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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.

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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
1 viewing
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
Infinite number of sets with an intersection property
Drytime   8
N 4 minutes ago by math90
Source: Romania TST 2013 Test 2 Problem 4
Let $k$ be a positive integer larger than $1$. Build an infinite set $\mathcal{A}$ of subsets of $\mathbb{N}$ having the following properties:

(a) any $k$ distinct sets of $\mathcal{A}$ have exactly one common element;
(b) any $k+1$ distinct sets of $\mathcal{A}$ have void intersection.
8 replies
Drytime
Apr 26, 2013
math90
4 minutes ago
2-var inequality
sqing   0
9 minutes ago
Source: Own
Let $ a,b> 0 ,a^3+ab+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq 8$$Let $ a,b> 0 ,a^3+ab(a+b)+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$
0 replies
+2 w
sqing
9 minutes ago
0 replies
Factorials divide
va2010   37
N 32 minutes ago by ND_
Source: 2015 ISL N2
Let $a$ and $b$ be positive integers such that $a! + b!$ divides $a!b!$. Prove that $3a \ge 2b + 2$.
37 replies
va2010
Jul 7, 2016
ND_
32 minutes ago
IMO Shortlist 2011, Number Theory 2
orl   24
N 34 minutes ago by ezpotd
Source: IMO Shortlist 2011, Number Theory 2
Consider a polynomial $P(x) =  \prod^9_{j=1}(x+d_j),$ where $d_1, d_2, \ldots d_9$ are nine distinct integers. Prove that there exists an integer $N,$ such that for all integers $x \geq N$ the number $P(x)$ is divisible by a prime number greater than 20.

Proposed by Luxembourg
24 replies
orl
Jul 11, 2012
ezpotd
34 minutes ago
Inequality in triangle
Nguyenhuyen_AG   3
N 43 minutes ago by Nguyenhuyen_AG
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]
3 replies
Nguyenhuyen_AG
Today at 6:17 AM
Nguyenhuyen_AG
43 minutes ago
Problem 1
randomusername   73
N an hour ago by ND_
Source: IMO 2015, Problem 1
We say that a finite set $\mathcal{S}$ of points in the plane is balanced if, for any two different points $A$ and $B$ in $\mathcal{S}$, there is a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that $\mathcal{S}$ is centre-free if for any three different points $A$, $B$ and $C$ in $\mathcal{S}$, there is no points $P$ in $\mathcal{S}$ such that $PA=PB=PC$.

(a) Show that for all integers $n\ge 3$, there exists a balanced set consisting of $n$ points.

(b) Determine all integers $n\ge 3$ for which there exists a balanced centre-free set consisting of $n$ points.

Proposed by Netherlands
73 replies
randomusername
Jul 10, 2015
ND_
an hour ago
x is rational implies y is rational
pohoatza   44
N an hour ago by ezpotd
Source: IMO Shortlist 2006, N2, VAIMO 2007, Problem 6
For $ x \in (0, 1)$ let $ y \in (0, 1)$ be the number whose $ n$-th digit after the decimal point is the $ 2^{n}$-th digit after the decimal point of $ x$. Show that if $ x$ is rational then so is $ y$.

Proposed by J.P. Grossman, Canada
44 replies
pohoatza
Jun 28, 2007
ezpotd
an hour ago
Multiplicative function
Tales   37
N an hour ago by ezpotd
Source: IMO Shortlist 2004, number theory problem 2
The function $f$ from the set $\mathbb{N}$ of positive integers into itself is defined by the equality \[f(n)=\sum_{k=1}^{n} \gcd(k,n),\qquad n\in \mathbb{N}.\]
a) Prove that $f(mn)=f(m)f(n)$ for every two relatively prime ${m,n\in\mathbb{N}}$.

b) Prove that for each $a\in\mathbb{N}$ the equation $f(x)=ax$ has a solution.

c) Find all ${a\in\mathbb{N}}$ such that the equation $f(x)=ax$ has a unique solution.
37 replies
Tales
Mar 23, 2005
ezpotd
an hour ago
NICE INEQUALITY
Kyleray   3
N an hour ago by sqing
Let's $a,b,c>0$. Prove:
$$(\frac{a}{b+c}+\frac{b}{c+a})(\frac{b}{c+a}+\frac{c}{a+b})(\frac{c}{a+b}+\frac{a}{b+c})\geq \frac{(a+b+c)^2}{3(ab+bc+ca)}$$$\text{P/S: No mapple, please :(}$
3 replies
Kyleray
Mar 11, 2021
sqing
an hour ago
Tough inequality
TUAN2k8   4
N an hour ago by cazanova19921
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
4 replies
TUAN2k8
May 28, 2025
cazanova19921
an hour ago
Easy Diophantne
anantmudgal09   20
N 2 hours ago by Adywastaken
Source: India Practice TST 2017 D1 P2
Find all positive integers $p,q,r,s>1$ such that $$p!+q!+r!=2^s.$$
20 replies
anantmudgal09
Dec 9, 2017
Adywastaken
2 hours ago
Converse of a classic orthocenter problem
spartacle   43
N 2 hours ago by ihategeo_1969
Source: USA TSTST 2020 Problem 6
Let $A$, $B$, $C$, $D$ be four points such that no three are collinear and $D$ is not the orthocenter of $ABC$. Let $P$, $Q$, $R$ be the orthocenters of $\triangle BCD$, $\triangle CAD$, $\triangle ABD$, respectively. Suppose that the lines $AP$, $BQ$, $CR$ are pairwise distinct and are concurrent. Show that the four points $A$, $B$, $C$, $D$ lie on a circle.

