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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 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
Putnam 2012 A3
Kent Merryfield   9
N 11 minutes ago by AngryKnot
Let $f:[-1,1]\to\mathbb{R}$ be a continuous function such that

(i) $f(x)=\frac{2-x^2}{2}f\left(\frac{x^2}{2-x^2}\right)$ for every $x$ in $[-1,1],$

(ii) $ f(0)=1,$ and

(iii) $\lim_{x\to 1^-}\frac{f(x)}{\sqrt{1-x}}$ exists and is finite.

Prove that $f$ is unique, and express $f(x)$ in closed form.
9 replies
Kent Merryfield
Dec 3, 2012
AngryKnot
11 minutes ago
JBMO 2013 Problem 1
Igor   29
N an hour ago by LeYohan
Source: Proposed by Serbia
Find all ordered pairs $(a,b)$ of positive integers for which the numbers $\dfrac{a^3b-1}{a+1}$ and $\dfrac{b^3a+1}{b-1}$ are both positive integers.
29 replies
Igor
Jun 23, 2013
LeYohan
an hour ago
Number theory
truongngochieu   0
2 hours ago
Prove that \[\tau\big(\varphi(n)\big) \geq \varphi\big(\tau(n)\big)\]with all integer n.
0 replies
truongngochieu
2 hours ago
0 replies
Interesting Succession
AlexCenteno2007   3
N 2 hours ago by evgeniy___
The sequence $\{a_n\}$ of integers is defined by
\[
-\frac{1}{2} < a_{n+1} - \frac{a_n^2}{a_{n-1}} \leq \frac{1}{2}
\]with $a_1 = 2$, $a_2 = 7$, prove that $a_n$ is odd for all values of $n \geq 2$.
3 replies
AlexCenteno2007
Yesterday at 6:21 PM
evgeniy___
2 hours ago
Evenish colorings
a_507_bc   7
N 2 hours ago by TigerOnion
Source: Australia MO 2024 P4
Consider a $2024 \times 2024$ grid of unit squares. Two distinct unit squares are adjacent if they share a common side. Each unit square is to be coloured either black or white. Such a colouring is called $\textit{evenish}$ if every unit square in the grid is adjacent to an even number of black unit squares. Determine the number of $\textit{evenish}$ colourings.
7 replies
a_507_bc
Feb 24, 2024
TigerOnion
2 hours ago
number theory
Hoapham235   0
2 hours ago
Let $x >  y$ be positive integer such that \[ \text{LCM}(x+2, y+2)+\text{LCM}(x, y)=2\text{LCM}(x+1, y+1).\]Prove that $x$ is divisible by $y$.
0 replies
Hoapham235
2 hours ago
0 replies
AMM problem section
Khalifakhalifa   1
N 2 hours ago by Khalifakhalifa
Does anyone have access to the current AMM edition? I’d like to see the problems section. If so, could someone please share it with me via PM?
1 reply
Khalifakhalifa
Yesterday at 11:17 AM
Khalifakhalifa
2 hours ago
GMA Problem
mihaig   4
N 2 hours ago by mihaig
Source: Own
Let $a_1,a_2, a_3,a_4\geq0$ be reals.
Prove
$$a_1+a_2+3a_3+3a_4+\sqrt{\left(a_{1}-a_{3}\right)^2+\left(a_{2}-a_{4}\right)^2}\geq\sum_{k=1}^{4}{\sqrt{a_k^2+a_{k+1}^2}},$$where $a_5=a_1.$ When do we have equality?
4 replies
mihaig
Jul 13, 2025
mihaig
2 hours ago
Interesting inequality
sqing   2
N 2 hours ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$ \frac {2a+b} {2b+a}+ \frac {3b+a} {3a+b} \geq  \frac {4\sqrt{6}} {5}    $$$$ \frac {2a+b} {2b+a}+ \frac {5b+a} {5a+b} \geq  \frac {4\sqrt{2}} {3}    $$$$  \frac {3a+b} {3b+a}+ \frac {11 b+a} {11 a+b}   \geq  \frac {\sqrt{15}} {2}    $$$$ \frac {3a+b} {3b+a}+ \frac {9b+a} {9a+b} \geq  \frac {8\sqrt{10}} {13}    $$
2 replies
sqing
4 hours ago
sqing
2 hours ago
JBMO 2013 Shortlist
IstekOlympiadTeam   8
N 2 hours ago by LeYohan
Source: Secret
$\boxed{N1}$ find all positive integers $n$ for which $1^3+2^3+\cdots+{16}^3+{17}^n$ is a perfect square.
8 replies
IstekOlympiadTeam
Apr 16, 2015
LeYohan
2 hours ago
Interesting inequality
sqing   3
N 2 hours ago by sqing
Source: Own
Let $ a,b >0  . $ Prove that
$$ \frac {2a+b} {2b+a}+ \frac {7b+a} {7a+b} \geq  \frac {24} {13}    $$$$ \frac {3a+b} {3b+a}+ \frac {17b+a} {17a+b}  \geq  \frac {48} {25}    $$$$  \frac {4a+b} {4b+a}+ \frac {31b+a} {31a+b}\geq  \frac {80} {41}    $$$$ \frac {5a+b} {5b+a}+ \frac {49b+a} {49a+b} \geq  \frac {120} {61}    $$
3 replies
sqing
4 hours ago
sqing
2 hours ago
At least One pair with square of distance multiple of 2016
Johann Peter Dirichlet   4
N 2 hours ago by MathLuis
Source: 38th Brazilian MO (2016) - First Day, Problem 2
Find the smallest number \(n\) such that any set of \(n\) ponts in a Cartesian plan, all of them with integer coordinates, contains two poitns such that the square of its mutual distance is a multiple of \(2016\).
4 replies
Johann Peter Dirichlet
Nov 23, 2016
MathLuis
2 hours ago
an integral
Svyatoslav   0
4 hours ago
How do we prove analytically that
$$\int_0^{\pi/2}\frac{\ln(1+\cos x)-x}{\sqrt{\sin x}}\,dx=0\quad?$$The sourse: Quora

