Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!

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
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
Inspired by qrxz17
sqing   2
N 12 minutes ago by MathsII-enjoy
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
2 replies
sqing
an hour ago
MathsII-enjoy
12 minutes ago
Prove DK and BC are perpendicular.
yunxiu   63
N 14 minutes ago by sknsdkvnkdvf
Source: 2012 European Girls’ Mathematical Olympiad P1
Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.

Netherlands (Merlijn Staps)
63 replies
yunxiu
Apr 13, 2012
sknsdkvnkdvf
14 minutes ago
Geometry with fix circle
falantrng   34
N 16 minutes ago by Aiden-1089
Source: RMM 2018 Problem 6
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.
34 replies
falantrng
Feb 25, 2018
Aiden-1089
16 minutes ago
Set of perfect powers is irreducible
Assassino9931   2
N 20 minutes ago by navid
Source: Al-Khwarizmi International Junior Olympiad 2025 P4
For two sets of integers $X$ and $Y$ we define $X\cdot Y$ as the set of all products of an element of $X$ and an element of $Y$. For example, if $X=\{1, 2, 4\}$ and $Y=\{3, 4, 6\}$ then $X\cdot Y=\{3, 4, 6, 8, 12, 16, 24\}.$ We call a set $S$ of positive integers good if there do not exist sets $A,B$ of positive integers, each with at least two elements and such that the sets $A\cdot B$ and $S$ are the same. Prove that the set of perfect powers greater than or equal to $2025$ is good.

(In any of the sets $A$, $B$, $A\cdot B$ no two elements are equal, but any two or three of these sets may have common elements. A perfect power is an integer of the form $n^k$, where $n>1$ and $k > 1$ are integers.)

Lajos Hajdu and Andras Sarkozy, Hungary
2 replies
Assassino9931
May 9, 2025
navid
20 minutes ago
Stop Projecting your insecurities
naman12   53
N 22 minutes ago by EeEeRUT
Source: 2022 USA TST #2
Let $ABC$ be an acute triangle. Let $M$ be the midpoint of side $BC$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Suppose that the common external tangents to the circumcircles of triangles $BME$ and $CMF$ intersect at a point $K$, and that $K$ lies on the circumcircle of $ABC$. Prove that line $AK$ is perpendicular to line $BC$.

Kevin Cong
53 replies
naman12
Dec 12, 2022
EeEeRUT
22 minutes ago
Roots of unity
Henryfamz   1
N 28 minutes ago by Mathzeus1024
Compute $$\sec^4\frac\pi7+\sec^4\frac{2\pi}7+\sec^4\frac{3\pi}7$$
1 reply
Henryfamz
May 13, 2025
Mathzeus1024
28 minutes ago
Shortest number theory you might've seen in your life
AlperenINAN   11
N 28 minutes ago by Assassino9931
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.
11 replies
AlperenINAN
May 11, 2025
Assassino9931
28 minutes ago
Inspired by qrxz17
sqing   1
N 34 minutes ago by lbh_qys
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
1 reply
sqing
an hour ago
lbh_qys
34 minutes ago
AZE JBMO TST
IstekOlympiadTeam   10
N 41 minutes ago by Assassino9931
Source: AZE JBMO TST
Prove that there are not intgers $a$ and $b$ with conditions,
i) $16a-9b$ is a prime number.
ii) $ab$ is a perfect square.
iii) $a+b$ is also perfect square.
10 replies
IstekOlympiadTeam
May 2, 2015
Assassino9931
41 minutes ago
Iran TST Starter
M11100111001Y1R   3
N 43 minutes ago by dgrozev
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
3 replies
M11100111001Y1R
May 27, 2025
dgrozev
43 minutes ago
An interesting functional equation
giannis2006   3
N 43 minutes ago by GreekIdiot
Source: Own
Find all functions $f:R^+->R^+$ such that:
$f(xf(y))=xy-xf(x)+f(x)^2$ for all $x,y>0$

