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
Easy P4 combi game with nt flavour
Maths_VC   1
N 2 hours ago by p.lazarov06
Source: Serbia JBMO TST 2025, Problem 4
Two players, Alice and Bob, play the following game, taking turns. In the beginning, the number $1$ is written on the board. A move consists of adding either $1$, $2$ or $3$ to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is $10$, then in a move a player can't choose the number $2$, but he can choose either $1$ or $3$). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
1 reply
Maths_VC
May 27, 2025
p.lazarov06
2 hours ago
Central sequences
EeEeRUT   14
N 2 hours ago by HamstPan38825
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
14 replies
EeEeRUT
Apr 16, 2025
HamstPan38825
2 hours ago
Elementary Problems Compilation
Saucepan_man02   32
N 3 hours ago by atdaotlohbh
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
32 replies
Saucepan_man02
May 26, 2025
atdaotlohbh
3 hours ago
Random Points = Problem
kingu   5
N 3 hours ago by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
kingu
Apr 27, 2024
happypi31415
3 hours ago
Combo resources
Fly_into_the_sky   1
N 4 hours ago by Fly_into_the_sky
Ok so i never did combinatorics in my life :oops: and i am willing to be able to do P1/P4 combos (or even more)
So yeah how can i start from scratch?
Remark:i don't want compuational combo resources :noo:
1 reply
Fly_into_the_sky
4 hours ago
Fly_into_the_sky
4 hours ago
Very odd geo
Royal_mhyasd   2
N 4 hours ago by Royal_mhyasd
Source: own (i think)
nevermind
2 replies
Royal_mhyasd
Yesterday at 6:10 PM
Royal_mhyasd
4 hours ago
Polynomial Application Sequences and GCDs
pieater314159   46
N 4 hours ago by cursed_tangent1434
Source: ELMO 2019 Problem 1, 2019 ELMO Shortlist N1
Let $P(x)$ be a polynomial with integer coefficients such that $P(0)=1$, and let $c > 1$ be an integer. Define $x_0=0$ and $x_{i+1} = P(x_i)$ for all integers $i \ge 0$. Show that there are infinitely many positive integers $n$ such that $\gcd (x_n, n+c)=1$.

Proposed by Milan Haiman and Carl Schildkraut
46 replies
pieater314159
Jun 19, 2019
cursed_tangent1434
4 hours ago
c^a + a = 2^b
Havu   10
N 4 hours ago by Havu
Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$.
10 replies
Havu
May 10, 2025
Havu
4 hours ago
Own made functional equation
JARP091   0
5 hours ago
Source: Own (Maybe?)
\[
\text{Find all functions } f : \mathbb{R} \to \mathbb{R} \text{ such that:} \\
f(a^4 + a^2b^2 + b^4) = f\left((a^2 - f(ab) + b^2)(a^2 + f(ab) + b^2)\right)
\]
0 replies
JARP091
5 hours ago
0 replies
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   16
N 5 hours ago by JARP091
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
16 replies
OgnjenTesic
May 22, 2025
JARP091
5 hours ago
equal segments on radiuses
danepale   8
N 5 hours ago by zuat.e
Source: Croatia TST 2016
Let $ABC$ be an acute triangle with circumcenter $O$. Points $E$ and $F$ are chosen on segments $OB$ and $OC$ such that $BE = OF$. If $M$ is the midpoint of the arc $EOA$ and $N$ is the midpoint of the arc $AOF$, prove that $\sphericalangle ENO + \sphericalangle OMF = 2 \sphericalangle BAC$.
8 replies
danepale
Apr 25, 2016
zuat.e
5 hours ago
Inequality
SunnyEvan   8
N 5 hours ago by arqady
Let $a$, $b$, $c$ be non-negative real numbers, no two of which are zero. Prove that :
$$ \sum \frac{3ab-2bc+3ca}{3b^2+bc+3c^2} \geq \frac{12}{7}$$
8 replies
SunnyEvan
Apr 1, 2025
arqady
5 hours ago
Inequality conjecture
RainbowNeos   2
N 5 hours ago by RainbowNeos
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
2 replies
RainbowNeos
May 29, 2025
RainbowNeos
5 hours ago
2- player game on a strip of n squares with two game pieces
parmenides51   2
N 5 hours ago by Gggvds1
Source: 2023 Austrian Mathematical Olympiad, Junior Regional Competition , Problem 3
Alice and Bob play a game on a strip of $n \ge  3$ squares with two game pieces. At the beginning, Alice’s piece is on the first square while Bob’s piece is on the last square. The figure shows the starting position for a strip of $ n = 7$ squares.
IMAGE
The players alternate. In each move, they advance their own game piece by one or two squares in the direction of the opponent’s piece. The piece has to land on an empty square without jumping over the opponent’s piece. Alice makes the first move with her own piece. If a player cannot move, they lose.

For which $n$ can Bob ensure a win no matter how Alice plays?
For which $n$ can Alice ensure a win no matter how Bob plays?

