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
IOQM P22 2024
SomeonecoolLovesMaths   3
N 23 minutes ago by SomeonecoolLovesMaths
In a triangle $ABC$, $\angle BAC = 90^{\circ}$. Let $D$ be the point on $BC$ such that $AB + BD = AC + CD$. Suppose $BD : DC = 2:1$. if $\frac{AC}{AB} = \frac{m + \sqrt{p}}{n}$, Where $m,n$ are relatively prime positive integers and $p$ is a prime number, determine the value of $m+n+p$.
3 replies
SomeonecoolLovesMaths
Sep 8, 2024
SomeonecoolLovesMaths
23 minutes ago
AP calc?
Thayaden   30
N an hour ago by Pengu14
How are we all feeling on AP calc guys?
30 replies
Thayaden
May 20, 2025
Pengu14
an hour ago
Cute NT Problem
M11100111001Y1R   4
N 2 hours ago by RANDOM__USER
Source: Iran TST 2025 Test 4 Problem 1
A number \( n \) is called lucky if it has at least two distinct prime divisors and can be written in the form:
\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]where \( p_1, \dots, p_k \) are distinct prime numbers that divide \( n \). (Note: it is possible that \( n \) has other prime divisors not among \( p_1, \dots, p_k \).) Prove that for every prime number \( p \), there exists a lucky number \( n \) such that \( p \mid n \).
4 replies
M11100111001Y1R
Today at 7:20 AM
RANDOM__USER
2 hours ago
USAMO 2003 Problem 4
MithsApprentice   72
N 2 hours ago by endless_abyss
Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$.
72 replies
MithsApprentice
Sep 27, 2005
endless_abyss
2 hours ago
Easy but unusual junior ineq
Maths_VC   1
N 2 hours ago by blug
Source: Serbia JBMO TST 2025, Problem 2
Real numbers $x, y$ $\ge$ $0$ satisfy $1$ $\le$ $x^2 + y^2$ $\le$ $5$. Determine the minimal and the maximal value of the expression $2x + y$
1 reply
Maths_VC
3 hours ago
blug
2 hours ago
Bosnia and Herzegovina JBMO TST 2009 Problem 1
gobathegreat   1
N 2 hours ago by FishkoBiH
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2009
Lengths of sides of triangle $ABC$ are positive integers, and smallest side is equal to $2$. Determine the area of triangle $P$ if $v_c = v_a + v_b$, where $v_a$, $v_b$ and $v_c$ are lengths of altitudes in triangle $ABC$ from vertices $A$, $B$ and $C$, respectively.
1 reply
gobathegreat
Sep 17, 2018
FishkoBiH
2 hours ago
USAMO 2001 Problem 2
MithsApprentice   53
N 3 hours ago by lksb
Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.
53 replies
MithsApprentice
Sep 30, 2005
lksb
3 hours ago
A=b
k2c901_1   89
N 3 hours ago by reni_wee
Source: Taiwan 1st TST 2006, 1st day, problem 3
Let $a$, $b$ be positive integers such that $b^n+n$ is a multiple of $a^n+n$ for all positive integers $n$. Prove that $a=b$.

Proposed by Mohsen Jamali, Iran
89 replies
k2c901_1
Mar 29, 2006
reni_wee
3 hours ago
Strange angle condition and concyclic points
lminsl   129
N 3 hours ago by Aiden-1089
Source: IMO 2019 Problem 2
In triangle $ABC$, point $A_1$ lies on side $BC$ and point $B_1$ lies on side $AC$. Let $P$ and $Q$ be points on segments $AA_1$ and $BB_1$, respectively, such that $PQ$ is parallel to $AB$. Let $P_1$ be a point on line $PB_1$, such that $B_1$ lies strictly between $P$ and $P_1$, and $\angle PP_1C=\angle BAC$. Similarly, let $Q_1$ be the point on line $QA_1$, such that $A_1$ lies strictly between $Q$ and $Q_1$, and $\angle CQ_1Q=\angle CBA$.

Prove that points $P,Q,P_1$, and $Q_1$ are concyclic.