The most difficult version of this problem is the following:
Find all functions $f:R^+->R^+$ such that:
$f(xf(y+f(x)))=xy+f(x)^2$ for all $x,y>0$
3 replies
giannis2006
Jun 8, 2023
GreekIdiot
43 minutes ago
A long non-classical problem
M11100111001Y1R   1
N an hour ago by dgrozev
Source: Iran TST 2025 Test 3 Problem 2
Suppose \( n \in \mathbb{N} \) is a natural number. A function \( f(x, y) \) is called \textit{\( n \)-friendly} if for fewer than 1\% of the integers \( k \) with \( -n \leq k \leq n \), the equation \( f(x, y) = k \) has a solution in natural numbers \( (x, y) \) such that \( \frac{y_0}{x_0} \in \left[\frac{1}{100}, 100\right] \), where \( (x_0, y_0) \) is a solution. Suppose \( f(x, y) \leq g(x, y) \), where \( g(x, y) \) is a polynomial with real coefficients, negative leading coefficients, and total degree greater than 2, and for every real number \( x \), we have \( g(x, y) \to \infty \) as \( \frac{y}{x} \in \left[\frac{1}{100}, 100\right] \). Prove that for sufficiently large \( n \), the function \( f \) is not \( n \)-friendly.
1 reply
M11100111001Y1R
May 27, 2025
dgrozev
an hour ago
Another FE
M11100111001Y1R   1
N an hour ago by Mathzeus1024
Source: Iran TST 2025 Test 2 Problem 3
Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for all $x,y>0$ we have:
$$f(f(f(xy))+x^2)=f(y)(f(x)-f(x+y))$$
1 reply
M11100111001Y1R
2 hours ago
Mathzeus1024
an hour ago
Problem 9
SlovEcience   0
an hour ago
Let the sequence $(x_n)$ be defined by
\[
x_1 = 2,\quad x_{n+1} = x_n + \frac{n}{x_n},\quad \text{for all } n \geq 1.
\]Prove that the sequences \( y_n = \frac{x_n}{n} \) and \( z_n = x_n - n \) have finite limits, and find those limits.
0 replies
SlovEcience
an hour ago
0 replies
IMO Shortlist 2012, Number Theory 1
lyukhson   45
N Apr 27, 2025 by sansgankrsngupta
Source: IMO Shortlist 2012, Number Theory 1
Call admissible a set $A$ of integers that has the following property:
If $x,y \in A$ (possibly $x=y$) then $x^2+kxy+y^2 \in A$ for every integer $k$.
Determine all pairs $m,n$ of nonzero integers such that the only admissible set containing both $m$ and $n$ is the set of all integers.

Proposed by Warut Suksompong, Thailand
45 replies
lyukhson
Jul 29, 2013
sansgankrsngupta
Apr 27, 2025
IMO Shortlist 2012, Number Theory 1
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO Shortlist 2012, Number Theory 1
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cj13609517288
1924 posts
#34
Y by
The answer is when $\boxed{\gcd(m,n)=1}$. Obviously this is necessary as otherwise everything in $A$ has to be a multiple of $\gcd(m,n)$.

Now we will prove that this is sufficient. By Bezout's, there exists integers $a$ and $b$ such that
\[ax^2-by^2=1.\]Now let $k_1=(2y^2-2x^2+1)a$ and $k_2=(2y^2-2x^2+1)b$. Then
\[m=x^2+k_1x^2+x^2=((2y^2-2x^2+1)a+2)x^2\in A\]and
\[n=y^2+k_2y^2+y^2=((2y^2-2x^2+1)b+2)y^2\in A.\]But $m=n+1$, so
\[m^2+(-2)(m)(n)+n^2=(m-n)^2=1\in A.\]Now for any positive integer $z$
\[(1)^2+(z-2)(1)(1)+(1)^2=z\in A\,\blacksquare\]
This post has been edited 1 time. Last edited by cj13609517288, Jun 8, 2023, 1:38 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Inconsistent
1455 posts
#35 • 1 Y
Y by ihatemath123
All coprime pairs. If not, just take a prime ideal containing both. If $m, n$ are coprime, then all multiples of $m^2, n^2$ lie in the set from $(m, m)$ and $(n, n)$. Thus by Bezout's theorem, there exists $s, t \in A$ such that $s - t = 1$. However by $k = -2$ we must have $1 = (s-t)^2 \in A$, then from $(1, 1)$ we have all integers must lie in $A$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
S.Das93
709 posts
#36
Y by
Storage
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pikapika007
298 posts
#37
Y by
All integers $(m, n)$ such that $\gcd(m,n) = 1$ work. Clearly $m,n$ not satisfying this condition do not work; then all numbers in $A$ must be a multiple of $\gcd(m,n)$, contradiction.