(Karl Czakler)
2 replies
parmenides51
Mar 26, 2024
Gggvds1
5 hours ago
BMO 2021 problem 3
VicKmath7   20
N Apr 27, 2025 by Grasshopper-
Source: Balkan MO 2021 P3
Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b]=2021^c$. If $|a-b|$ is a prime number, prove that the number $(a+b)^2+4$ is composite.

Proposed by Serbia
20 replies
VicKmath7
Sep 8, 2021
Grasshopper-
Apr 27, 2025
BMO 2021 problem 3
G H J
G H BBookmark kLocked kLocked NReply
Source: Balkan MO 2021 P3
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
VicKmath7
1391 posts
#1 • 2 Y
Y by jhu08, PineApplePen
Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b]=2021^c$. If $|a-b|$ is a prime number, prove that the number $(a+b)^2+4$ is composite.

Proposed by Serbia
This post has been edited 1 time. Last edited by VicKmath7, Jan 1, 2023, 2:17 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
laikhanhhoang_3011
637 posts
#2 • 1 Y
Y by jhu08
look like it is not difficult but small cases confused us
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
InvertedDiabloNemesisXD
6 posts
#3 • 3 Y
Y by jhu08, Danie1, Arabian_Math
Case Bash solution
Click to reveal hidden text
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
BarisKoyuncu
577 posts
#4 • 2 Y
Y by jhu08, Iora
WLOG $a-b=p$ where $p$ is a prime number.
i) $p|b$
Let $b=pd$. Then, $(a,b)=(pd+p,pd)=p(d+1,d)=p$ and $[a,b]=[pd+p,pd]=p[d+1,d]=pd(d+1)$. Hence, $p+pd(d+1)=2021^c\Rightarrow p|2021^c\Rightarrow p|2021\Rightarrow p=43,47$.
i.a) $p=43$
$2021^c=p(d^2+d+1)=43(d^2+d+1)\Rightarrow d^2+d+(1-43^{c-1}\cdot 47^c)=0$. Hence, the number $\triangle_d 1-4(1-43^{c-1}\cdot 47^c)=4\cdot 43^{c-1}\cdot 47^c-3$ must a perfect square. But, $4\cdot 43^{c-1}\cdot 47^c-3\equiv -3\pmod{47}$ and $47\equiv 2\pmod{3}$. So, this number cannot be a perfect square. Contradiction.
i.b) $p=47$
$2021^c=p(d^2+d+1)=47(d^2+d+1)\Rightarrow d^2+d+(1-43^c\cdot 47^{c-1})=0$. Hence, the number $\triangle_d 1-4(1-43^c\cdot 47{c-1}c)=4\cdot 43^c\cdot 47^{c-1}-3$ must a perfect square. If $c\ge 2$, again $4\cdot 43^c\cdot 47^{c-1}-3\equiv -3\pmod{47}$. Contradiction. Thus, $c=1$. Then, $\triangle_d=4\cdot 43^c\cdot 47^{c-1}-3=4\cdot 43-3=169=13^2$. Hence, $d=\dfrac{-1\pm\sqrt{\triangle_d}}{2}=\dfrac{-1\pm 13}{2}=\{-7,6\}$. Since $d>0$, we find that $d=6$. Then, $b=pd=47\cdot 6=282$ and $a=b+p=282+47=329\Rightarrow (a+b)^2+4=611^2+4\equiv 0\pmod{5}$. Clearly, $611^2+4>5$, so it is composite.
ii) $p\not |b$
Then, $(a,b)=(b+p,b)=(b,p)=1$ and $[a,b]=[b+p,b]=(b+p)b$. Hence, $1+(b+p)b=2021^c\Rightarrow b^2+pb+(1-2021^c)=0$. Hence, the number $\triangle_d=p^2-4(1-2021^c)=p^2+4\cdot 2021^c-4$ must be a perfect square. Let $p^2+4\cdot 2021^c-4=t^2$ where $t\in \mathbb{Z^+}$. Then, $b=\dfrac{-p\pm \sqrt{t^2}}{2}$ and since $b>0$, we find that $b=\dfrac{t-p}{2}$. Then, $(a+b)^2+4=\left(\dfrac{t-p}{2}+\dfrac{t+p}{2}\right)^2+4=t^2+4$. Suppose that $t^2+4=q$ where $q$ is a prime number. Hence, $p^2+4\cdot 2021^c=t^2+4=q$.
ii.a) $c$ is even.
Let $c=2c_1$. Then, $p^2+(2\cdot 2021^{c_1})^2=q=t^2+2^2$. But, each prime number can be written in $1$ or $0$ different way as the sum of $2$ perfect squares. Thus, $\{p,2\cdot 2021^{c_1}\}=\{t,2\}$. Clearly, $2\cdot 2021^{c_1}>2$ so $p=2$. Then $q=p^2+4\cdot 2021^c\equiv 0\mod{2}\Rightarrow q=2$ but $p^2+4\cdot 2021^c>$. Contradiction.
ii.b) $c$ is odd.
