Happy Memorial Day! Please note that AoPS Online is closed May 24-26th.

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
Serbian selection contest for the IMO 2025 - P3
OgnjenTesic   2
N 9 minutes ago by korncrazy
Source: Serbian selection contest for the IMO 2025
Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that:
- $f$ is strictly increasing,
- there exists $M \in \mathbb{N}$ such that $f(x+1) - f(x) < M$ for all $x \in \mathbb{N}$,
- for every $x \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$ such that
\[
            f(y) = \frac{f(x) + f(x + 2024)}{2}.
        \]Proposed by Pavle Martinović
2 replies
OgnjenTesic
May 22, 2025
korncrazy
9 minutes ago
A sharp one with 4 var
mihaig   0
an hour ago
Source: Own
Let $a,b,c,d\geq0$ satisfying
$$\left(a+b+c+d-1\right)^2+7\leq\frac83\cdot\left(ab+bc+ca+ad+bd+cd\right).$$Prove
$$\frac1{a+1}+\frac1{b+1}+\frac1{c+1}+\frac1{d+1}\leq2.$$
0 replies
mihaig
an hour ago
0 replies
A geometry problem
Lttgeometry   0
an hour ago
Triangle $ABC$ has two isogonal conjugate points $P$ and $Q$. The circle $(BPC)$ intersects circle $(AP)$ at $R \neq P$, and the circle $(BQC)$ intersects circle $(AQ)$ at $S\neq Q$. Prove that $R$ and $S$ are isogonal conjugates in triangle $ABC$.
Note: Circle $(AP)$ is the circle with diameter $AP$, Circle $(AQ)$ is the circle with diameter $AQ$.
0 replies
Lttgeometry
an hour ago
0 replies
A sharp one with 3 var
mihaig   9
N an hour ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
9 replies
mihaig
May 13, 2025
mihaig
an hour ago
Van der Corput Inequality
EthanWYX2009   0
an hour ago
Source: en.wikipedia.org/wiki/Van_der_Corput_inequality
Let $V$ be a real or complex inner product space. Suppose that ${\displaystyle v,u_{1},\dots ,u_{n}\in V} $ and that ${\displaystyle \|v\|=1}.$ Then$${\displaystyle \displaystyle \left(\sum _{i=1}^{n}|\langle v,u_{i}\rangle |\right)^{2}\leq \sum _{i,j=1}^{n}|\langle u_{i},u_{j}\rangle |.}$$
0 replies
EthanWYX2009
an hour ago
0 replies
Hard Functional Equation in the Complex Numbers
yaybanana   6
N 2 hours ago by jasperE3
Source: Own
Find all functions $f:\mathbb {C}\rightarrow \mathbb {C}$, s.t :

$f(xf(y)) + f(x^2+y) = f(x+y)x + f(f(y))$

for all $x,y \in \mathbb{C}$
6 replies
yaybanana
Apr 9, 2025
jasperE3
2 hours ago
FE with conditions on $x,y$
Adywastaken   3
N 2 hours ago by jasperE3
Source: OAO
Find all functions $f:\mathbb{R_{+}}\rightarrow \mathbb{R_{+}}$ such that $\forall y>x>0$,
\[
f(x^2+f(y))=f(xf(x))+y
\]
3 replies
Adywastaken
Friday at 6:18 PM
jasperE3
2 hours ago
Point satisfies triple property
62861   36
N 2 hours ago by cursed_tangent1434
Source: USA Winter Team Selection Test #2 for IMO 2018, Problem 2
Let $ABCD$ be a convex cyclic quadrilateral which is not a kite, but whose diagonals are perpendicular and meet at $H$. Denote by $M$ and $N$ the midpoints of $\overline{BC}$ and $\overline{CD}$. Rays $MH$ and $NH$ meet $\overline{AD}$ and $\overline{AB}$ at $S$ and $T$, respectively. Prove that there exists a point $E$, lying outside quadrilateral $ABCD$, such that
[list]
[*] ray $EH$ bisects both angles $\angle BES$, $\angle TED$, and
[*] $\angle BEN = \angle MED$.
[/list]

