(p) is a power of 2
truongphatt2668   5
N an hour ago by tom-nowy
Source: ???
Find all polynomial $P(x) \in \mathbb{R}[x]$ such that:
$$P(p_i) = 2^{a_i}$$with $p_i$ is an $i$ th prime and $a_i$ is an arbitrary positive integer.
5 replies
truongphatt2668
Thursday at 1:05 PM
tom-nowy
an hour ago
Interesting problem from a friend
v4913   10
N an hour ago by OronSH
Source: I'm not sure...
Let the incircle $(I)$ of $\triangle{ABC}$ touch $BC$ at $D$, $ID \cap (I) = K$, let $\ell$ denote the line tangent to $(I)$ through $K$. Define $E, F \in \ell$ such that $\angle{EIF} = 90^{\circ}, EI, FI \cap (AEF) = E', F'$. Prove that the circumcenter $O$ of $\triangle{ABC}$ lies on $E'F'$.
10 replies
1 viewing
v4913
Nov 25, 2023
OronSH
an hour ago
No more topics!
Prove that the number of $a$ is o(p)
Seungjun_Lee   13
N Apr 25, 2025 by ihategeo_1969
Source: 2024 FKMO P6
Prove that there exists a positive integer $K$ that satisfies the following condition.

Condition: For any prime $p > K$, the number of positive integers $a \le p$ that $p^2 \mid a^{p-1} - 1$ is less than $\frac{p}{2^{2024}}$
13 replies
Seungjun_Lee
Mar 24, 2024
ihategeo_1969
Apr 25, 2025
Prove that the number of $a$ is o(p)
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 FKMO P6
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Seungjun_Lee
526 posts
#1 • 1 Y
Y by GeoKing
Prove that there exists a positive integer $K$ that satisfies the following condition.

Condition: For any prime $p > K$, the number of positive integers $a \le p$ that $p^2 \mid a^{p-1} - 1$ is less than $\frac{p}{2^{2024}}$
This post has been edited 3 times. Last edited by Seungjun_Lee, Mar 24, 2024, 5:15 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
qwedsazxc
167 posts
#3 • 2 Y
Y by GeoKing, jameskuang1112
Too easy for a P6 (in fact, I solved this instead of P4... I suck at geo) I will come to uploading my solution when I've got time.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lminsl
544 posts
#4 • 8 Y
Y by EthanWYX2009, Kingsbane2139, ljwn357, ys-lg, GeoKing, selberg_atle, bhan2025, lkjin7932
Weird sol, but seems to work anyway:

Fix any $\epsilon > 0$, and we will show that the number of solutions is $\le \epsilon p$. Let $S$ be the set of solutions to $x^{p-1} \equiv 1 \pmod {p^2}$ with $1 \le x \le p$. Let $T$ denote the set of solutions $x$ with $1 \le x \le p^2$. Assume $|S| > \epsilon p$ happens infinitely often.

It is well known that (say, from the existence of a primitive root modulo $p^2$) $T$ contains exactly $p-1$ elements. Now note that if $x, y \in S$, then $xy \in T$, as $1 \le xy \le p^2$ and $(xy)^{p-1} \equiv 1 \pmod {p^2}$. For each $u \in T$, let $c_u$ be the number of solutions to $u = xy$ for $x, y \in S$. Then we have
\[ \left(\sum_{s \in S} s \right)^2 = \sum_{t \in T} tc_t \le \sum_{t \in T} p^2c_t. \]Now note that the left hand side is $\ge \frac{|S|^2(|S|-1)^2}{4}$, so there exists a constant $C (= 10^{-10})$ and $t \in T$ such that $c_t \ge C \epsilon^4 p$. But this means that $\tau(t) \ge C\epsilon^4 \sqrt{t}$ holds infinitely often, which contradicts this. We’re done.
This post has been edited 2 times. Last edited by lminsl, Mar 24, 2024, 7:00 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EthanWYX2009
868 posts
#5 • 3 Y
Y by ys-lg, GeoKing, selberg_atle
The above proof gives $\mathcal O(p^{3/4}),$ what is the best bound?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
qwedsazxc
167 posts
#6 • 1 Y
Y by GeoKing
$a$ must be in the form of $g^{xp}$ mod $p^2$ while not exceeding $n$, and let such $a$s be $a_1,a_2,\ldots,a_k$. Clearly all $a_ia_j$ is also in the form of $g^{xp}$ mod $p^2$. Now write all $k^2$ values of $a_ia_j$. Then each number (call it $t$) can be written out at most $\sigma_0(t)$ times. It isn't hard to show that these are less than $100t^{0.49}$. (Better bounds exist but this is enough to solve the problem.) Thus same numbers can be written at max $100p^{0.98}$ times, where there are only $p-1$ numbers of the form $g^{xp}$ mod $p^2$. Thus $k^2<100p^{1.98}$ and $k<10p^{0.99}$. Set $K$ as $(10\times2^{2024})^{100}$ and we're done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Seungjun_Lee
526 posts
#7 • 1 Y
Y by GeoKing
EthanWYX2009 wrote:
The above proof gives $\mathcal O(p^{3/4}),$ what is the best bound?