If $p\neq 3$, then $t^2+4=p^2+4\cdot 2021^c\equiv (-1)^c\equiv -1\pmod{3}\Rightarrow t^2\equiv 2\pmod{3}$. Contradiction. So $p=3$. Then, $t^2+4=9+4\cdot 2021^c\equiv 9\pmod{47}\Rightarrow t^2\equiv 5\pmod{47}$. Contradiction.
This post has been edited 1 time. Last edited by BarisKoyuncu, Sep 8, 2021, 5:34 PM
Reason: .
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
grupyorum
1435 posts
#6 • 1 Y
Y by jhu08
BarisKoyuncu wrote:
WLOG $a-b=p$ where $p$ is a prime number.
i) $p|b$
Let $b=pd$. Then, $(a,b)=(pd+p,pd)=p(d+1,d)=p$ and $[a,b]=[pd+p,pd]=p[d+1,d]=pd(d+1)$. Hence, $p+pd(d+1)=2021^c\Rightarrow p|2021^c\Rightarrow p|2021\Rightarrow p=43,47$.
i.a) $p=43$
$2021^c=p(d^2+d+1)=43(d^2+d+1)\Rightarrow d^2+d+(1-43^{c-1}\cdot 47^c)=0$. Hence, the number $\triangle_d 1-4(1-43^{c-1}\cdot 47^c)=4\cdot 43^{c-1}\cdot 47^c-3$ must a perfect square. But, $4\cdot 43^{c-1}\cdot 47^c-3\equiv -3\pmod{47}$ and $47\equiv 2\pmod{3}$. So, this number cannot be a perfect square. Contradiction.
i.b) $p=47$
$2021^c=p(d^2+d+1)=47(d^2+d+1)\Rightarrow d^2+d+(1-43^c\cdot 47^{c-1})=0$. Hence, the number $\triangle_d 1-4(1-43^c\cdot 47{c-1}c)=4\cdot 43^c\cdot 47^{c-1}-3$ must a perfect square. If $c\ge 2$, again $4\cdot 43^c\cdot 47^{c-1}-3\equiv -3\pmod{47}$. Contradiction. Thus, $c=1$. Then, $\triangle_d=4\cdot 43^c\cdot 47^{c-1}-3=4\cdot 43-3=169=13^2$. Hence, $d=\dfrac{-1\pm\sqrt{\triangle_d}}{2}=\dfrac{-1\pm 13}{2}=\{-7,6\}$. Since $d>0$, we find that $d=6$. Then, $b=pd=47\cdot 6=282$ and $a=b+p=282+47=329\Rightarrow (a+b)^2+4=611^2+4\equiv 0\pmod{5}$. Clearly, $611^2+4>5$, so it is composite.
ii) $p\not |b$
Then, $(a,b)=(b+p,b)=(b,p)=1$ and $[a,b]=[b+p,b]=(b+p)b$. Hence, $1+(b+p)b=2021^c\Rightarrow b^2+pb+(1-2021^c)=0$. Hence, the number $\triangle_d=p^2-4(1-2021^c)=p^2+4\cdot 2021^c-4$ must be a perfect square. Let $p^2+4\cdot 2021^c-4=t^2$ where $t\in \mathbb{Z^+}$. Then, $b=\dfrac{-p\pm \sqrt{t^2}}{2}$ and since $b>0$, we find that $b=\dfrac{t-p}{2}$. Then, $(a+b)^2+4=\left(\dfrac{t-p}{2}+\dfrac{t+p}{2}\right)^2+4=t^2+4$. Suppose that $t^2+4=q$ where $q$ is a prime number. Hence, $p^2+4\cdot 2021^c=t^2+4=q$.
ii.a) $c$ is even.
Let $c=2c_1$. Then, $p^2+(2\cdot 2021^{c_1})^2=q=t^2+2^2$. But, each prime number can be written in $1$ or $0$ different way as the sum of $2$ perfect squares. Thus, $\{p,2\cdot 2021^{c_1}\}=\{t,2\}$. Clearly, $2\cdot 2021^{c_1}>2$ so $p=2$. Then $q=p^2+4\cdot 2021^c\equiv 0\mod{2}\Rightarrow q=2$ but $p^2+4\cdot 2021^c>$. Contradiction.
ii.b) $c$ is odd.
If $p\neq 3$, then $t^2+4=p^2+4\cdot 2021^c\equiv (-1)^c\equiv -1\pmod{3}\Rightarrow t^2\equiv 2\pmod{3}$. Contradiction. So $p=3$. Then, $t^2+4=9+4\cdot 2021^c\equiv 9\pmod{47}\Rightarrow t^2\equiv 5\pmod{47}$. Contradiction.

You can excise a fair amount of work here. Let $d={\rm gcd}(a,b)$ with $a=da_1$ and $b=db_1$, $(a_1,b_1)=1$. Assume w.l.o.g. $a_1>b_1$ (clearly $a\ne b$). We then have $d(a_1-b_1)=p$ for a prime $p$, thus $d\in\{1,p\}$. Now, if $d=p$ then we obtain that
\[
d\left(b_1^2+b_1+1\right)=43^c\cdot 47^c.
\]If $47\mid b_1^2+b_1+1$, then $47\mid (2b_1+1)^2+3$, but $(-3/47)=-1$ as $47\equiv -1\pmod{6}$. Hence, in this case, $47^c\mid d = p$, thus $c=1$ and $d=47$ is the only possibility. With this we find $b_1^2+b_1+1=43$, for which $(a_1,b_1)=(7,6)$ is obtained; and for this solution, $(a+b)^2+4>5$ is divisible by $5$, hence is the conclusion.