Proposed by Anton Trygub, Ukraine
129 replies
lminsl
Jul 16, 2019
Aiden-1089
3 hours ago
Simple inequality
sqing   12
N 3 hours ago by Rayvhs
Source: MEMO 2018 T1
Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that$$\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$$
12 replies
sqing
Sep 2, 2018
Rayvhs
3 hours ago
Random concyclicity in a square config
Maths_VC   2
N 3 hours ago by Maths_VC
Source: Serbia JBMO TST 2025, Problem 1
Let $M$ be a random point on the smaller arc $AB$ of the circumcircle of square $ABCD$, and let $N$ be the intersection point of segments $AC$ and $DM$. The feet of the tangents from point $D$ to the circumcircle of the triangle $OMN$ are $P$ and $Q$ , where $O$ is the center of the square. Prove that points $A$, $C$, $P$ and $Q$ lie on a single circle.
2 replies
Maths_VC
4 hours ago
Maths_VC
3 hours ago
Serbian selection contest for the IMO 2025 - P3
OgnjenTesic   3
N 3 hours ago by atdaotlohbh
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
3 replies
OgnjenTesic
May 22, 2025
atdaotlohbh
3 hours ago
Inequalities
sqing   26
N 3 hours ago by ytChen
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
26 replies
1 viewing
sqing
May 13, 2025
ytChen
3 hours ago
Pls help I'm stuck
UtkarshSri   1
N 5 hours ago by vanstraelen
So I was solving this problem
Cbrt(x+1) - cbrt(x-1) = pow((x²-1),1/6)


Here, the square root is defined in the domain x≤-1 OR x≥1
I substituted a= cbrt(x+1) and b= cbrt(x-1)
And then the equation turned out to be
a-b= sqrt(ab)
Here, since the RHS is +ve, the LHS must be too, so a>b (Done so as to prevent extraneous roots)
Now, I can write the domain in terms of a and b as ab≥0
Squaring, we get a²+b²-3ab=0
On solving we get a/b = (3±sqrt(5))/2
Now, we know that a>b from the above condition, so the only root is a/b= (3+sqrt(5))/2 and on substituting for x, and applying componendo and dividendo after cubing, I get x= sqrt(5)/2

Now, my question is, since the equation is even i.e f(x)=f(-x), it's symmetric about the y-axis, so x= -sqrt(5)/2 must also satisfy the equation, and indeed it does, so where did I made some irreversible step that caused this loss of the root?
when I plugged the equation to wolframalpha, it showed that the "principle root" has the former solⁿ only where's for a "real valued root", it has both.I also found that on solving the equation
b-a= sqrt(ab), I get the latter root.But I mean where did that extra -ve sign come from?
I'm familiar that f(x)=sqrt(g(x)) is equivalent to f²(x)=g(x) AND f(x)≥0 AND g(x)≥0 but the g(x)≥0 condition is not necessary as f²(x) is already ≥0
and also that abs(f(x))=abs(g(x)) is equivalent to f²(x)=g²(x)

PS: Pardon me I don't know latex well, I'm tried to make this as comprehensive as possible and made sure there's no loss of equivalence in any step.
I hope someone clarifies this for me, it's really bothering me for a while.
1 reply
UtkarshSri
Today at 9:57 AM
vanstraelen
5 hours ago
Hard number theory
td12345   7
N Apr 10, 2025 by td12345
Let $p$ be a prime number. Find the number of elements of the set
\[
M_p = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2p^{n} x} \in \mathbb{Q} \right\}. 
\]
Edit: I edited with a variation of the same problem because I intend to give this one to an exam for my pupils.
7 replies
td12345
Apr 9, 2025
td12345
Apr 10, 2025
Hard number theory
G H J
G H BBookmark kLocked kLocked NReply
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
td12345
233 posts
#1
Y by
Let $p$ be a prime number. Find the number of elements of the set
\[
M_p = \left\{ x \in \mathbb{Z}^* \,\middle|\, \sqrt{x^2 + 2p^{n} x} \in \mathbb{Q} \right\}. 
\]
Edit: I edited with a variation of the same problem because I intend to give this one to an exam for my pupils.
This post has been edited 1 time. Last edited by td12345, Apr 14, 2025, 6:17 PM
Reason: variation of the problem
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathprodigy2011
344 posts
#2
Y by
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
td12345
233 posts
#3
Y by
Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.
This post has been edited 1 time. Last edited by td12345, Apr 10, 2025, 1:57 AM
Reason: typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathprodigy2011
344 posts
#4
Y by
td12345 wrote:
Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

aren't these solutions also 2 times perfect squares? I expanded to get 2(2^2023-1)^2. And when I let x=2m^2; and we if i made the same difference of squares; i would get (a-m)(a+m) = 2^2025. Letting a-m=2 and a+m=2^2024 wll give us m=2^2023-1 which results in the solution you have mentioned
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
td12345
233 posts
#5
Y by
mathprodigy2011 wrote:
td12345 wrote:
Regarding the counting of $M_2$, didn't we miss 2 there? I understand where $1013+1013$ comes from but I think when $gcd(x,x+2q^{2025})=2$ we get 2 extra: $x = 2^{2 \cdot 2025-3}+2-2^{2025}, x = -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. Can you please double check if you counted these in your solution ?

