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

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

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Brilliant guessing game on triples
Assassino9931   1
N 3 minutes ago by Sardor_lil
Source: Al-Khwarizmi Junior International Olympiad 2025 P8
There are $100$ cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from $1$ to $100$. In a move, Madina is allowed to choose any $3$ cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?

Shubin Yakov, Russia
1 reply
Assassino9931
Yesterday at 9:46 AM
Sardor_lil
3 minutes ago
in n^2-9 has 6 positive divisors than GCD (n-3, n+3)=1
parmenides51   7
N 8 minutes ago by AylyGayypow009
Source: Greece JBMO TST 2016 p3
Positive integer $n$ is such that number $n^2-9$ has exactly $6$ positive divisors. Prove that GCD $(n-3, n+3)=1$
7 replies
parmenides51
Apr 29, 2019
AylyGayypow009
8 minutes ago
Combi Geo
Adywastaken   1
N 19 minutes ago by jainam_luniya
Source: NMTC 2024/8
A regular polygon with $100$ vertices is given. To each vertex, a natural number from the set $\{1,2,\dots,49\}$ is assigned. Prove that there are $4$ vertices $A, B, C, D$ such that if the numbers $a, b, c, d$ are assigned to them respectively, then $a+b=c+d$ and $ABCD$ is a parallelogram.
1 reply
1 viewing
Adywastaken
Yesterday at 3:58 PM
jainam_luniya
19 minutes ago
Calculus
youochange   12
N 20 minutes ago by FriendPotato
Find the area enclosed by the curves $e^x,e^{-x},x^2+y^2=1$
12 replies
youochange
Yesterday at 2:38 PM
FriendPotato
20 minutes ago
The familiar right angle from the orthocenter
buratinogigle   2
N 21 minutes ago by jainam_luniya
Source: Own, HSGSO P6
Let $ABC$ be a triangle inscribed in a circle $\omega$ with orthocenter $H$ and altitude $BE$. Let $M$ be the midpoint of $AH$. Line $BM$ meets $\omega$ again at $P$. Line $PE$ meets $\omega$ again at $Q$. Let $K$ be the orthogonal projection of $E$ on the line $BC$. Line $QK$ meets $\omega$ again at $G$. Prove that $GA\perp GH$.
2 replies
buratinogigle
3 hours ago
jainam_luniya
21 minutes ago
a deep thinking topic. either useless or extraordinary , not yet disovered
jainam_luniya   5
N 22 minutes ago by jainam_luniya
Source: 1.99999999999....................................................................1. it this possible or not we can debate
it can be a new discovery in world or NT
5 replies
jainam_luniya
38 minutes ago
jainam_luniya
22 minutes ago
Divisibilty...
Sadigly   4
N 41 minutes ago by jainam_luniya
Source: Azerbaijan Junior NMO 2025 P2
Find all $4$ consecutive even numbers, such that the sum of their squares divides the square of their product.
4 replies
+1 w
Sadigly
Yesterday at 9:07 PM
jainam_luniya
41 minutes ago
ioqm to imo journey
jainam_luniya   2
N 42 minutes ago by jainam_luniya
only imginative ones are alloud .all country and classes or even colleges
2 replies
jainam_luniya
an hour ago
jainam_luniya
42 minutes ago
Inequality
Sadigly   5
N 43 minutes ago by jainam_luniya
Source: Azerbaijan Junior MO 2025 P5
For positive real numbers $x;y;z$ satisfying $0<x,y,z<2$, find the biggest value the following equation could acquire:


$$(2x-yz)(2y-zx)(2z-xy)$$
5 replies
Sadigly
May 9, 2025
jainam_luniya
43 minutes ago
D'B, E'C and l are congruence.
cronus119   7
N an hour ago by Tkn
Source: 2022 Iran second round mathematical Olympiad P1
Let $E$ and $F$ on $AC$ and $AB$ respectively in $\triangle ABC$ such that $DE || BC$ then draw line $l$ through $A$ such that $l || BC$ let $D'$ and $E'$ reflection of $D$ and $E$ to $l$ respectively prove that $D'B, E'C$ and $l$ are congruence.
7 replies
cronus119
May 22, 2022
Tkn
an hour ago
a set of $9$ distinct integers
N.T.TUAN   17
N an hour ago by hlminh
Source: APMO 2007
Let $S$ be a set of $9$ distinct integers all of whose prime factors are at most $3.$ Prove that $S$ contains $3$ distinct integers such that their product is a perfect cube.
17 replies
N.T.TUAN
Mar 31, 2007
hlminh
an hour ago
Asymmetric FE
sman96   13
N an hour ago by youochange
Source: BdMO 2025 Higher Secondary P8
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that$$f(xf(y)-y) + f(xy-x) + f(x+y) = 2xy$$for all $x, y \in \mathbb{R}$.
13 replies
sman96
Feb 8, 2025
youochange
an hour ago
Divisibility NT
reni_wee   1
N an hour ago by Pal702004
Source: Iran 1998
Suppose that $a$ and $b$ are natural numbers such that
$$p = \frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$$is a prime number. Find all possible values of $a$,$b$,$p$.
1 reply
reni_wee
3 hours ago
Pal702004
an hour ago
Drawing equilateral triangle
xeroxia   0
an hour ago
Equilateral triangle $ABC$ is given. Let $M_a$ and $M_c$ be the midpoints of $BC$ and $AB$, respectively.
A point $D$ on segment $BM_c$ is given. Draw equilateral $\triangle DEF$ such that $E$ is on $BC$ and $F$ is on $AM_a$.
0 replies
xeroxia
an hour ago
0 replies
gcd (a^n+b,b^n+a) is constant
EthanWYX2009   81
N May 4, 2025 by GingerMan
Source: 2024 IMO P2
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
81 replies
EthanWYX2009
Jul 16, 2024
GingerMan
May 4, 2025
gcd (a^n+b,b^n+a) is constant
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 IMO P2
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EthanWYX2009
864 posts
#1 • 17 Y
Y by MathIQ., aaaa_27, NO_SQUARES, GeoKing, VIATON, aidan0626, kamatadu, Sedro, Rounak_iitr, sevket12, Eka01, eduD_looC, farhad.fritl, MS_asdfgzxcvb, cubres, Ibrahim_K, Jackson0423
Determine all pairs $(a,b)$ of positive integers for which there exist positive integers $g$ and $N$ such that
$$\gcd (a^n+b,b^n+a)=g$$holds for all integers $n\geqslant N.$ (Note that $\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y.$)

Proposed by Valentio Iverson, Indonesia
This post has been edited 5 times. Last edited by EthanWYX2009, Jul 19, 2024, 5:26 AM
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hotmonkey1
2192 posts
#2 • 1 Y
Y by cubres
spoilered question
This post has been edited 1 time. Last edited by hotmonkey1, Jul 16, 2024, 1:23 PM
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IndoMathXdZ
692 posts
#3 • 89 Y
Y by ATGY, hotmonkey1, CT17, ChubbyTomato426, lucas3617, aaaa_27, Assassino9931, kingu, CyclicISLscelesTrapezoid, Aryan-23, Sylvestra, BlazingMuddy, InternetPerson10, math90, Davi Medeiros, GuvercinciHoca, Quidditch, Jalil_Huseynov, bin_sherlo, Supertinito, math_comb01, navi_09220114, Funcshun840, Stuffybear, nmoon_nya, ehuseyinyigit, GeoMetrix, Seicchi28, trk08, hamon, thdnder, GeoKing, Mathological03, Sedro, eg4334, Mogmog8, lpieleanu, szpolska, ihatemath123, GorgonMathDota, aidan0626, EpicBird08, NO_SQUARES, talkon, timon92, OronSH, khina, justJen, mathfan2020, MathisWow, centslordm, megarnie, tricky.math.spider.gold.1, ohiorizzler1434, iamnotgentle, MS_Kekas, crocodilepradita, gghx, kamatadu, Filipjack, Capryon, pepat, avisioner, bachkieu, Supercali, MathIQ., RobertRogo, WinterSecret, TheMathCruncher_007, RevolveWithMe101, sarjinius, eduD_looC, erringbubble, MAKEANALITGREATAGAIN2018, khan.academy, Kingsbane2139, oVlad, MathPassionForever, Nartku, Kosiu, somebodyyouusedtoknow, farhad.fritl, GreenTea2593, DensSv, MS_asdfgzxcvb, cubres, giangtruong13, DroneChaudhary, cosinesine
Feels surreal to say that this is the first Indonesia problem on IMO, proposed by yours truly (Valentio Iverson from Indonesia). Pleasantly surprised that it appears as P2! Hope people enjoy this problem as much as I do.