This brings us to the case $(a,b)=1$, $a-b=p$, which is handled exactly as demonstrated by Baris. (Let me also add that one way to prove the also contradiction, $t^2\equiv 5\pmod{47}$, in the very last step is to use the quadratic reciprocity: $(5/47)(47/5)=1$ whereas $(57/5)=(2/5)=-1$.)
This post has been edited 1 time. Last edited by grupyorum, Sep 8, 2021, 5:48 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
steppewolf
351 posts
#7 • 1 Y
Y by jhu08
Proposed by Serbia
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
VicKmath7
1391 posts
#8 • 1 Y
Y by mijail
Woah this was very hard and nice NT, involving QRs. This is similar to the above soln.
Case 1. $gcd(a,b)$ is not $1$. Then $a=p(x+1)$ and $b=px$. Thus $p(x^2+x+1)=2021^c$. Hence $p=43$ or $p=47$.
Case 1.1 $p=47$. The number we want to be composite is $A=4.47^{c+1}.43^c-3.47^2+4$. Note that if $c$ is even, then $A$ is divisible by $3$, done. If $c$ is odd, then $A$ is $(-1)2^{c+1}.(-2)^c+2=2(2^{2c}+1) (mod 5)$ which is divisible by $5$ for odd $c$.
Case 1.2 $p=43$. We prove that this is actually impossible. Note that $(2x+1)^2=4.47^c.43^{c-1}+3$, so $-3$ is a QR modulo $47$, but that's impossible due to quadratic reciprocity.
Case 2. $gcd(a,b)=1$. Thus $1+ab=2021^c$ and $a-b=p$ and we want $A=p^2+4.2021^c$ to be composite.
Case 2.1 $p>3$. Then $c$ can't be odd, otherwise $A$ is divisible by $3$. So suppose $c$ is even. We have that $b^2+bp+1-2021^c=0$ and it's discriminant is $p^2+4.2021^c-4=d^2$. Thus the prime $A$ is representable as sum of two squares in two ways. But that's impossible (view this as a lemma: if $p=a^2+b^2=c^2+d^2$, then $p^2=(ac-bd)^2+(ad+bc)^2=(ad-bc)^2+(ac+bd)^2$ but note that $a^2c^2=b^2d^2 (mod p)$, and now we easily see contradiction).
Case 2.2 $p=3$ ($2$ is impossible, obviously). We have similarly that $b^2+3b+1-2021^c=0$ so it's discriminant is $4.2021^c+5=d^2$, but now finish again with quadratic reciprocity modulo $47$.
So we're done.
This post has been edited 3 times. Last edited by VicKmath7, Sep 9, 2021, 6:57 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
P2nisic
406 posts
#9
Y by
VicKmath7 wrote:
Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b]=2021^c$. If $|a-b|$ is a prime number, prove that the number $(a+b)^2+4$ is composite.

Let $d=(a,b)$ and $a=dx$,$b=dy$ and suppose that $d$ different from $1$ then:
As $d|x-y|=prime$ we have $x=y+1$ and $d=prime$.
Now at the first equation we have:
$d(y^2+y+1)=2021^c$
If $c>=2$ then $47|y^2+y+1$ by the well known lemma:Let $q=prime=2(mod3)$ then if $q|c^2+cd+d^2$ we have $q|c$ and $q|d$.So $47|1$ contradiction.
If $c=1$ we have $d=47$ and $y=6$ which gives$47^2(6+7)^2+4=0(mod5)$

So $d=1$ and we have:$ab+1=2021^c$ (1)and $|a-b|=p$.(2)
We consider two cases:

If $c=1(mod2)$ then (1) $mod3$ gives $a=b(mod3)$ using condition (2) we have $a=b+3$ so equation (1) became:
$b^2+3b+1=2021^c$ or $(2b+3)^2-5=4*2021^c$
But $(5/43)=(43/5)=(3/5)=-1$ so no solution.

If $c=0(mod2)$ set $c=2d$ then we have:
$ab+1=2021^{2d}$ or $4ab+4=2021^{2d}*4$
or$(a+b)^2-(a-b)^2+4=2021^{2d}*4$
or$(a+b)^2+4=2021^{2d}*4+p^2$.

Suppose that $(a+b)^2+4=prime$ then it is well known that every prime in the form $4k+1$ can be written as a sum of two square in a unique way.
This mean that $p=2$ but it is obvious that $p=odd$ so contradiction.
This post has been edited 1 time. Last edited by P2nisic, Sep 8, 2021, 8:53 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sbealing
308 posts
#10
Y by
Let $p=|a-b|$.