Numeric evaluation
0 replies
Svyatoslav
4 hours ago
0 replies
OMOUS-2025 (Team Competition) P10
enter16180   3
N 5 hours ago by khina
Source: Open Mathematical Olympiad for University Students (OMOUS-2025)
Let $f: \mathbb{N} \rightarrow \mathbb{N}$ and $g: \mathbb{N} \rightarrow\{A, G\}$ functions are given with following properties:
(a) $f$ is strict increasing and for each $n \in \mathbb{N}$ there holds $f(n)=\frac{f(n-1)+f(n+1)}{2}$ or $f(n)=\sqrt{f(n-1) \cdot f(n+1)}$.
(b) $g(n)=A$ if $f(n)=\frac{f(n-1)+f(n+1)}{2}$ holds and $g(n)=G$ if $f(n)=\sqrt{f(n-1) \cdot f(n+1)}$ holds.

Prove that there exist $n_{0} \in \mathbb{N}$ and $d \in \mathbb{N}$ such that for all $n \geq n_{0}$ we have $g(n+d)=g(n)$
3 replies
enter16180
Apr 18, 2025
khina
5 hours ago
Putnam 2003 B1
btilm305   13
N Apr 13, 2025 by clarkculus
Do there exist polynomials $a(x)$, $b(x)$, $c(y)$, $d(y)$ such that \[1 + xy + x^2y^2= a(x)c(y) + b(x)d(y)\] holds identically?
13 replies
btilm305
Jun 23, 2011
clarkculus
Apr 13, 2025
Putnam 2003 B1
G H J
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btilm305
439 posts
#1 • 4 Y
Y by Mathuzb, itslumi, Adventure10, Mango247
Do there exist polynomials $a(x)$, $b(x)$, $c(y)$, $d(y)$ such that \[1 + xy + x^2y^2= a(x)c(y) + b(x)d(y)\] holds identically?
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Kent Merryfield
18574 posts
#2 • 7 Y
Y by Leooooo, Kezer, Binomial-theorem, Adventure10, Mango247, Doru2718, and 1 other user
The answer is no, that is not possible.

Let $m$ be the maximum degree of any of the polynomials $a, b, c,$ or $d.$ (In practice, it is a little silly to imagine that $m$ could be anything other than $2,$ but it causes no harm in the proof). Write $a(x) = \sum_{k=0}^m   a_kx^k,$ with similar notation for $b, c,$ and $d.$ Then
$a(x)c(y) + b(x)d(y) =  \sum_{j=0}^m  \sum_{j=0}^m  (a_jc_k + b_jd_k)x^jy^k.$ This has the structure of a matrix multiplication. In particular, take the $(m + 1) \times 2$ matrix whose columns are the coefficients $a_j$ and $b_j$ and multiply it by the $2 \times (m + 1)$ matrix whose rows are the coefficients $c_k$ and $d_k.$ The result is an $(m + 1) \times (m + 1)$ matrix whose $(j, k)$ entry is $a_jc_k + b_jd_k$ - that is, the coefficient of $x^jy^k$ in the product polynomial. Since the set of these monomials $x^jy^k$ is linearly independent in the set of polynomials, the only way for this to equal a particular polynomial is for each and every of the $(m + 1)^2$ coefficients to be identical. But the polynomial $1 + xy + x^2y^2$ corresponds to the matrix with $1$s in each of the first three elements of the main diagonal and zeros everywhere else. That is, it has the $3 \times 3$ identity matrix in the upper left corner and is otherwise zero. But this is a matrix of rank $3.$ The product of an $(m + 1) \times 2$ matrix by a $2 \times (m + 1)$ matrix must have rank no greater than $2.$ So the identity is not possible.
Z K Y
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BoyosircuWem
479 posts
#3 • 2 Y
Y by Adventure10, Mango247
Here is an interesting thought:
Is there a partial differential equation whose solutions are precisely the functions of the form $p(x)q(y) + r(x)s(y)$?
Z K Y
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Kent Merryfield
18574 posts
#4 • 2 Y
Y by Adventure10, Mango247
An odd doubt occurs to me concerning this problem, and it centers on the words "holds identically."