@math
mathprodigy2011 wrote:
Click to reveal hidden text I assumed Z with an asterick to be all positive integers cuz ive never seen that before.

aren't these solutions also 2 times perfect squares? I expanded to get 2(2^2023-1)^2. And when I let x=2m^2; and we if i made the same difference of squares; i would get (a-m)(a+m) = 2^2025. Letting a-m=2 and a+m=2^2024 wll give us m=2^2023-1 which results in the solution you have mentioned

Yes, they lead to the same one when computing the square root, but the set counts the \(x\) not the set of the square root expression no? I got a different number from yours and I don't know where I went wrong, I will lay down the \(x\) I found as it goes over your claimed result and maybe one of us can figure out:

For \( M_2\):
- \(x\) can be \( (2^{2025-k}-2^{k-1})^2 \) where \(k \in \{1,2,\dots, 2025\}\) . This leads to \( [\frac{2025+1}{2}]\) values of \(x\).
- \(x\) can be \( -(2^{2 \cdot 2025-2k}+2^{2k-2}+2^{2025})\) where \(k \in \{1,2,\dots, 2025\}\) . This leads to \( [\frac{2025+1}{2}]\) values of \(x\).
- \(x\) can be $2^{2 \cdot 2025-3}+2-2^{2025},  -2^{2 \cdot 2025 - 3} -2- 2^{2025}$. This leads to \(2\) extra values for \(x\) distinct from the above.

For \(M_{2027}\):
- \(x\) can be \( \frac{2027^k+2027^{2 \cdot 2025 - k}}{2} -2027^{2025} \) where \(k \in \{0, 1,2,\dots, 2 \cdot 2025\}\) . This leads to \( 2025\) distinct non-zero values of \(x\).
- \(x\) can be \( \frac{-2027^k-2027^{2 \cdot 2025 - k}}{2} -2027^{2025} \) where \(k \in \{0, 1,2,\dots, 2 \cdot 2025\}\) . This leads to \( 2026\) distinct non-zero values of \(x\).
I double checked with W|A to see if they lead to perfect squares under the square root and they do, so unless I double counted, the answer should be $2026+2+2025 + 2026$?


Edit: ok there is a miscount for \( M_2\) because when \(k=1013\) it leads to \(0\) for the very first family and \(x\) can't be \(0\). So for \(M_2\) we shall have $1012+1013+2=2027$. But for \(M_{2027}\) count seems solid? I got $2024+2+2025+2026$
This post has been edited 8 times. Last edited by td12345, Apr 10, 2025, 8:34 PM
Reason: tpyo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
td12345
233 posts
#6
Y by
Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?
This post has been edited 2 times. Last edited by td12345, Apr 10, 2025, 8:40 PM
Reason: question
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathprodigy2011
344 posts
#7
Y by
td12345 wrote:
Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?

ohhhh so its not just positive integers? i just never have seen z with an asterick so thats probably where i went wrong
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
td12345
233 posts
#8
Y by
mathprodigy2011 wrote:
td12345 wrote:
Ok the 2 solutions I mentioned above as being separate (for \(M_2\)) actually come from another family:

- \(x\) can be $2^{2 \cdot 2025-(2k+1)}+2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.
- \(x\) can be $-2^{2 * 2025-(2k+1)}-2^{2k-1}-2^{2025}$, where $k$ goes from 1 to 2024. This leads to \(1012\) extra values for \(x\) distinct from the above.

So in total for \(M_2\) we should have \(1012 +1013+1012+1012=4049\). Which would make the total \(8100\). Can somebody else confirm this answer?

ohhhh so its not just positive integers? i just never have seen z with an asterick so thats probably where i went wrong

Yes it's integers that are not 0. $\mathbb{N}$ is natural numbers (positive integers) and \( \mathbb{N}^*\) is natural numbers without \(0\). Would you like to give it another try just to see if it's really $8100$?
This post has been edited 1 time. Last edited by td12345, Apr 10, 2025, 9:30 PM
Reason: question
Z K Y
N Quick Reply
G
H
=
a