Otherwise, $(x, y) = (m,m)$ and $(x,y) = (n,n)$ show that all multiples of $m^2, n^2$ are in $A$; by Bezout's, there exists integers $a,b$ so that
\[am^2 - bn^2 = 1;\]taking $(x,y) = (am^2, bn^2)$ and $k = -2$ shows that $1 \in A$; now it is trivial to check that $(1, 1)$ gives that all integers are in $A$.
This post has been edited 1 time. Last edited by pikapika007, Jun 21, 2023, 5:38 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
minusonetwelth
225 posts
#38
Y by
Nice problem:

First note that if $(m,n)=(1,1)$ or $(m,n)=(-1,-1)$ results in $A=\mathbb{Z}$, as $1^2+k\cdot1\cdot1+1^2=2+k$ covers all the integers. Now, let $\gcd(m,n)=s>1$ so that $m=xs$ and $s=ys$. Then each element of $A$ is divisble by $s^2$, so in particular, $1\notin A$, and thus $A\neq\mathbb{Z}$.

Claim: If $m,n\in A$ and $\gcd(m,n)=1$, then $A=\mathbb{Z}$.

Proof: By taking $m=m$, $m^2+km^2+m^2=(k+2)m^2$ all multiples of $m^2$ are in $A$. Similarly, all multiples of $n^2$ are in $A$. By Bezout's theorem, there exist integers $x,y$ such that $xm^2-yn^2=1$. So there are two elements in $A$ which are consecutive. Let them be $r$ and $r+1$. Then,
\[r^2-2r(r+1)+(r+1)^2=r^2-2r^2-2r+r^2+2r+1=1\in A\]and thus, by our first observation, $A=\mathbb{Z}$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lian_the_noob12
173 posts
#39
Y by
$\color{magenta}\boxed{\textbf{SOLUTION N1}}$

If $\gcd(m,n) > 1$, Then $A =$ the multiples of $\gcd(m,n)$ works
So, Suppose $\gcd(m,n) = 1$. Let $P(x,y,k)$ be the statement. Then,
$\bullet P(m,m,k)$ and $P(n,n,k)$ $\implies$ all multiples of $m^2$ and $n^2$ are in $A$.
$\bullet P(am^2,bn^2,2)$ gives $$a^2 \cdot m^4 + 2ab \cdot m^2n^2 + b^2 n^4 = (am^2+bn^2)^2 \in A$$Hence, $1 \in A$.
$\bullet P(1,1,k) \implies  A = {\mathbb Z} \blacksquare$.
This post has been edited 2 times. Last edited by lian_the_noob12, Feb 10, 2024, 12:58 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ywgh1
139 posts
#40
Y by
We claim the answer is for all pairs such that $gcd(m,n)=1$
Assume that $gcd(m.n) > 1$, and assume that $gcd(m,n)=d$ then we can see that the set $A$ will be equal to the multiples of $d$, namely
$$A= \{...-3d,-2d,-d,0,d,2d,3d,....\}$$
Now we show that $gcd(m.n)=1 $ works.

Claim 1: If $j$ is in $A$ then all multiples of $j^2$ are as well in $A$:


Plugging in $(m,m)$ in the equation
\[x^2+kxy+y^2 (1) \]We will have all multiples of $m^2$, similarly for $n^2$.

Claim 2: If $gcd(m,n)=1 $ then $1 \in A$:

We know that $gcd(m^2,n^2)=1$.
Here we apply Bezout lemma, so there exists $a,b$ such that
$am^2+bn^2=1$, so plugging $x=am^2$,$y=bn^2$ and $k=-2$ we have :
$$(am^2)^2+2(am^2)(bn^2)+(bn^2)^2= (am^2+bn^2)^2=1$$Meaning that $1 \in A $ as desired.