Remark about problem difficulty (spoilers)
Remark about problem creation
This post has been edited 3 times. Last edited by IndoMathXdZ, Jul 16, 2024, 6:27 PM
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Physicsknight
643 posts
#4 • 2 Y
Y by ehuseyinyigit, cubres
$\text{Very nice problem Valentino}$
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mathhotspot
70 posts
#6 • 2 Y
Y by ehuseyinyigit, cubres
Good advanced diophantine practice!
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thdnder
198 posts
#7 • 9 Y
Y by BlazingMuddy, Funcshun840, Iveela, crocodilepradita, kamatadu, AlexCenteno2007, MathIQ., Bazo1, cubres
I claim that the only such pair is $(a, b) = (1, 1)$. Indeed, $(a, b) = (1, 1)$ satisfies the problem condition.

Claim: $ab + 1$ is a power of 2.

Proof. Suppose the exists a prime divisor $2 < p$ that divides $ab + 1$. Then, taking $n \equiv p-2 (p-1)$ and $n > N$, we see that $p \mid a^n + b$ and $p \mid b^n + a$. Therefore, $p \mid g$. Now, taking $n \equiv 0 (p-1)$, we get that $p \mid g  = \gcd(a^n + b, b^n + a)$, so by FLT, $p \mid a + 1, p \mid b + 1 \implies p \mid 2$, a contradiction. $\square$

Now suppose $ab + 1 = 2^k$ for some $k \ge 1$. Taking $\nu_2(n)$ large enough and $n > N$, we see that $\nu_2(a^n - 1) > \nu_2(b + 1)$, so $\nu_2(a^n + b) = \nu_2(a^n - 1 + b + 1) = \nu_2(b + 1)$. Similarly, $\nu_2(b^n + a) = \nu_2(a + 1)$. Hence, $\nu_2(g) = \min(\nu_2(a + 1), \nu_2(b + 1))$.

Take a positive integer $n$ such that $\nu_2(n + 1)$ large and $n > N$, we see that $\nu_2(a^{n+1} - 1) > k$. Let $M = \nu_2(a^{n + 1} - 1)$, then $a^n + b \equiv \frac{1}{a} + b \equiv \frac{ab + 1}{a} (2^M)$, so $\nu_2(a^n + b) = k$. Analogously, $\nu_2(b^n + a) = k$. This implies $\nu_2(g) = k$. However, $\nu_2(g) = \min(\nu_2(a + 1), \nu_2(b + 1))$, this forces to $k = 1$, as wanted. $\blacksquare$
This post has been edited 2 times. Last edited by thdnder, Jul 16, 2024, 2:01 PM
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bsf714
61 posts
#8 • 4 Y
Y by JanHaj, VicKmath7, Haris1, cubres
First note that if $a=b$, we have $(a^n + a, a^n + a) = a^n + a$ which is eventually constant if and only if $a=1$. We may now assume that $a<b$ due to symmetry.

Let $d = (a, b)$ with $a = dx$, $b = dy$ and $(x, y) = 1$. We have $$G_n := (a^n + b, b^n + a) = d(d^{n-1}x^n + y, d^{n-1}y^n + x).$$Further, we have $$\frac{G_n}{d} = (d^{n-1}x^n 
+ y, d^{n-1}y^n + x) = (d^{n-1}x^n 
+ y, x^{n+1} - y^{n+1})$$since $$y^n(d^{n-1}x^n + y) - x^n(d^{n-1}y^n + x) = y^{n+1} - x^{n+1},$$noting that the coefficients $x^n$ and $y^n$ are coprime to $G_n/d$ (indeed, if $p$ is a prime dividing both $x$ and $G_n/d$, then $p\mid d^{n-1}x^n + y$ implies that $p\mid y$, but $x$ and $y$ are coprime; similarly if $p$ is a prime dividing both $y$ and $G_n/d$).

Now, let $q$ be a positive integer parameter, assumed to be coprime to $x, y$ and $d$, as well as such that $q\nmid x-y$. We would like to force $q\mid d^{n-1}x^n + y$ and $q\mid x^{n+1} - y^{n+1}$ for infinitely many $n$, as well as $q\nmid x^{n+1} - y^{n+1}$ for infinitely many $n$. The latter relation gives $(x/y)^{n+1}\equiv 1\pmod q$ and we can force this to happen by choosing $n+1 = k\phi(q)$ for any positive integer $k$, and force it not to happen by choosing $n+1 = k\phi(q) + 1$, since $q\nmid x-y$. It remains to show that such a positive integer $q$ can be chosen with the additional property that $q\mid d^{n-1}x^n + y$ when $n+1 = k\phi(q)$. We need $$d^{k\phi(q)-2}x^{k\phi(q)-1} + y \equiv 0 \pmod q \iff d^{-2}x^{-1}+y\equiv 0\pmod q\iff q\mid d^2xy + 1.$$Note that such $q$ is automatically coprime to $x, y$ and $d$. To ensure that $q\nmid x-y$, we can take $q=d^2xy + 1$. Indeed, if $x\equiv y\pmod q$, we have $d^2xy + 1\mid x-y$, but $y>x$ by the first paragraph and obviously $d^2xy + 1 > y > y - x$.