Proposed by Evan Chen
36 replies
62861
Jan 22, 2018
cursed_tangent1434
2 hours ago
Inspired by 2025 Xinjiang
sqing   1
N 2 hours ago by sqing
Source: Own
Let $ a,b,c >0  . $ Prove that
$$  \left(1+\frac {a} { b}\right)\left(2+\frac {a}{ b}+\frac {b}{ c}\right) \left(1+\frac {a}{b}+\frac {b}{ c}+\frac {c}{ a}\right)  \geq 12+8\sqrt 2 $$
1 reply
sqing
Yesterday at 5:32 PM
sqing
2 hours ago
Inspired by 2025 Beijing
sqing   4
N 2 hours ago by pooh123
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
4 replies
sqing
Yesterday at 4:56 PM
pooh123
2 hours ago
Find the minimum
sqing   27
N 3 hours ago by sqing
Source: Zhangyanzong
Let $a,b$ be positive real numbers such that $a^2b^2+\frac{4a}{a+b}=4.$ Find the minimum value of $a+2b.$
27 replies
sqing
Sep 4, 2018
sqing
3 hours ago
Sharygin 2025 CR P15
Gengar_in_Galar   7
N 3 hours ago by Giant_PT
Source: Sharygin 2025
A point $C$ lies on the bisector of an acute angle with vertex $S$. Let $P$, $Q$ be the projections of $C$ to the sidelines of the angle. The circle centered at $C$ with radius $PQ$ meets the sidelines at points $A$ and $B$ such that $SA\ne SB$. Prove that the circle with center $A$ touching $SB$ and the circle with center $B$ touching $SA$ are tangent.
Proposed by: A.Zaslavsky
7 replies
Gengar_in_Galar
Mar 10, 2025
Giant_PT
3 hours ago
Inequality olympiad algebra
Foxellar   1
N 3 hours ago by sqing
Given that \( a, b, c \) are nonzero real numbers such that
\[
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b},
\]let \( M \) be the maximum value of the expression
\[
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}.
\]Determine the sum of the numerator and denominator of the simplified fraction representing \( M \).
1 reply
Foxellar
Yesterday at 10:01 PM
sqing
3 hours ago
Inspired by RMO 2006
sqing   2
N 3 hours ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
2 replies
sqing
Yesterday at 3:24 PM
sqing
3 hours ago
A number theory problem from the British Math Olympiad
Rainbow1971   13
N Apr 3, 2025 by ektorasmiliotis
Source: British Math Olympiad, 2006/2007, round 1, problem 6
I am a little surprised to find that I am (so far) unable to solve this little problem:

[quote]Let $n$ be an integer. Show that, if $2 + 2 \sqrt{1+12n^2}$ is an integer, then it is a perfect square.[/quote]

I set $k := \sqrt{1+12n^2}$. If $2 + 2 \sqrt{1+12n^2}$ is an integer, then $k (=\sqrt{1+12n^2})$ is at least rational, so that $1 + 12n^2$ must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs $(n, k)$ for which this happens are $(0,1), (2,7), (28,97)$ and $(390, 1351)$, and that, for every such pair $(n,k)$, the "next" such pair can be calculated as
$$
\begin{bmatrix}
7 & 2 \\
24 & 7 
\end{bmatrix} \begin{bmatrix}
n \\
k 
\end{bmatrix}
.$$The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?

Thank you.




13 replies
Rainbow1971
Mar 28, 2025
ektorasmiliotis
Apr 3, 2025
A number theory problem from the British Math Olympiad
G H J
G H BBookmark kLocked kLocked NReply
Source: British Math Olympiad, 2006/2007, round 1, problem 6
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rainbow1971
35 posts
#1
Y by
I am a little surprised to find that I am (so far) unable to solve this little problem:
Quote:
Let $n$ be an integer. Show that, if $2 + 2 \sqrt{1+12n^2}$ is an integer, then it is a perfect square.