Now by Claim 1 we have that all multiples of $1$ are in $A$, so $A$
is the set of all integers.
This post has been edited 1 time. Last edited by Ywgh1, Feb 14, 2024, 8:06 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dolphinday
1329 posts
#41
Y by
Clearly if $m$ and $n$ are not relatively prime then all elements in $A$ will be divisible by $\gcd(m, n)$ which is undesirable(since we won't be able to get numbers that are not divisible by $\gcd(m, n)$.
So then $\gcd(m, n) = 1$. We can plug in $(m, m)$ and $(n, n)$ and vary $k$ to get multiples of $m^2$ and $n^2$ respectively. Since Bezout's lemma states there exists $a$ and $b$ so that $ax + by = \gcd(x, y)$ we can plug in $(am^2, bn^2)$ with $k = 2$ to get $(am^2 + bn^2) \in A$ with $a$ and $b$ chosen so that $am^2 + bn^2 = 1$. Now that $1 \in A$ we can plug in $(1, 1)$ and vary $k$ to get all integers, done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
KevinYang2.71
428 posts
#42 • 2 Y
Y by blueberryfaygo_55, LostDreams
We claim $\boxed{\gcd(m,n)=1}$ is necessary and sufficent. It is necessary since the admissible set generated by $m$ and $n$ is a subset of $(m,n)$ (the ideal).

Suppose $\gcd(m,n)=1$. Let $A$ be the admissible set generated by $m$ and $n$. Then $xm^2,yn^2\in A$ for all integers $x$ and $y$. By Bezout, there exists integers $x$ and $y$ such that $xm^2+yn^2=1$. Then $(xm^2)^2+2(xm^2)(yn^2)+(yn^2)^2=1$ so $1\in A$. The conclusion follows. $\square$
This post has been edited 1 time. Last edited by KevinYang2.71, May 18, 2024, 7:55 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ezpotd
1303 posts
#43
Y by
I claim the answer is only the pairs $m,n$ with greatest common divisor $1$.

To prove nothing else works, note that the set of all numbers divisible by the greatest common divisor of $m,n$ is a working solution.

To prove all such pairs work, consider all numbers of the form $2m^2 + km^2, 2n^2 + kn^2$ are in the set. We can then find two of these numbers with difference $1$ by Bezout's theorem, then set $k = - 2$ and we get that $1$ is part of the set. We are clearly done from there, as all numbers of the form $1^2 + k \cdot 1 \cdot 1 + 1^2 = 2 + k$ are part of the set, which is just all integers.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EpicBird08
1755 posts
#44
Y by
We claim that the answer is all $(m,n)$ such that $\boxed{\gcd(m,n) = 1}.$

First, if $\gcd(m,n) = d > 1,$ then obviously every element of $A$ is divisible by $d,$ so we can make $A$ not contain every positive integer.
On the other hand, if $\gcd(m,n) = 1,$ by using $m,m$ in the given condition, we see that every multiple of $m^2$ is in $A,$ and similarly every multiple of $n^2$ is also in $A.$ Since $\gcd(m,n) = 1,$ we also have $\gcd(m^2,n^2) = 1,$ so by Bezout we may find integers $k,l$ such that $km^2 + ln^2 = 1.$ Since $km^2, ln^2 \in A,$ we see that $$(km^2)^2 + 2(km^2)(ln^2) + (ln^2)^2 = (km^2+ln^2)^2 = 1 \in A.$$Then taking $1,1$ in the given condition, we see that every integer is in $A.$

Thus all $m,n$ such that $\gcd(m,n) = 1$ work, and these are the only solutions, as claimed.
This post has been edited 1 time. Last edited by EpicBird08, Oct 15, 2024, 11:49 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#45
Y by
We claim that the only solutions are coprime pairs. Clearly if they are not coprime, every integer generated will be divisible by their GCD.

Now, we show that coprime pairs $(m, n)$ work. Letting $x=y=m$ yields all multiples of $m^2$ are in $A,$ and similarly all multiples of $n^2$ are in $A.$ By Bezout's Theorem, it follows that there are $a, b$ such that $am^2-bn^2=1.$ Thus letting $x=am^2, y=bn^2, k=-2$ yields $(am^2-bn^2)^2 \in A \implies  1 \in A.$ Now letting $x=y=1$ gives $k+2 \in A$ for all integers $k$ and we are done. QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Sedro
5855 posts
#46
Y by
The answer is all relatively prime $m$ and $n$. If there is some prime $p\mid m,n$, then the set of all multiples of $p$ contains $m$ and $n$, is admissible, but does not contain all integers. We now show that if $\gcd(m,n)=1$, the only admissible set containing both $m$ and $n$ is the set of all integers.