It might not be true, but I heard from my friend that we can prove the $o (p^{\frac{1}{2} + \varepsilon})$ bound
This post has been edited 1 time. Last edited by Seungjun_Lee, Aug 29, 2024, 6:30 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
starchan
1610 posts
#8 • 6 Y
Y by David-Vieta, GeoKing, math_comb01, Pranav1056, mxlcv, RM1729
solution
This post has been edited 2 times. Last edited by starchan, Mar 25, 2024, 8:31 AM
Reason: latex error
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ylmath123
105 posts
#9 • 1 Y
Y by GeoKing
starchan wrote:
solution

What is Erdos Szemedei theory???
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Siddharth03
177 posts
#10 • 3 Y
Y by lminsl, GeoKing, bhan2025
lminsl wrote:
Weird sol, but seems to work anyway:

Fix any $\epsilon > 0$, and we will show that the number of solutions is $\le \epsilon p$. Let $S$ be the set of solutions to $x^{p-1} \equiv 1 \pmod {p^2}$ with $1 \le x \le p$. Let $T$ denote the set of solutions $x$ with $1 \le x \le p^2$. Assume $|S| > \epsilon p$ happens infinitely often.

It is well known that (say, from the existence of a primitive root modulo $p^2$) $T$ contains exactly $p-1$ elements. Now note that if $x, y \in S$, then $xy \in T$, as $1 \le xy \le p^2$ and $(xy)^{p-1} \equiv 1 \pmod {p^2}$. For each $u \in T$, let $c_u$ be the number of solutions to $u = xy$ for $x, y \in S$. Then we have
\[ \left(\sum_{s \in S} s \right)^2 = \sum_{t \in T} tc_t \le \sum_{t \in T} p^2c_t. \]Now note that the left hand side is $\ge \frac{|S|^2(|S|-1)^2}{4}$, so there exists a constant $C (= 10^{-10})$ and $t \in T$ such that $c_t \ge C \epsilon^4 p$. But this means that $\tau(t) \ge C\epsilon^4 \sqrt{t}$ holds infinitely often, which contradicts this. We’re done.

A slightly different finish which gives a bound of $o(n^{\frac{1}{2}+\epsilon})$, for any $\epsilon > 0$:

By counting the number of triples $(x,y,u)$ where $x,y \in S, u\in T$ and $xy = u$, we get that
$$|S|^2 \leq \sum_{t\in T} d(s) \leq (p-1) o(p^{2\epsilon}) \implies |S| = o(p^{\frac{1}{2}+\epsilon})$$
This post has been edited 2 times. Last edited by Siddharth03, Mar 25, 2024, 3:25 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathhotspot
70 posts
#11
Y by
@starchan, Archit isn't it more reasonable to use density to improve bounding conditions? I mean the bound O(p^3/4) can be improved i think! Woll, can i improve by using probability ?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathhotspot
70 posts
#12
Y by
@EthanWYX2009 Can't the bound be improved using probability?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tintarn
9044 posts
#14
Y by
So, let me try to clear things up a little bit:
1) There are exactly $p-1$ solutions to $a^{p-1} \equiv 1 \pmod{p^2}$ when we consider $a$ modulo $p^2$. These are exactly the numbers of the form $a=g^{kp}$ where $g$ is a primitive root modulo $p^2$ and they form a subgroup of $(\mathbb{Z}/p^2\mathbb{Z})^{\times}$ of order $p-1$. Moreover, there is a unique such $a$ in each invertible residue class modulo $p$.
2) The problem really is about the following: Could it happen that these $p-1$ values $a$ cluster abnormally among the "small" $a \le p$?
3) As above, the reason why this is impossible is as follows: Suppose that $k$ of the values $a$ are "small" i.e. $\le p$. Then all their $k^2$ products $aa'$ are also solutions (by the subgroup property) and since $a,a' \le p$, their residues modulo $p^2$ are equal to their product.
But using the bound $\tau(n) \ll n^{\varepsilon}$ we see that each such number $\le p^2$ can occur at most $p^{2\varepsilon}$ times as such a product. Hence we get at least $\gg \frac{k^2}{p^{2\varepsilon}}$ distinct solutions $a$ in total. But we have only $p-1$, so we get $k \ll p^{1/2+\varepsilon}$.
4) If we use this with the classical $\varepsilon=\frac{1}{2}$ we get $k \ll p^{3/4}$ as noted in #5. But as noted in #10, the bound for $\tau(n)$ and hence our total bound is really true for any $\varepsilon>0$.
5) What can we expect? By basic probabilistic heuristics, it is clear that we should epect $k \ll p^{\varepsilon}$ to be true for any $\varepsilon>0$. But it is also quite clear that this will be impossibly hard to prove. Even getting some exponent less than $\frac{1}{2}$ could be quite difficult. Note that even an optimal version of the Erdös-Szemeredi Sum-Product Theorem would not help us, since we have essentially already used that the product set is as large as possible in our set-up (for this reason, using the sum-product theorem for this problem is really a complete overkill...).
This post has been edited 1 time. Last edited by Tintarn, Mar 27, 2024, 10:15 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EthanWYX2009
868 posts
#15
Y by
I tried this problem again today with the goal $\mathcal O(p^{1/2+\varepsilon})$ (I have bad memory, usually can't remember solutions after 3 months), not so hard as I thought.

The key is seperating $A$(set of solution) into $B,C$ where $B$ contains all number less than $\sqrt p$ and $C$ contains all number bigger than $\sqrt p.$ We only need to estimate $|C|.$ For $c_1,c_2\in C$ $(c_1c_2)^{p-1} \equiv 1 \pmod {p^2}.$ By $p<c_1c_2<p^2,$ using Hensel Lemma $c_1c_2\mod p\notin A.$

Note that $m=c_1c_2$ is calculated at most $\tau (m)<Cm^{\delta}$ where $\delta>0$ is real. Therefore $$\binom{|C|}2\le C(p^2)^{\delta}\cdot p$$Taking $\delta =\varepsilon/2$ we get $|C|\in \mathcal O(p^{1/2+\varepsilon}),$ done$.\Box$
This post has been edited 1 time. Last edited by EthanWYX2009, Sep 27, 2024, 2:00 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihategeo_1969
236 posts
#16
Y by
Let $S$ denote all $a$ such that $1 \le a <p$ and $p^2 \mid a^{p-1}-1$. Now let $X$ denote all $x$ such that $1 \le x< p^2$ and $p^2 \mid x^{p-1}-1$. By Lagrange/Frobenius/Hensel/primitive roots or whatever you could think of, we have $|X|=p-1$.

Count number of triples $T=(a_1,a_2,a_1a_2)$ where $a_1$, $a_2 \in S$. See that $X$ is closed under multiplication and since $a_1a_2<p^2$ we get the last spot must be in $X$. Hence we have \[|S|^2=T \le \sum_{x \in X} d(x) \le \sum_{x \in X} O(x^{0.01}) \le O \left((p^2-1)^{0.01} \right) (p-1) \implies |S| \le O \left(p^{\frac 12+0.01} \right)=o(p)\]And we are done.
This post has been edited 1 time. Last edited by ihategeo_1969, Apr 27, 2025, 5:56 AM
Z K Y
N Quick Reply
G
H
=
a