I set $k := \sqrt{1+12n^2}$. If $2 + 2 \sqrt{1+12n^2}$ is an integer, then $k (=\sqrt{1+12n^2})$ is at least rational, so that $1 + 12n^2$ must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs $(n, k)$ for which this happens are $(0,1), (2,7), (28,97)$ and $(390, 1351)$, and that, for every such pair $(n,k)$, the "next" such pair can be calculated as
$$
\begin{bmatrix}
7 & 2 \\
24 & 7 
\end{bmatrix} \begin{bmatrix}
n \\
k 
\end{bmatrix}
.$$The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?

Thank you.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
vsamc
3789 posts
#2 • 1 Y
Y by centslordm
Try to write the general solution $(n, k)$ using pell equation theory, and then consider $2+2k$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rainbow1971
35 posts
#3
Y by
Thank you for your suggestion. I am aware that there is a Pell equation here, but I am not sure if I can make any progress using the corresponding theory. That theory also tells us how to generate further solutions from a given solution, in a way that is quite parallel to the matrix-vector multiplication in my post. I cannot see anything beyond that which would be helpful.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pooh123
72 posts
#4 • 1 Y
Y by Rainbow1971
The solution is surprisingly simple:

Let \( 2 + 2\sqrt{1+12n^2} = 2m \) (because it is even), where \( m \) is a positive integer.

Dividing both sides by 2 and subtracting 1 from both sides, we get

\[
m - 1 = \sqrt{1+12n^2}
\]
Squaring both sides, we get

\[
(m-1)^2 = 1 + 12n^2
\]
so

\[
12n^2 = m(m-2).
\]
Since the left side is even, \( m \) is even. Let \( m = 2k \), where \( k \) is a positive integer.

The equation becomes

\[
12n^2 = 2k(2k-2)
\]
or

\[
3n^2 = k(k-1).
\]
Since \( (k, k-1) = 1 \) and the left side is divisible by 3, exactly one of \( k \) and \( k-1 \) is divisible by 3.

We consider two cases:

Case 1: \( k \) is divisible by 3, so \( k-1 \) is a square, which is impossible because \( k-1 \equiv 2 \pmod{3} \).

Case 2: \( k-1 \) is divisible by 3, so \( k \) is a square. Let \( k = t^2 \), where \( t \) is a positive integer.

Then \[2 + 2\sqrt{1+12n^2} = 2m = 4k = (2t)^2\], which is a square.
This post has been edited 1 time. Last edited by pooh123, Mar 29, 2025, 1:49 PM
Reason: Typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ektorasmiliotis
109 posts
#5
Y by
another problem from a competition called Kurshak(i think)
if 2 + 2sqrt(28n^2+1) is integer,show that is a perfect square
i solved it with pell,i think its the same with this problem
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rainbow1971
35 posts
#6
Y by
Thank you, pooh123, your solution is indeed a nice one! A question to ektorasmiliotis: What exactly do you mean by "I solved it with pell"?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ektorasmiliotis
109 posts
#7
Y by
Rainbow1971 wrote:
Thank you, pooh123, your solution is indeed a nice one! A question to ektorasmiliotis: What exactly do you mean by "I solved it with pell"?

using pell's equation : https://en.wikipedia.org/wiki/Pell%27s_equation
for your exercise :
M=2+2sqrt(12n^2+1)
so sqrt(12n^2+1)=M/2 -1, Set X= M/2 -1
X^2-12n^2=1 with (Xo,No)=(7,2)
So Xn=1/2(((7+2sqrt(12))^n +(7-2sqrt(12))^n)
see that 7+2sqrt(12)=(2+sqrt(3))^2 ........
M=((2+sqrt(3))^n+(2-3sqrt(3))^n)^2,so M is always a perfect square by Binomial theorem
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rainbow1971
35 posts
#8
Y by
Thank you for the clarification. There is one line in it which surprises me:

$$X_n=\tfrac{1}{2} \big( (7+2\sqrt{12})^n + (7-2\sqrt{12})^n)\big).$$
I did not know that it is so easy to calculate (the first component of) the $n$th solution pair, particularly without recursion, and I cannot find any remark in that direction in the Wikipedia article. However, numerical calculation confirms your claim. Can you give me a source that would explain the line above?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
vsamc
3789 posts
#9 • 1 Y
Y by centslordm
I'm surprised you knew about Conway's topograph method but not the general formula. If $d > 1$ is squarefree, and $x^2 -dy^2 = 1$ has primitive (meaning "smallest", based on, say, $x$) solution $(x_1, y_1)$ in positive integers, then the $n$'th largest solution $(x_n, y_n)$ satisfies $x_n + y_n\sqrt{d} = (x_1+y_1\sqrt{d})^n$. The reason this even works is because $x^2-dy^2 = (x+y\sqrt{d})(x-y\sqrt{d})$, so $(x_n+y_n\sqrt{d})(x_n-y_n\sqrt{d}) = (x_1+y_1\sqrt{d})^n(x_1-y_1\sqrt{d})^n = (x_1^2-dy_1^2)^n = 1$. You can find the proof that this is all solutions on wikepedia (sketch: multiply a solution $x+y\sqrt{d}$ by $x_1-y_1\sqrt{d}$ and induct down)
This post has been edited 1 time. Last edited by vsamc, Mar 29, 2025, 6:55 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rainbow1971
35 posts
#10
Y by
Lots of surprises in this thread... Thank you, vsamc. You mention that
$$x_n + y_n\sqrt{d} = (x_1+y_1\sqrt{d})^n,$$which is fine, but calculating $x_n$ still seems to require expanding the right-hand side of the equation and collecting the terms that are free of square roots. ektorasmiliotis, however, mentioned a way to calculate $x_n$ directly. Maybe ektorasmiliotis' method follows from what you, vsamc, wrote, but so far I don't see that...
This post has been edited 1 time. Last edited by Rainbow1971, Mar 29, 2025, 7:11 PM
Reason: minor error
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
vsamc
3789 posts
#11 • 2 Y
Y by Rainbow1971, centslordm
$x_n + y_n\sqrt{d} = (x_1+y_1\sqrt{d})^n$, it follows by taking the "radical conjugate" of both sides that
$x_n - y_n\sqrt{d} = (x_1-y_1\sqrt{d})^n$
so $x_n = \frac{(x_1+y_1\sqrt{d})^n + (x_1-y_1\sqrt{d})^n)}{2}$ and $y_n = \frac{(x_1+y_1\sqrt{d})^n - (x_1-y_1\sqrt{d})^n}{2\sqrt{d}}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Rainbow1971
35 posts
#12
Y by
Ah, now I see the connection. Thanks a lot.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ektorasmiliotis
109 posts
#13
Y by
vsamc wrote:
I'm surprised you knew about Conway's topograph method but not the general formula. If $d > 1$ is squarefree, and $x^2 -dy^2 = 1$ has primitive (meaning "smallest", based on, say, $x$) solution $(x_1, y_1)$ in positive integers, then the $n$'th largest solution $(x_n, y_n)$ satisfies $x_n + y_n\sqrt{d} = (x_1+y_1\sqrt{d})^n$. The reason this even works is because $x^2-dy^2 = (x+y\sqrt{d})(x-y\sqrt{d})$, so $(x_n+y_n\sqrt{d})(x_n-y_n\sqrt{d}) = (x_1+y_1\sqrt{d})^n(x_1-y_1\sqrt{d})^n = (x_1^2-dy_1^2)^n = 1$. You can find the proof that this is all solutions on wikepedia (sketch: multiply a solution $x+y\sqrt{d}$ by $x_1-y_1\sqrt{d}$ and induct down)

and the primitive solution is not (1,0),(-1,0)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ektorasmiliotis
109 posts
#14
Y by
another solution ,it's much easier i think..

sqrt(12n^2+1)=2k+1
12n^2+1=4k^2+4k+1
3n^2=k(k+1) but (k,k+1)=1
1)if k=x^2 and k+1=3y^2,then x^2-3y^2=-1 and we use mod 4,no solutions
2)if k=b^2 and k+1=a^2
M=2+2sqrt(12n^2+1)=2 +2(2k+1)=2 +4k +2=4(k+1)=4a^2=(2a)^2

**I just saw post 4, we have similar solutions.
This post has been edited 1 time. Last edited by ektorasmiliotis, Apr 3, 2025, 8:30 AM
Z K Y
N Quick Reply
G
H
=
a