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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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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
Sum and product of digits
Sadigly   0
2 minutes ago
Source: Azerbaijan NMO 2018
For a positive integer $n$, define $f(n)=n+P(n)$ and $g(n)=n\cdot S(n)$, where $P(n)$ and $S(n)$ denotes the product and sum of the digits of $n$, respectively. Find all solutions to $f(n)=g(n)$
0 replies
+1 w
Sadigly
2 minutes ago
0 replies
Pentagon with given diameter, ratio desired
bin_sherlo   1
N 4 minutes ago by AnSoLiN
Source: Türkiye 2025 JBMO TST P7
$ABCDE$ is a pentagon whose vertices lie on circle $\omega$ where $\angle DAB=90^{\circ}$. Let $EB$ and $AC$ intersect at $F$, $EC$ meet $BD$ at $G$. $M$ is the midpoint of arc $AB$ on $\omega$, not containing $C$. If $FG\parallel DE\parallel CM$ holds, then what is the value of $\frac{|GE|}{|GD|}$?
1 reply
bin_sherlo
2 hours ago
AnSoLiN
4 minutes ago
Somewhat ugly equation
Sadigly   1
N 10 minutes ago by ehuseyinyigit
Source: Azerbaijan Senior NMO 2019
Solve the following equation $$\sqrt{\frac{x^2}3-ax+a^2}+\sqrt{\frac{x^2}3-bx+b^2}=\sqrt{a^2-ab+b^2}$$where $a;b\in\mathbb{R^+}$
1 reply
Sadigly
39 minutes ago
ehuseyinyigit
10 minutes ago
Expressing polynomial as product of two polynomials
Sadigly   0
10 minutes ago
Source: Azerbaijan Senior NMO 2021
Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$, where $a_1,a_2...,a_n\in\mathbb{Z}$. Prove that $P(x)$ could be expressed as product of two non-constant polynomials with integer coefficients.
0 replies
Sadigly
10 minutes ago
0 replies
Prove or disprove the existence of such a
Sadigly   2
N 11 minutes ago by straight
Source: Azerbaijan NMO
A positive number $a$ is given, such that $a$ could be expressed as difference of two perfect squares ($a=\frac1{n^2}-\frac1{m^2}$). Is it possible for $2a$ to be expressed as difference of two perfect squares?
2 replies
Sadigly
an hour ago
straight
11 minutes ago
Inequality
Sadigly   1
N 11 minutes ago by RevolveWithMe101
Source: Azerbaijan Senior NMO 2019
Prove that for any $a;b;c\in\mathbb{R^+}$, we have $$(a+b)^2+(a+b+4c)^2\geq \frac{100abc}{a+b+c}$$When does the equality hold?
1 reply
Sadigly
34 minutes ago
RevolveWithMe101
11 minutes ago
Selecting 3 people from a 30 man crowd
Sadigly   0
14 minutes ago
Source: Azerbaijan Senior NMO 2021
There are $30$ contestants and each contestant has $6$ friends each. $3$ people is selected from these $30$ contestants, and it is called $good~triple$, if either all three are mutual friends, or none of them are friends with each other. How many $good~triples$ are there?

(Note: If contestant $A$ is friends with $B$, then $B$ is friends with $A$. Similarly, if $A$ is not friends with $B$, then $B$ is not friends with $A$)
0 replies
Sadigly
14 minutes ago
0 replies
It just wants you to factorize 47 factorial
Sadigly   0
20 minutes ago
Source: Azerbaijan Senior NMO 2021
At least how many numbers must be deleted from the product $1 \times 2 \times \dots \times 46 \times 47$ in order to make it a perfect square?
0 replies
Sadigly
20 minutes ago
0 replies
hard inequality omg
tokitaohma   2
N 25 minutes ago by arqady
1. Given $a, b, c > 0$ and $abc=1$
Prove that: $ \sqrt{a^2+1} + \sqrt{b^2+1} + \sqrt{c^2+1} \leq \sqrt{2}(a+b+c) $

2. Given $a, b, c > 0$ and $a+b+c=1 $
Prove that: $ \dfrac{\sqrt{a^2+2ab}}{\sqrt{b^2+2c^2}} + \dfrac{\sqrt{b^2+2bc}}{\sqrt{c^2+2a^2}} + \dfrac{\sqrt{c^2+2ca}}{\sqrt{a^2+2b^2}} \geq \dfrac{1}{a^2+b^2+c^2} $
2 replies
tokitaohma
4 hours ago
arqady
25 minutes ago
Quadratic trinomial swaps values
Sadigly   0
28 minutes ago
Source: Azerbaijan Senior NMO 2017
$P(x)$ is a quadratic trinomial such that there exists $a\neq b$ real numbers that satisfies $P(a)=b$ and $P(b)=a$. Prove that $a,b$ are the only numbers satisfying $P(x)=y$ and $P(y)=x$
0 replies
Sadigly
28 minutes ago
0 replies
Constructing equilateral triangles on 2D/3D planes
Sadigly   0
36 minutes ago
Source: Azerbaijan Senior NMO 2019
Is it possible to construct a equilateral triangle such that:

$\text{a)}$ Coordinates of this triangle are integers in two dimensional plane?

$\text{b)}$ Coordinates of this triangle are integers in three dimensional plane?
0 replies
1 viewing
Sadigly
36 minutes ago
0 replies
Shortest number theory you might've seen in your life
AlperenINAN   3
N 37 minutes ago by NO_SQUARES
Source: Turkey JBMO TST 2025 P4
Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)$ is a perfect square, then $pq + 1$ is also a perfect square.
3 replies
AlperenINAN
2 hours ago
NO_SQUARES
37 minutes ago
Inequality involving semiperimeter
Sadigly   2
N 39 minutes ago by ehuseyinyigit
Source: Azerbaijan Junior NMO 2019
Prove that, for any triangle with side lengths $a,b,c$, the following inequality holds $$\frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}\geq\frac9{8p}$$($p$ denotes the semiperimeter of a triangle)
2 replies
Sadigly
an hour ago
ehuseyinyigit
39 minutes ago
Cyclic Quads and Parallel Lines
gracemoon124   18
N an hour ago by busy-beaver
Source: 2015 British Mathematical Olympiad?
Let $ABCD$ be a cyclic quadrilateral. Let $F$ be the midpoint of the arc $AB$ of its circumcircle which does not contain $C$ or $D$. Let the lines $DF$ and $AC$ meet at $P$ and the lines $CF$ and $BD$ meet at $Q$. Prove that the lines $PQ$ and $AB$ are parallel.
18 replies
gracemoon124
Aug 16, 2023
busy-beaver
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
1916 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
1325 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
1271 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
1752 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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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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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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