Case 1: $p \vert a,b$ We have:
$$2021^c=(a,b)+[a,b]=p+\frac{ab}{p} \Rightarrow ab=p \left(2021^{c}-p\right) $$$$\Longrightarrow (a+b)^2=(a-b)^2+4ab=p^2+4p\left(2021^{c}-p\right)=p \left(4 \times 2021^{c}-3p\right)$$As $p \vert a+b$ we have $p \vert 4 \times 2021^{c}$ and hence $p \in \{2,43,47\}$. In the case $p=2$, the quantity in question is even and $>2$ so composite. For $p=47$ observe:
$$(a+b)^2+4 \equiv 2 \left(4-1\right)+4 \equiv 0 \pmod{5}$$and as $(a+b)^2+4>5$ it follows it is composite. Finally, for $p=43$ observe that $2021^{c} \in \{7,11\} \pmod{19}$ thus:
$$(a+b)^2=43 \left(4 \times 2021^{c}-3 \times 43\right) \in \{8,12\} \pmod{19}$$and by a direct check neither of these are quadratic residues modulo $19$ thus this case cannot occur.

Case 2: $p \nmid a,b$ We have:
$$2021^c=(a,b)+[a,b]=1+ab \Rightarrow (a+b)^2+4=(a-b)^2+4\left(1+ab\right)=p^2+4 \times 2021^{c}$$Firstly observe if $p=2$ then the quantity is even and $>2$ so composite. Now consider $p>2$.

Case 2.1: $c$ is evenIn this case, as $p>2$ and $2 \times 2021^{c/2}>2$, it follows $(a+b)^2+2^2$ is composite else we would have a prime written as a sum of two squares in two distinct ways.

Case 2.2: $p=3$ In this case observe:
$$(a+b)^2=4 \times 2021^{c}+3^2-4 \equiv 5 \pmod{43}$$but by LQR as $\mathrm{LHS}$ is a perfect square we have:
$$1=\left(\frac{5}{43}\right)=\left(\frac{43}{5}\right)=\left(\frac{3}{5}\right)=-1$$which is a contradiction.

Case 2.3: $p>3$, $c$ odd Here we have:
$$(a+b)^2+4 \equiv 4 \times (-1)^{c}+1 \equiv 0 \pmod{3}$$so $\mathrm{LHS}$ is divisible by $3$ and $>3$ therefore composite.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
square_root_of_3
78 posts
#11 • 1 Y
Y by MathsLion
I just wonder how the person who came up with this problem thought of this. I wonder at what point did they say 'let's put a-b to be a prime'. Did they first come up with the solution for the case where $(a,b)=1$ and $a-b$ is prime and then just added the particular case to make it longer? Or did they try solving the general $(a,b)+[a,b]=2021^c$ and then managed to just do the first two small cases?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
oVlad
1746 posts
#12 • 3 Y
Y by Pitagar, steppewolf, Mango247
Why so bashy? :noo: Anyways, my only goal while solving this was to shorten the solution as much as possible. I think I succeeded:

Let $b=a+p.$ We then have two cases:

Case One: Assume that $p$ divides $a.$ In other words, let $a=pk$ and $b=p(k+1).$ Our condition is then equivalent to \[\Phi_3(k)=k^2+k+1=\frac{2021^c}{p}\]It's well known that for any prime number $q$ and positive integer $n,$ only prime numbers congruent to $0$ or 1 modulo $q$ can divide $\Phi_q(n).$

Thus, since $47\equiv 2\bmod 3$ then $47\nmid \Phi_3(k)$ so $47\nmid 2021^c/p.$ Therefore, $c=1$ and $p=47.$ After computing, this yields $k=6.$ Just bash $(a+b)^2+4.$

Case Two: Assume that $p$ does not divide $a.$ Then, our condition rewrites as \[1+a(a+p)=2021^c\iff (2a+p)^2+4=4\cdot 2021^c+p^2.\]
Assume that $p>3.$ Clearly, $3$ cannot divide $(2a+p)^2+4$ so $4\cdot 2021^c+p^2\equiv (-1)^c+1\not\equiv 0\bmod 3$ which implies that $c$ is even.

Hence, $(a+b)^2+4=(2a+p)^2+4$ can be written as the sum of $2$ squares in two ways, so it must be composite.