This finishes the solution, for we have shown that $dq\mid G_n$ for infinitely many $n$ as well as $dq\nmid G_n$ for infinitely many $n$, thus $G_n$ cannot be eventually constant.
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Tintarn
9042 posts
#9 • 21 Y
Y by thdnder, BlazingMuddy, CBMaster, Assassino9931, KST2003, Sedro, Fardad, CahitArf, crocodilepradita, pingupignu, adityaguharoy, sami1618, mastermind.hk16, Begli_I., tag-, Pratik12, MS_asdfgzxcvb, cubres, IndoMathXdZ, X.Luser, Kingsbane2139
Let $q=ab+1$. Then $q \mid a^n+b,b^n+a$ whenever $n \equiv -1 \pmod{\varphi(q)}$, hence $q \mid g$, but then $q \mid a^{n+1}-a^n$ for large $n$ and hence $q \mid a-1$ (since $q$ is clearly coprime to $a$), thus $a=1$ and similarly $b=1$, hence $a=b=1$, which indeed works.
This post has been edited 2 times. Last edited by Tintarn, Jul 23, 2024, 1:15 PM
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BlazingMuddy
282 posts
#10 • 5 Y
Y by abdirasulov, Nartku, ngduchieu1903, Pratik12, cubres
The proposer noted that the main idea is looking at what $ab + 1$ does. Here is another solution, inspired by @above's first step.

Edit: I forgot something. The problem has a funny backstory; I'll let the proposer post it if he's willing to :)

First note that $ab + 1$ is coprime with $a$ and $b$. Picking $n$ big enough with $n \equiv -1 \pmod{\phi(ab + 1)}$ gives $ab + 1 \mid a^n + b$ and $ab + 1 \mid b^n + a$, so $ab + 1 \mid g$. In particular, picking $n$ big enough with $n \equiv 1 \pmod{\phi(ab + 1)}$ gives $ab + 1 \mid a + b$. This forces either $a = 1$ or $b = 1$.

Finally, WLOG $b = 1$. Then $\gcd(a^n + 1, a + 1)$ is eventually constant. For $n$ odd, it is equal to $a + 1$, so $a + 1 \mid a^n + 1$ for all $n$ big enough. Picking $n$ even gives $a + 1 \mid 2 \implies a = 1$, and thus $g = 2$.


Indonesia has made another history! Congratulations to IndoMathXdZ!
This post has been edited 1 time. Last edited by BlazingMuddy, Jul 16, 2024, 3:09 PM
Reason: Add backstory
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math_comb01
662 posts
#11 • 1 Y
Y by cubres
Cute!
Notice that substituting $n=\varphi(ab+1)t-1$ yields that $ab+1 \mid g$ then $ab+1 \mid a^n+b$ now take $n \equiv 1 (\mod \varphi(ab+1))$ to get either $a=1$ or $b=1$, WLOG $b=1$ then $gcd(a^n+1,a+1)$ is constant for $n \geq N$ for odd $n$ it is $a+1$ while for even it divides $2$, so $a=1$, therefore $a=b=1$ is the only solution.
EDIT: sniped by @above
This post has been edited 1 time. Last edited by math_comb01, Jul 16, 2024, 2:44 PM
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Funcshun840
22 posts
#12 • 1 Y
Y by cubres
What is the MOHS of this problem?
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vsamc
3789 posts
#13 • 4 Y
Y by centslordm, KevinYang2.71, persamaankuadrat, cubres
Solution
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EeEeRUT
70 posts
#14 • 1 Y
Y by cubres
Let $\gcd(a,b) = k, a= pk$ and $b=qk$, where $p$ and $q$ are coprime.