Claim: If $1\in A$ and $A$ is admissible, then $A$ is the set of all integers.

Proof: Since $A$ is admissible, $1^2+k+1^2 = k+2 \in A$ for every integer $k$, as desired. $\blacksquare$

Claim: If an admissible set $A$ contains two consecutive integers, then $A$ is the set of all integers.

Proof: Suppose the two consecutive integers are $r$ and $r+1$. Since $A$ is admissible, $(r+1)^2 + kr(r+1) + r^2\in A$ for any integer $k$. Setting $k=-2$ implies that $1\in A$, which proves the claim. $\blacksquare$

Claim: If $s\in A$, then every integer multiple of $s^2$ is in $A$.

Proof: Since $A$ is admissible, $s^2+ks^2+s^2 = (k+2)s^2\in A$ for every integer $k$, as desired. $\blacksquare$

Suppose that $\gcd(m,n)=1$ and $A$ is an admissible set containing $m$ and $n$. Then, every every multiple of $m^2$ and $n^2$ is in $A$. Since $m$ and $n$ are relatively prime, by Bezout, there exist a multiple of $m^2$ (call it $m'$) and a multiple of $n^2$ (call it $n'$) such that $|m'-n'| = 1$. Since $A$ contains two consecutive integers, it must contain all integers; this completes the proof. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
g0USinsane777
48 posts
#47
Y by
Ans: All pairs $(m,n)$ of integers with $\gcd(m,n)=1$.
Proof that other pairs do not work. First, say that $\gcd(m,n)=d>1$, then notice that the set containing all the integer multiples of $d$ is an admissible set since if any two integer multiples of $d$, say $s$ and $t$ are in an admissible set then so is $s^2 + kst + t^2 = d^2l$, since it is multiple of $d$.

Now, we prove that if two co-prime integers, $m$ and $n$ are in an admissible set $A$, then the only possible set is the set of integers.
Lemma: $1 \in A$
Notice, that $am^2$ for $a \in \mathbb{Z}$ is also in $A$, this is because $(m,m) \in A$ which implies that $(k+2)m^2 \in A$.
Similarly, $bn^2$ for $b \in \mathbb{Z}$ also belongs to $A$.
Now, by bezout's theorem we can find integers $a$ and $b$ such that $am^2+bn^2 = 1$ since $\gcd(m^2,n^2)=1$
Also, for some $x,y$ belonging to $A$, $(x+y)^2 \in A$ by putting $k=2$ in the condition.
So, since $am^2, bn^2 \in A$, then $(am^2+bn^2)^2=1^2=1 \in A$ proving the lemma.

Now, the lemma is enough to finish since we have gotten that $1 \in A$, then putting $x=y=1$ in the condition we get that any $k \in \mathbb{Z}$ belongs to $A$, proving that the only set containing $m,n$ with $\gcd(m,n)=1$ is the set of integers.
This post has been edited 1 time. Last edited by g0USinsane777, Apr 19, 2025, 3:54 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sansgankrsngupta
151 posts
#48
Y by
OG! We claim that the answer is only and all pairs $(m,n)$ such that $gcd(m,n)=1$
Proof that other pairs don't work: Let $gcd(m,n)= d>1$. Then $A=\{dk|k \in \mathbb{Z}\}$ works well , while we note that $A \neq \mathbb{Z}$
Proof that $gcd(m,n)=1$ works:
Claim: There is an integer $t$ such that $t \in A, t \equiv 1 \pmod{n^2}$
Proof: set $x=y=m$ and you get that all integers of the form $m^2(k+2)$ are there; select $k$ such that $k \equiv m^{-2} -2 \pmod{n^2}$ to get $t$.
Let $t=n^2l+1$ , set $x=y=n , k=l-2$, then you get that $n^2l \in A$. Now, set $k=-2, x=t, y= n^2l $ to get that $1 \in A$ taking $x=y=1$ gives that any integer is in $A$. Hence $A=Z$ is the only possible admissible set(obviously $Z$ is an admissible set for any pair $(m,n)$).
Hence, the pairs which satisfy are the ones which we claimed at the start; as desired.
Z K Y
N Quick Reply
G
H
=
a