I always envisioned this problem as being about real polynomials, in which case the issue I'm worried about doesn't arise. But what if these are polynomials over a finite field? Note that a key point in my argument above is the linear independence of the monomials $x^jy^k.$ If by "holds identically" you intend that these be the same elements of the ring $\mathbb{F}[x,y],$ then that independence is a given, and the argument still stands. But what if by "holds identically" you intend that these take on the same values as functions for all $(x,y)?$ Then the monomials $x^jy^k$ might not be independent. If $\mathbb{F}=\mathbb{F}_2,$ the field with two elements, then the set of all functions in $(x,y)$ is a 4-dimensional vector space - and if that is so, then the argument collapses.

That's taking it in a rather different direction than BoyosircuWem's speculation.
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mavropnevma
15142 posts
#5 • 5 Y
Y by FlakeLCR, MSTang, MissionModi2019, Adventure10, and 1 other user
The old clear difference between a polynomial (as an object) and a polynomial function (as a recipe). Surely $X$ and $X^2$ as polynomials are distinct, even in $\mathbb{F}_2[X]$, while as polynomial functions defined on $\mathbb{F}_2$ they coincide. For me, any time the word polynomial is used, we deal with "objects", and so expressions like "equality identically holds" are just an over-emphasizing that we have equality, the same as one talks about the identically null polynomial, instead of just saying the zero polynomial.
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SCP
1502 posts
#6 • 2 Y
Y by Adventure10, Mango247
Kent Merryfield wrote:
The product of an $(m + 1) \times 2$ matrix by a $2 \times (m + 1)$ matrix must have rank no greater than $2$.

Why the $ (m+1)\times (m+1) $ matrix has a rank of max $ 2$ / why did that quote holds?
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Kent Merryfield
18574 posts
#7 • 3 Y
Y by SCP, Adventure10, Mango247
Two theorems of linear algebra, both very basic, and which hold for matrices over any field.

If $A$ is $m\times n,$ then $\text{rank}(A)\le\min(m,n).$

$\text{rank}(AB)\le\min(\text{rank}(A),\text{rank}(B)).$
Z K Y
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viperstrike
1198 posts
#8 • 2 Y
Y by Adventure10, Mango247
An elementary solution but does this work?
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va2010
1276 posts
#9 • 9 Y
Y by tapir1729, kapilpavase, Mathuzb, Imayormaynotknowcalculus, itslumi, Pluto1708, CyclicISLscelesTrapezoid, Adventure10, Mango247
This problem has a very fast solution:

Click to reveal hidden text
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Kezer
986 posts
#10 • 3 Y
Y by methylbenzene, Adventure10, Mango247
I'm almost embarrassed to post my bashy solution after seeing actual nice solutions on this thread. It does seem different than the other proofs here, though.

Suppose $a(x) = a_0+a_1x+\dots+a_nx^n$, then we can quickly conclude that $b(x) = b_0+b_1x+b_2x^2 - \frac{c(y)}{d(y)} \left(x^3+x^4+\dots+x^n \right)$ for every $y$ and $d(y) \neq 0$. So $c = d$. Setting $x=0$ shows that they are a constant which quickly yields a contradiction.

So the degrees of $a,b,c,d$ are all $\leq 2$. (same argument for $c,d$) Setting $a(x) = a_0 + a_1x + a_2x^2, b(x) = b_0 + b_1x+b_2x^2$ and so on gives us a lot of equations to work with by simply comparing the coefficients of the sides of the equation. \[ a_0c_1+b_0d_1 = a_0c_2+b_0d_2 = a_1c_0+b_1d_0 = a_1c_2+b_1d_2 = a_2c_0+b_2d_0=a_2c_1+b_2d_1 = 0 \]If some of $c_0, c_1, c_2$ were $0$, then we can quickly reach a contradiction - it'll eventually yield $c= 0$ giving $1+xy+x^2y^2 = b(x)d(y)$. Set $x=0$, that shows that $d$ is constant, contradiction.

