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

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

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)!

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

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


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


More specifically:

For new threads:


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

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


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

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


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


For answers to already existing threads:


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

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



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


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

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Functional equation
Math-wiz   25
N 6 minutes ago by Adywastaken
Source: IMOC SL A1
Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x,y\in\mathbb{R}$,
$$f(xy+f(x))=f(xf(y))+x$$
25 replies
Math-wiz
Dec 15, 2019
Adywastaken
6 minutes ago
Nice numer theory
GeoArt   5
N 21 minutes ago by Primeniyazidayi
$p$ is a prime number, $m, x, y$ are natural numbers ($m, x, y > 1$). It is known that $\frac{x^p + y^p}{2}$ $=$ $(\frac{x+y}{2} )^m$. Prove that $p = m$.
5 replies
+1 w
GeoArt
Jan 7, 2021
Primeniyazidayi
21 minutes ago
Prove XBY equal to angle C
nataliaonline75   2
N an hour ago by starchan
Let $M$ be the midpoint of $BC$ on triangle $ABC$. Point $X$ lies on segment $AC$ such that $AX=BX$ and $Y$ on line $AM$ such that $XY//AB$. Prove that $\angle XBY = \angle ACB$.
2 replies
nataliaonline75
Yesterday at 2:47 PM
starchan
an hour ago
Unique number to make a square of a rational
Zavyk09   2
N an hour ago by Zavyk09
Source: Homework
Find all positive integers $n$ there exists a unique positive integers $m$ such that $\frac{n+m}{m}$ is a square of a rational number.
2 replies
Zavyk09
2 hours ago
Zavyk09
an hour ago
Classical NT using modular arithmetic
electrovector   7
N an hour ago by Blackbeam999
Source: 2022 Turkey TST P1 Day 1 + 2023 Dutch BxMO TST, Problem 5
Find all pairs of prime numbers $(p,q)$ for which
\[2^p = 2^{q-2} + q!.\]
7 replies
electrovector
Mar 13, 2022
Blackbeam999
an hour ago
Inequality
lgx57   9
N an hour ago by lgx57
Source: Own
$a,b,c>0,ab+bc+ca=1$. Prove that

$$\sum \sqrt{8ab+1} \ge 5$$
(I don't know whether the equality holds)
9 replies
+1 w
lgx57
Saturday at 3:14 PM
lgx57
an hour ago
old one but good one
Sunjee   2
N an hour ago by ehuseyinyigit
If $x_1,x_2,...,x_n $ are positive numbers, then prove that
$$\frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+ \frac{x_n}{1+x_1^2+\cdots+x_n^2}\geq \sqrt{n}$$
2 replies
Sunjee
3 hours ago
ehuseyinyigit
an hour ago
Inspired by giangtruong13
sqing   0
an hour ago
Source: Own
Let $ a,b>0  .$ Prove that$$ \frac{a}{b}+\frac{b}{a}+\frac{a^3}{2b^3+kab^2}+\frac{2b^3}{a^3+b^3+kab^2} \geq \frac{2k+7}{k+2}$$Where $ k\geq 0. $
0 replies
sqing
an hour ago
0 replies
we can find one pair of a boy and a girl
orl   18
N 2 hours ago by bin_sherlo
Source: Vietnam TST 2001 for the 42th IMO, problem 3
Some club has 42 members. It’s known that among 31 arbitrary club members, we can find one pair of a boy and a girl that they know each other. Show that from club members we can choose 12 pairs of knowing each other boys and girls.
18 replies
orl
Jun 26, 2005
bin_sherlo
2 hours ago
geometry
blug   0
2 hours ago
In trapezius $ABCD$, segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle B$ and $\angle D$ intersect at $P$. Circumcircles of $ABP$ and $CDP$ meet again at $Q$. Angle bisector of $\angle D$ cuts $AB$ at $S$. Prove that
$$1. QD=QS,$$$$2. \angle DCQ=\angle BCQ,$$$$3. \angle BAQ=\angle QAD.$$
0 replies
blug
2 hours ago
0 replies
Eulerline problem
Retemoeg   0
2 hours ago
Source: Extension from a problem I read in a book
Show that the isogonal conjugate of the isotomic conjugate of the orthocenter lies on the Euler line.
0 replies
Retemoeg
2 hours ago
0 replies
Inequality
MathsII-enjoy   6
N 2 hours ago by MathsII-enjoy
A interesting problem generalized :-D
6 replies
MathsII-enjoy
Saturday at 1:59 PM
MathsII-enjoy
2 hours ago
Number Theory
fasttrust_12-mn   8
N 2 hours ago by ErTeeEs06
Source: Pan African Mathematics Olympiad P1
Find all positive intgers $a,b$ and $c$ such that $\frac{a+b}{a+c}=\frac{b+c}{b+a}$ and $ab+bc+ca$ is a prime number
8 replies
fasttrust_12-mn
Aug 15, 2024
ErTeeEs06
2 hours ago
My Unsolved FE on R+
ZeltaQN2008   4
N 3 hours ago by mashumaro
Source: IDK
Give $a>0$. Find all funcitions $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that for all any $x,y\in (0,\infty):$
$$f(xf(y)+a)=yf(x+y+a)$$
4 replies
ZeltaQN2008
4 hours ago
mashumaro
3 hours ago
Interesting Function
Kei0923   4
N Apr 30, 2025 by CrazyInMath
Source: 2024 JMO preliminary p8
Function $f:\mathbb{Z}_{\geq 0}\rightarrow\mathbb{Z}$ satisfies
$$f(m+n)^2=f(m|f(n)|)+f(n^2)$$for any non-negative integers $m$ and $n$. Determine the number of possible sets of integers $\{f(0), f(1), \dots, f(2024)\}$.
4 replies
Kei0923
Jan 9, 2024
CrazyInMath
Apr 30, 2025
Interesting Function
G H J
G H BBookmark kLocked kLocked NReply
Source: 2024 JMO preliminary p8
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Kei0923
95 posts
#1
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Function $f:\mathbb{Z}_{\geq 0}\rightarrow\mathbb{Z}$ satisfies
$$f(m+n)^2=f(m|f(n)|)+f(n^2)$$for any non-negative integers $m$ and $n$. Determine the number of possible sets of integers $\{f(0), f(1), \dots, f(2024)\}$.
This post has been edited 1 time. Last edited by Kei0923, Jan 9, 2024, 8:41 AM
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nguyenloc1712
62 posts
#3
Y by
answer :maybe:
sol
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sami1618
902 posts
#5
Y by
The answer is $\boxed{992}$. Let $P(m,n)$ be the given assertion. Notice $P(0,0)$ gives $f(0)^2=2f(0)$ so either $f(0)=0$ or $f(0)=2$.