So, we get that $$\gcd(k^{n-1}p^{n+1} + pq, k^{n-1}q^{n+1} + pq) = g_1$$$$\gcd(p^{n+1}-q^{n+1},k^{n-1}q^{n+1} + pq) = g_2$$for some $g_1,g_2 \in \mathbb{N}$

Note that $g_2$ is fixed, thus there exist prime $t$ such that $t \mid g_2$

Consider $p^{n+1} \equiv q^{n+1} \pmod{t}$ $$p^{n+2} \equiv p^{n+1}q \equiv q^{n+1} \pmod{t}$$$$p \equiv q \pmod t$$Consider $k^{n-1}q^{n+1} + pq \equiv 0 \pmod t$

Let $M \leqslant n = \phi{(t)} \Phi + c$,we get that $$k^{c-1}q^{c+1} + pq \equiv 0 \pmod t$$Take $c =1$, $$q^2 + pq \equiv 0 \pmod t$$$$p + q \equiv 0 \pmod t$$thus, $p=q$, which follows $a=b$

So, $a^n + a$ has to be a fixed constant, which gives us only $a=1$.

Consequently $(a,b) = (1,1)$
This post has been edited 1 time. Last edited by EeEeRUT, Jul 16, 2024, 3:16 PM
Reason: Bracket
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shanelin-sigma
165 posts
#15 • 2 Y
Y by Funcshun840, cubres
my solution
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Assassino9931
1342 posts
#16 • 6 Y
Y by GeoKing, Orestis_Lignos, Maths_Girl, AlexCenteno2007, EvansGressfield, cubres
Splendid problem! What makes this problem hard is that one can write 30000 things which seem useful but then when logically realizing what one needs to prove, they get thrown in the bin. Solved with Maths_Girl and I admit that personally would probably not have solved it completely in contest conditions.

Step 1: Working with a prime p not dividing a and b is easier - what do we get? Firstly, we show that there does not exist a prime $p \geq 3$ which does not divide $a$ and $b$, but divides $a^n + b$ and $b^n + a$ for all large $n$. Suppose otherwise, then $(-1)^n a^{n^2} + a \equiv 0 \pmod p$. Take $n$ to be even, then $a^{n^2-1} \equiv -1 \pmod p$, so the order of $a$ mod $p$ divides $2n^2 - 2$ for all large $n$. In particular, it divides $2(n+2)^2 - 2 = 2n^2 + 8n + 6$, so it divides $8n+8$ for large even $n$, hence divides $8n+8$ and $8(n+2) + 8 = 8n+24$, i.e. it divides $8$. However, $2n^2-2$ is divisible by $2$, but not by $4$ for all large even $n$, so $a^2 = 1 \pmod p$. Now from $a^{n^2-1} \equiv -1 \pmod p$ for large even $n$ we get $a\equiv -1 \pmod p$. Analogously $b\equiv -1 \pmod p$, but then $n$ odd in $a^n + b$ and $b^n + a$ imply that $p$ must divide $(-1)^n - 1 = -2$, contradiction!

Step 2: If we show that there is a prime $p\geq 3$ such that $a^n + b$ and $b^n + a$ are divisible by $p$ for infinitely many $n$, then $(a,b)$ does not work, since $g$ is divisible by $p$ infinitely often, but not always by Step 1. (Realized this is really needed by playing with $a=4$, $b=2$.) We look for a congruence of the form $n\equiv 
\ ? \pmod p$ where $?$ is a constant in order to apply Fermat's little theorem, for a suitable $p$ dividing an expression of $a$ and $b$. Here $? = 0,1,2$ did not seem to work, but $? = -1$ works! Indeed, $a^{-1} + b$ and $b^{-1} + a$ are divisible by $p$ if and only if $ab+1$ is (and note that if $p$ divides $ab+1$, then $p$ does not divide $a$ and $b$). So if $ab+1$ has a prime divisor $p\geq 3$, then we have obtained a contradiction.

Step 3: Take care of $ab+1$ being a power of $2$. Fortunately, we only need that $ab+1$ is divisible by $4$ (unless $a=b=1$, which satisfies the problem conditions). Indeed, we may assume $a\equiv 3 \pmod 4$ and $b \equiv 1 \pmod 4$ and now if $n$ is even, then $a^n + b$ is divisible by $2$, but not by $4$ (and $b^n + a$ is divisible by $4$), while if $n$ is odd, then both $a^n + b$ and $b^n + a$ are divisible by $4$, so $\gcd(a^n+b,b^n+a)$ is infinitely often divisible by $4$ and not divisible by $4$, contradiction.

EDIT: Reminded now that the trick with $n\equiv -1 \pmod p$ famously appears in ELMO SL 2014 N7 (and to some rather unrelated extent, in IMO 2005/4).
This post has been edited 6 times. Last edited by Assassino9931, Jul 16, 2024, 3:56 PM
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