So now we will get \begin{align*} a_0 &= -b_0 \frac{d_1}{c_1} = -b_0 \frac{d_2}{c_2} \\ a_1 &= -b_1 \frac{d_0}{c_0} = -b_1 \frac{d_2}{c_2} \\ a_2 &= -b_2 \frac{d_0}{c_0} = -b_2 \frac{d_1}{c_1} \end{align*}In particular $\frac{d_0}{c_0} = \frac{d_1}{c_1} = \frac{d_2}{c_2}$. Summing gives \[ a(x) = -\frac{d_0}{c_0}(b_0+b_1x+b_2x^2) = -\frac{d_0}{c_0}b(x). \]Similarly $c(x) = -\frac{b_0}{a_0}d(x)$. Hence, we have to solve \[ 1 + xy + x^2y^2 = \left(-\frac{d_0}{c_0}-\frac{b_0}{a_0} \right) b(x)c(y) =: \lambda b(x)c(y) \]It becomes easy now. Set $x=0$, that shows that $c$ is constant. Contradiction.
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yayups
1614 posts
#11 • 1 Y
Y by Adventure10
Storage
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erdoswiles
60 posts
#12 • 4 Y
Y by Arkajit_Ganguly, Pluto1708, Adventure10, Mango247
Since $a(1)c(1) + b(1)d(1)  = 3$, we can assume WLOG that $b(1)d(1) \not = 0$ so that $b(1), d(1) \not =0$. Now, note that \begin{align} \label{1} a(\omega)c(\omega) + b(\omega)d(\omega) = 1 + w^2 + w^4 = 0 \\ a(1)c(\omega) + b(1)d(\omega) = 1 + w + w^2 = 0 \\ a(\omega)c(1) + b(\omega)d(1) = 0,\end{align}so that $d(\omega) = -a(1)c(\omega)/b(1)$ and $b(\omega) = -a(\omega)c(1)/d(1)$. If we put, this into $(1)$, we get $$a(\omega)c(\omega)\Big(1 + \frac{a(1)c(1)}{b(1)d(1)}\Big) = 0.$$Thus we must have $a(\omega)c(\omega) = 0$, so that at least $a(\omega) = 0$ or $c(\omega) = 0$. WLOG assume $a(\omega) = 0$, then through $(3)$ we see that $b(\omega) = 0$ and therefore $$ 0 = a(\omega)c(-1) + b(\omega)d(-1) = 1 -\omega + \omega^2 = -2\omega, $$implying such polynomials don't exist.
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WolfusA
1900 posts
#13 • 2 Y
Y by Adventure10, Mango247
Kezer wrote:
Suppose $a(x) = a_0+a_1x+\dots+a_nx^n$, then we can quickly conclude that $b(x) = b_0+b_1x+b_2x^2 - \frac{c(y)}{d(y)} \left(x^3+x^4+\dots+x^n \right)$ for every $y$ and $d(y) \neq 0$.
Didn't you loose somewhere $1+xy+x^2y^2$ part?
my sol
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clarkculus
254 posts
#14 • 1 Y
Y by centslordm
The answer is no. Let $A(0)=a_0$ and etc., and let the functional equation be $P(x,y)$.

If $b_0=0$, $P(0,y)$ implies $C(y)=\frac{1}{a_0}$, so $D(y)$ must be nonconstant and $A(x)=a_0$. However, this implies $(b_1x+b_2x^2)(d_0+d_1y+d_2y^2)=xy+x^2y^2$, so $b_1d_1=b_2d_2=1$. In particular, this implies $b_1,d_2\neq0$, which contradicts the $xy^2$ coefficient $b_1d_2=0$. $d_0=0$ leads to a similar contradiction.

If $b_0,d_0\neq0$, $P(0,y)$ and $P(x,0)$ give $D(y)=\frac{1-a_0C(y)}{b_0}$ and $B(x)=\frac{1-c_0A(x)}{d_0}$, so we can rewrite $P(x,y)$ as $1+xy+x^2y^2=k_1A(x)C(y)+k_2A(x)+k_3C(y)+k_4$ for real $k_i$. If either of $A$ or $C$ are constant, then it is impossible to have a $xy$ term on the RHS side, but if both are nonconstant, then as above, $k_1a_1c_1=k_1a_2c_2=1$ while $k_1a_1c_2=0$, contradiction.

(when you do a non lin alg solution in a lin alg unit :( )
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