If $p=3$ then $(2a+p)^2\equiv 5\bmod{43}$ which is a contradiction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IndoMathXdZ
694 posts
#13 • 2 Y
Y by steppewolf, mijail
Balkan MO 2021/3 wrote:
Let $a, b$ and $c$ be positive integers satisfying the equation $(a, b) + [a, b]=2021^c$. If $|a-b|$ is a prime number, prove that the number $(a+b)^2+4$ is composite.
WLOG $a > b$. Since $(a,b) \mid |a - b|$, which is a prime number, then there are two possible cases.
Case 01. $(a,b) = p$ for some prime number $p$.
Since $|a - b| = (a,b) = p$. This implies $(a,b) = (px + p, px)$ for some $x \in \mathbb{N}$. Therefore, we get
\[ p(x^2 + x + 1) = 2021^c \]First, we claim that $c = 1$. Otherwise, $x^2 + x + 1 \equiv 0 \pmod{47}$. However, $-3$ is not a QR modulo $47$. Furthermore, this implies that $p = 47$, which gives us $x^2 + x + 1 = 43$, and this gives $x = 6$ as a solution. Just check that
\[ (a + b)^2 + 4 = p^2(2x + 1)^2 + 4 = 47^2 \cdot 13^2 + 4 \equiv 0 \pmod{5} \]and $a + b > 1$, which implies $(a + b)^2 + 4$ is composite.
Case 02. $(a,b) = 1$.
We then have
\[ (a + b)^2 + 4 = (a - b)^2 + 4(ab + 1) = (a - b)^2 + 4 \cdot 2021^c = p^2 + 4 \cdot 2021^c \]We first claim that $|a - b| \not= 3$. Otherwise, then $b^2 + 3b + 1 = 2021^c \equiv 0 \pmod{47}$, and one can check that $5$ is not a QR modulo $47$. We claim that $c$ must be even. Indeed, if $c$ is odd, then $p^2 + 4 \cdot 2021^c \equiv 0 \pmod{3}$.
Now, note that $(a + b)^2 + 2^2$ can be represented as $p^2 + (2 \cdot 2021^{c/2})^2$ as well, and we could quickly check that $p \not= 2$, or otherwise it's composite because it's divisible by $4$. We'll finish off by the following claim and conclude that $(a + b)^2 + 4$ must in fact be composite.

Claim. Every prime $1$ modulo $4$ has a unique representation as a sum of squares.
Proof. Suppose otherwise, that $p = a^2 + b^2 = c^2 + d^2$ for some $a,b,c,d \in \mathbb{Z}$. Then,
\[ (a + bi)(a - bi) = (c + di)(c - di) \]Note that $\mathbb{Z}[i]$ is a UFD, which implies that $a + bi$ and $c + di$ can't both be primes. WLOG $a + bi$ is not a prime. Then, there exists a nontrivial factorization $a + bi = (x + yi)(z + wi)$. Therefore,
\[ p = N(a + bi) = N(x + yi)N(z + wi) = (x^2 + y^2)(z^2 + w^2) \]contradicting the fact that $p$ is a prime.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lazizbek42
548 posts
#14
Y by
By Thue lemma c odd
mod 3 c Evan
This post has been edited 1 time. Last edited by lazizbek42, Dec 4, 2021, 5:14 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
CT17
1481 posts
#15
Y by
WLOG let $a < b$.

Case 1: $(a,b) = 43$. Let $a = 43a'$ and $b = 43(a'+1)$. Then we have

$$43 + 43a'(a'+1) = 2021^c\implies a'^2 + a' + 1 = \frac{2021^c}{43}$$
a contradiction by mod $47$.

Case 2: $(a,b) = 47$. Let $a = 47a'$ and $b = 47(a' + 1)$. Then we have

$$47 + 47a'(a'+1) = 2021^c\implies a'^2 + a' + 1 = \frac{2021^c}{47}$$
a contradiction by mod $47$ unless $c = 1$. When $c = 1$, we have $a' = 6$ so that

$$(a+b)^2 + 4 = (13\cdot 47)^2 + 4\equiv 0\pmod{5}$$
is composite, as desired.

Case 3: $(a,b) = 1$. Let $p = b - a$ so that $a^2 + ap + 1 = 2021^c$. Note that $p\neq 2$, as otherwise $a$ and $b$ would both be even. We have $2$ subcases.

Subcase 3.1: $c$ is odd. Then $2021^c - 1\equiv 1\pmod{3}$, so $a\equiv b\pmod{3}$. Hence $p = 3$, and $a^2 + 3a + 1\equiv 0\pmod{43}$. In particular, the discriminant $5$ of this quadratic must be a QR mod $43$, which we can verify is false with quadratic reciprocity.

Subcase 3.2: $c$ is even. Then we have

$$(a+b)^2 + 4 = (2a + p)^2 + 4 = 4a^2 + 4ap + p^2 + 4 = 4\cdot 2021^c + p^2 = \left(2\cdot 2021^{\frac{c}{2}}\right)^2 + p^2$$
By a well-known theorem, since $(a+b)^2 + 4$ is expressible as the sum of $2$ squares in $2$ different ways it is composite, as desired.
This post has been edited 1 time. Last edited by CT17, Apr 1, 2022, 2:39 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sttsmet
144 posts
#16 • 1 Y
Y by Mango247
Can anybody tell me the NAME of this well known theorem with the sum of two squares??
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
alinazarboland
172 posts
#17
Y by
I though it's gonna require a lot of case works but it didn't. Assume the statement doesn't hold:
If $gcd(a,b) \geq 2$ , then it would be $p \in \{43,47\}$. let $a=px , b=p(x+1)$ so $p(x^2+x+1)=2021^c$. But it's well-known that the polynomial $x^2+x+1$has no prime divisor in the form $3k+2$ , but $47$ is such number. So $p=47$ and $c=1$ and we should've:$x^2+x+1=43$ which means $x=6$ and $a=282$,$b=329$ so we can just compute the desired expression and it wouldn't be a prime number.

