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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
Disjoint Pairs
MithsApprentice   42
N 14 minutes ago by endless_abyss
Source: USAMO 1998
Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1\leq i\leq 999$) so that for all $i$, $|a_i-b_i|$ equals $1$ or $6$. Prove that the sum \[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|  \] ends in the digit $9$.
42 replies
MithsApprentice
Oct 9, 2005
endless_abyss
14 minutes ago
FE with gcd
a_507_bc   8
N 16 minutes ago by Tkn
Source: Nordic MC 2023 P2
Find all functions $f: \mathbb{N} \to \mathbb{N}$ such that $$\gcd(f(x),y)f(xy)=f(x)f(y)$$for all positive integers $x, y$.
8 replies
a_507_bc
Apr 21, 2023
Tkn
16 minutes ago
2014 JBMO Shortlist G1
parmenides51   19
N 25 minutes ago by tilya_TASh
Source: 2014 JBMO Shortlist G1
Let ${ABC}$ be a triangle with $m\left( \angle B \right)=m\left( \angle C \right)={{40}^{{}^\circ }}$ Line bisector of ${\angle{B}}$ intersects ${AC}$ at point ${D}$. Prove that $BD+DA=BC$.
19 replies
parmenides51
Oct 8, 2017
tilya_TASh
25 minutes ago
Stars and bars i think
RenheMiResembleRice   1
N an hour ago by NicoN9
Source: Diao Luo
Solve the following attached with steps
1 reply
RenheMiResembleRice
an hour ago
NicoN9
an hour ago
Sequence
Titibuuu   1
N an hour ago by Titibuuu
Let \( a_1 = a \), and for all \( n \geq 1 \), define the sequence \( \{a_n\} \) by the recurrence
\[
a_{n+1} = a_n^2 + 1
\]Prove that there is no natural number \( n \) such that
\[
\prod_{k=1}^{n} \left( a_k^2 + a_k + 1 \right)
\]is a perfect square.
1 reply
Titibuuu
6 hours ago
Titibuuu
an hour ago
Show that three lines concur
benjaminchew13   2
N an hour ago by benjaminchew13
Source: Revenge JOM 2025 P2
t $A B C$ be a triangle. $M$ is the midpoint of segment $B C$, and points $E$, $F$ are selected on sides $A B$, $A C$ respectively such that $E$, $F$, $M$ are collinear. The circumcircles $(A B C)$ and $(A E F)$ intersect at a point $P != A$. The circumcircle $(A P M)$ intersects line $B C$ again at a point $D != M$. Show that the lines $A D$, $E F$ and the tangent to $(A E F)$ at point $P$ concur.
2 replies
benjaminchew13
an hour ago
benjaminchew13
an hour ago
slightly easy NT fe
benjaminchew13   2
N an hour ago by benjaminchew13
Source: Revenge JOM 2025 P1
Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $$f(a) + f(b) + f(c) | a^2 + af(b) + cf(a)$$for all $a, b, c\in\mathbb{N}$
2 replies
benjaminchew13
an hour ago
benjaminchew13
an hour ago
Cheesy's math casino
benjaminchew13   1
N an hour ago by benjaminchew13
Source: Revenge JOM 2025 P4
There are $p$ people playing a game at Cheesy's math casino, where $p$ is an odd prime number. Let $n$ be a positive integer. A subset of length $s$ from the set of integers from $1$ to $n$ inclusive is randomly chosen, with an equal probability ($s <= n$ and is fixed). The winner of Cheesy's game is person $i$, if the sum of the chosen numbers are congruent to $i mod p$ for $0 <= i <= p - 1$.

For each $n$, find all values of $s$ such that no one will sue Cheesy for creating unfair games (i.e. all the winning outcomes are equally likely).
1 reply
benjaminchew13
an hour ago
benjaminchew13
an hour ago
2013 Japan MO Finals
parkjungmin   0
an hour ago
help me

we cad do it
0 replies
parkjungmin
an hour ago
0 replies
inequality
benjaminchew13   1
N an hour ago by benjaminchew13
Source: Revenge JOM 2025 P3
Let $n \ge 2$ be a positive integer and let $a_1$, $a_2$, ..., $a_n$ be a sequence of non-negative real numbers. Find the maximum value of $$3(a_1  + a_2 + \cdots + a_n) - (a_1^2 + a_2^2 + \cdots + a_n^2) - (a_1a_2\cdots a_n)$$in terms of $n$.
1 reply
benjaminchew13
an hour ago
benjaminchew13
an hour ago
IMO ShortList 1999, algebra problem 2
orl   11
N an hour ago by ezpotd
Source: IMO ShortList 1999, algebra problem 2
The numbers from 1 to $n^2$ are randomly arranged in the cells of a $n \times n$ square ($n \geq 2$). For any pair of numbers situated on the same row or on the same column the ratio of the greater number to the smaller number is calculated. Let us call the characteristic of the arrangement the smallest of these $n^2\left(n-1\right)$ fractions. What is the highest possible value of the characteristic ?
11 replies
orl
Nov 14, 2004
ezpotd
an hour ago
Coolabra
Titibuuu   2
N an hour ago by no_room_for_error
Let \( a, b, c \) be distinct real numbers such that
\[
a + b + c + \frac{1}{abc} = \frac{19}{2}
\]Find the maximum possible value of \( a \).
2 replies
Titibuuu
6 hours ago
no_room_for_error
an hour ago
Hard centroid geo
lucas3617   0
an hour ago
Source: Revenge JOM 2025 P5
A triangle $A B C$ has centroid $G$. A line parallel to $B C$ passing through $G$ intersects the circumcircle of $A B C$ at $D$. Let lines $A D$ and $B C$ intersect at $E$. Suppose a point $P$ is chosen on $B C$ such that the tangent of the circumcircle of $D E P$ at $D$, the tangent of the circumcircle of $A B C$ at $A$ and $B C$ concur. Prove that $G P = P D$.
0 replies
lucas3617
an hour ago
0 replies
Cute construction problem
Eeightqx   5
N an hour ago by HHGB
Source: 2024 GPO P1
Given a triangle's intouch triangle, incenter, incircle. Try to figure out the circumcenter of the triangle with a ruler only.
5 replies
Eeightqx
Feb 14, 2024
HHGB
an hour ago
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
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cj13609517288
1915 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
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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 $A$.
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S.Das93
709 posts
#36
Y by
Storage
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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
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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.
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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
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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
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dolphinday
1324 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.
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KevinYang2.71
427 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
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ezpotd
1267 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.
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EpicBird08
1751 posts
#44
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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
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Maximilian113
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#45
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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
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Sedro
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#46
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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. $\blacksquare$
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g0USinsane777
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#47
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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
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sansgankrsngupta
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#48
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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.
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