Case 1: $f(0)=0$
The assertion $P(m,0)$ gives $f(m)^2=0\Rightarrow f(m)=0$. Which can be checked to work.

Case 2: $f(0)=2$
The assertion $P(0,n)$ gives $f(n)^2=2+f(n^2)$. The assertion $P(m,0)$ gives $f(m)^2=f(2m)+2$. Combined we get $f(n^2)=f(2n)$. The assertion $P(0,1)$ gives $f(1)^2=2+f(1)$ so either $f(1)=2$ or $f(1)=-1$. If $f(1)=-1$ then $f(2)=-1$ and $P(1,1)$ gives a contradiction. Now assume $f(1)=2\Rightarrow f(2)=2$. We get from $P(m,1)$ that $f(m+1)^2=f(2m)+2\Rightarrow f(2m+2)=f(2m)\Rightarrow$ $f(2m)=2\Rightarrow f(m^2)=2$. Thus we get $f(n)^2=4\Rightarrow f(n)=\pm 2$. Thus $f(x)=2$ whenever $x$ is even or a square and $f(x)$ can be either $2$ or $-2$ when $x$ is not even or a square. It is easy to check that all such functions work.
This post has been edited 1 time. Last edited by sami1618, May 2, 2024, 11:47 AM
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JerryXi
2 posts
#8
Y by
I think the correct answer should be 991, because the number of possible sets in Case 2 should be 990. It seems like you may have made a small counting mistake. The number of integers from 3 to 2024 that are neither even nor perfect squares is 990.
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CrazyInMath
457 posts
#9
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$P(0, 0)\Longrightarrow f(0)=0, 2$
if $f(0)=0$, then $P(m, 0)\Longrightarrow f(m)^2=0\Longrightarrow f(m)=0$

Let $f(0)=2$ from now on
$P(0, 1)\Longrightarrow f(1)^2=2+f(1)\Longrightarrow f(1)=-1, 2$
if $f(1)=-1$ then $P(1, 1)\Longrightarrow f(2)^2=-2$, contradiction, so $f(1)=2$
$P(m, 1)\Longrightarrow f(m+1)^2=f(2m)+2$, $P(m, 0)\Longrightarrow f(m)^2=f(2m)+2$, so $f(m+1)=\pm f(m)$, and so $f(m)=\pm2$
and so the original FE is now $4=f(2m)+f(n^2)$, which means even numbers and perfect squares must have $f(a)=2$, the others can have $f(a)=2, -2$

The number of sets would be just $2$ as sets don't count duplicates. If it is multiset then the answer is $991$.
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