Now let $gcd(a,b)=1$ and $ab=2021^c -1$ , which means : $(a+b)^2 + 4 = p^2 + 4.2021^c$ where $p=|a-b|$. if $c$ was odd , we're done since it's divisible by $3$. if it was odd , it's well-known that every prime number in the form $4k+1$ can be UNIQUELY written as $x^2+y^2$ for positive integers $x,y$. So $\{2.2021^{c/2},p\}=\{a+b,2\}$ but by the definition of $p$ , we have $p+1 \le a+b$ so $p=2 , a+b=2.2021^{c/2}$ . this is impossible since one of $a-b , a+b$ for $a=b (mod 2)$ should be divisible by $4$ which is a contradiction and we're done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
alinazarboland
172 posts
#18 • 1 Y
Y by sttsmet
sttsmet wrote:
Can anybody tell me the NAME of this well known theorem with the sum of two squares??

I don't know the name but in post #8 , VicKmath7 explained it. let $p=a^2+b^2=c^2+d^2$ so
$$p^2=(ac-bd)^2+(ad+bc)^2=(ad-bc)^2+(ac+bd)^2 *$$and $a^2c^2=b^2d^2 (mod p)$ follows from the fact that if $p=a^2+b^2$ , $\frac{a}{b}$ (clearly $a,b$ are not zero modulo $p$) is the solution of $x^2 = -1 (modp)$ in Z_p so $\frac{a^2}{b^2}=\frac{d^2}{c^2}$. Which contradicts $*$
This post has been edited 1 time. Last edited by alinazarboland, Aug 22, 2022, 5:02 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ATGY
2502 posts
#19
Y by
brah interesting problem

WLOG, say $a > b$. Let $\gcd(a, b) = d$, $a = dx, b = dy, (x, y) = 1, \text{lcm}(a, b) = dxy$. Notice that $d \mid dxy \implies d \mid 2021^c$. Say:
$$p = (a - b) = dx - dy = d(x - y) \implies d = 1 \; \text{or} \; (x - y) = 1$$Case 1: $d = 1$. This means that $(x, y) = (a, b)$, so we have $1 + ab = 2021^c$. If $c$ is even:
$$(a + b)^2 + 4 = (a - b)^2 + 4ab + 4 = p^2 + 4(ab + 1) = p^2 + 4\cdot2021^c$$If $(a + b)^2 + 4$ was prime, it only can be uniquely represented as a sum of squares, which means $(a + b)^2 + 4 = p^2 + 4\cdot2021^c \implies p = 2$, however, that means it's even, contradiction.
If $c$ was odd, we have $2021^c \equiv 2\mod3 \implies ab \equiv 1 \mod3 \implies a \equiv b\mod3$. However, this means $a - b = 3$ since it's prime, so $a = b + 3 \implies (a + b)^2 = (2b + 3)^2 = 4b(b + 3) + 9 = 4\cdot2021^c + 5 = 4\cdot43^c\cdot47^c + 5$. This means $5$ is a quadratic residue mod $47$.
$$\left(\frac{5}{47}\right) = \left(\frac{47}{5}\right) = \left(\frac{2}{5}\right) = -1$$Contradiction.

Case 2: If $d \neq 1$, we have $x - y = 1 \implies x = y + 1$. Furthermore, $d$ is prime and $d \mid 2021^c \implies d = 43, 47$.
Subcase 2.1: $d = 43$. We have:
$$d + dxy = 43(xy + 1) = 2021^c \implies xy + 1 = 43^{c - 1}\cdot47^c$$We also have $(x + y)^2 = (2y + 1)^2 = 4y^2 + 4y + 1 = 4y(y + 1) + 1 = 4\cdot43^{c - 1}\cdot47^c - 3$, which means $-3$ is a quadratic residue mod 47. We have:
$$\left(\frac{-3}{47}\right) = \left(\frac{-1}{47}\right)\cdot\left(\frac{3}{47}\right) = \left(\frac{47}{3}\right) = \left(\frac{2}{3}\right) = -1$$Contradiction.
Subcase 2.2: $d = 47$. We have:
$$d + dxy = 47(xy + 1) = 2021^c \implies xy + 1 = 43^c\cdot47^{c - 1}$$Now, $(x + y)^2 = (2y + 1)^2 = 4y(y + 1) + 1 = 4\cdot43^c\cdot47^{c - 1} - 3$. For $c > 1$, we are done by the same step as earlier, however if $c = 1$, we have $(2y + 1)^2 = 169 \implies y = 6, x = 7, a = 47\cdot7, b = 47\cdot6, 5 \mid (a + b)^2 + 4$. Hence, we are done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MathLuis
1557 posts
#20
Y by
First WLOG $a>b$ ($a=b$ obviously cannot happen), then since $a-b$ is a prime and $d=(a,b) \mid a-b$ we have $d=a-b$ or $d=1$. Suppose FTSOC that $(a+b)^2+4=p$ was a prime.
Case 1: $d=1$
In this case $ab+1=2021^c$ it means that $ab=2021^c-1$, now if both $a,b$ were even then clearly $(a+b)^2+4$ cannot be a prime as it is divisbile by $4$. Now back to $a-b=q$ where $q$ is a prime we also have that $b^2+qb+1-2021^c=0$ which means by quadratic formula that:
$$b=\frac{-q+\sqrt{q^2+4 \cdot 2021^c-4}}{2} \implies q^2+4 \cdot (2021^c-1)=t^2$$Now clearly $a+b$ is odd so $q \ge 3$, and also notice that $b=\frac{t-q}{2}$ implies that $a=\frac{t+q}{2}$ so we in fact get $a+b=t$, so if we have that $p=t^2+4=q^2+4 \cdot 2021^c$ and $c$ is even then as $p \equiv 1 \pmod 4$ must have exactly one representation of the form $p=x^2+y^2$ (this can be proven using Thue Lemma), then we have that either $4=q^2$ or $4=4 \cdot 2021^c$, of course neither can happen therefore we get a contradiction!. And if $c$ is odd then if $q \ge 5$ we get that as $2021 \equiv -1 \pmod 3$ that $t^2 \equiv 2 \pmod 3$ which is a contradiction so $q=3$, but then $t^2=4 \cdot 2021^c+5 \equiv 5 \pmod 47$ and this is a contradiction as by QR's we have that $\left( \frac{5}{47} \right) = \left( \frac{47}{5} \right)= \left( \frac{2}{5} \right)=-1$ so no such $t$ should exist, contradiction!.
Case 2: $d=q$ prime.
In this case we have $a=qx+q$ and $b=qx$ and $[a,b]=q(x^2+x)$ so $2021^c=qx^2+qx+q=q(x^2+x+1)$ and so $q \mid 2021^c$ therefore $q=43,47$, as an extra notice that if some prime $r \mid x^2+x+1$ then we must have by orders that either $r=3$ and $x \equiv 1 \pmod 3$ or $r \equiv 1 \pmod 3$, but notice that $47 \equiv 2 \pmod 3$ so we must have $q=47$ and $c=1$ or else $47 \mid x^2+x+1$ and that can't happen, which means that we must have $43=x^2+x+1$ and it's clear that the only positive solution is $x=6$ therefore $a=47 \cdot 7=329$ and $b=47 \cdot 6=282$ and thus $p=611^2+4$ must be prime, but this can't happen as then $611^2+4 \equiv 1+4 \equiv 0 \pmod 5$ so $5 \mid p$ which would mean $p=5$, an obvious size contradiction!.
Therefore in either case we can't have that $(a+b)^2+4$ is prime, thus we are done :cool:.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
NuMBeRaToRiC
22 posts
#21
Y by
Is this wrong?
Can someone check it!
Let $a>b$, $a-b=p$ prime and $(a,b)=d$. For the contrary let $(a+b)^2+4=q$ prime. Then $d\mid (a,b)\mid {a-b}=p$, so $d=1$ or $p$.
Case 1: If $d=1$, then $ab+1=[a,b]+(a,b)=2021^c$. If $c$ even then $(a+b)^2+2^2=q=(a-b)^2+4(ab+1)=(a-b)^2+(2\cdot2021^\frac{c}{2})^2$, which is the contradiction, because a prime number has unique represantion as sum of two squares (I think its Fermat theorem). So if $c$ odd then $ab\equiv 1 \pmod 3$, i.e $a\equiv b\pmod 3$, so $a-b=p=3$. So our condition becomes $b^2+3b+1\equiv 0\pmod {43}$ (in fact $b^2+3b+1=2021^c$). In other word $(2b+3)^2\equiv 8\pmod {43}$, i.e $(\frac{8}{43})=1$, but $1=(\frac{8}{43})=(\frac{2^3}{43})=(\frac{2}{43})=(-1)^{\frac{43^2-1}{8}}=-1$, which is contradiction.
Case 2: If $d=p$, then $a=b+p$, $b=pb_1$ and $p(1+b_1(b_1+1))=(a,b)+[a,b]=2021^c$, so $b_1(b_1+1)+1=43^{c-1}47^c$ or $43^c47^{c-1}$. If $47\mid b_1(b_1+1)+1$, then $b_1(b_1+1)+1\equiv 0\pmod {47}$, i.e $(2b_1+1)^2\equiv -3\pmod {47}$. So $(\frac{-3}{47})=1$, but
$1=(\frac{-3}{47})=(\frac{-1}{47})(\frac{3}{47})=(-1)(\frac{47}{3})(-1)^{\frac{(3-1)(47-1)}{4}}=-1$,
which is contradiction. So $47\nmid b_1(b_1+1)+1$, i.e $b_1(b_1+1)+1=43$ ($c=1$), then $b_1=6$, $p=47$, $b=6\cdot47$, $a=7\cdot47$. And $(a+b)^2+4$ is not prime (because it divisible by 5).
So we are done!
This post has been edited 4 times. Last edited by NuMBeRaToRiC, May 2, 2025, 4:04 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Grasshopper-
10 posts
#24
Y by
sttsmet wrote:
Can anybody tell me the NAME of this well known theorem with the sum of two squares??

I think it's called "Fermat's Two Squares Theorem"
Z K Y
N Quick Reply
G
H
=
a