Andrew Gu
43 replies
spartacle
Dec 14, 2020
ihategeo_1969
2 hours ago
Symmetric points part 2
CyclicISLscelesTrapezoid   22
N 2 hours ago by ihategeo_1969
Source: USA TSTST 2022/6
Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $ABC$. The perpendicular bisector of $\overline{AH}$ intersects $\overline{AB}$ and $\overline{AC}$ at $X_A$ and $Y_A$ respectively. Let $K_A$ denote the intersection of the circumcircles of triangles $OX_AY_A$ and $BOC$ other than $O$.

Define $K_B$ and $K_C$ analogously by repeating this construction two more times. Prove that $K_A$, $K_B$, $K_C$, and $O$ are concyclic.

Hongzhou Lin
22 replies
CyclicISLscelesTrapezoid
Jun 27, 2022
ihategeo_1969
2 hours ago
Periodicity of factorials
Cats_on_a_computer   0
2 hours ago
Source: Thrill and challenge of pre-college mathematics
Let a_k denote the first non zero digit of the decimal representation of k!. Does the sequence a_1, a_2, a_3, … eventually become periodic?
0 replies
Cats_on_a_computer
2 hours ago
0 replies
number theory
Levieee   7
N Apr 19, 2025 by g0USinsane777
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).
7 replies
Levieee
Apr 18, 2025
g0USinsane777
Apr 19, 2025
number theory
G H J
G H BBookmark kLocked kLocked NReply
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Levieee
244 posts
#1
Y by
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).
This post has been edited 2 times. Last edited by Levieee, Apr 18, 2025, 7:47 PM
Z K Y
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Lil_flip38
58 posts
#2
Y by
I might be missing something, but why does \((n+a)(n+b)=k^2\) imply \(n+a\mid k\) and \(k\mid n+b\)? Why cant for example \(n+a=25\) and \(n+b=4\) and \(k=10\). The factors of \(k\) do not have to be distributed nicely between \(n+a\) and \(n+b\).
Z K Y
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DTforever
6 posts
#3
Y by
I think the condition implies both $n+a$ and $n+b$ are perfect squares.
Z K Y
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MathLuis
1557 posts
#4 • 1 Y
Y by Levieee
You need a slightly more sophisitcated finish after realising $(n+a)(n+b)$ is a square for infinitely many $n$ and it is that we have $n^2+(a+b)n+ab$ is a square for infinitely many $n$ and now setting some arbitrarily large $n$ and considering some $k$ for which $2k \ge a+b >2k-2$ we could have that it is a square bounded between $(n+k)^2$ and $(n+k-1)^2$ so it has to be $(n+k)^2$ which shows that $(a+b)^2=4ab$ by coefficient checking and matching and thus $(a-b)^2=0$ so $a=b$ which is a contradiction!.
Z K Y
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Safal
171 posts
#5
Y by
Levieee wrote:
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).

$k_1$ and $k_2$ need not be integers, they can be rational beacuse we can have that the denominator ot that rational divide $n+a$
This post has been edited 1 time. Last edited by Safal, Apr 18, 2025, 9:19 PM
Z K Y
The post below has been deleted. Click to close.
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Levieee
244 posts
#6
Y by
Safal wrote:
Levieee wrote:
Idk where it went wrong, marks was deducted for this solution
$\textbf{Question}$
Show that for a fixed pair of distinct positive integers \( a \) and \( b \), there cannot exist infinitely many \( n \in \mathbb{Z} \) such that
\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]
$\textbf{Solution}$

Let
\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]
Then,
\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]So:
\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]
Therefore,
\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]
Let
\[
(n + a)(n + b) = k^2.
\]Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).

Without loss of generality, we take the first case:
\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]
Thus,
\[
k_1 k_2 = \frac{n + b}{n + a}.
\]
Since \( k_1 k_2 \in \mathbb{N} \), we have:
\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]
For infinitely many \( n \), \( \frac{b - a}{n + a} \) must be an integer, which is not possible.

Therefore, there cannot be infinitely many such \( n \).

$k_1$ and $k_2$ need not be integers, they can be rational beacuse we can have that the denominator ot that rational divide $n+a$

yea but after it's reduced we can get it to be an integer , doesnt $ a \mid b$ imply for some $k$ $ak=b$ , k is an integer

@2above how do u say that $n^2 + ab + (a+b)n$
This post has been edited 2 times. Last edited by Levieee, Apr 19, 2025, 7:41 AM
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Levieee
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#7
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DTforever wrote:
I think the condition implies both $n+a$ and $n+b$ are perfect squares.

does the counter example provided by @Lil_flip38 work? or that counter example can be invalidated because $n = 4-b$, $b \ge 1$ so as n increases we can just show it cant happen
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g0USinsane777
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#8 • 1 Y
Y by Levieee
Levieee wrote:
Assume \( n + a \neq n + b \). Then we have:
\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]or it could also be that \( k \mid n + a \quad \text{and} \quad n + b \mid k \).
This part is certainly wrong, since $(n+a)(n+b)=k^2$ can imply $n+a \leq k$ and $n+b \geq k$ or vice-versa but not the divisibility condition above. You can find many counter examples to this.

Another finish from this part can be that let $\gcd(n+a,n+b)=d$. (of course, $n+a \neq n+b$)
So, $n+a = dx$ and $n+b=dy$, $\gcd(x,y)=1$
This gives us that, $d^2xy = k^2$
Which means that $xy=l^2$ and $x=u^2$ and $y=v^2$, $\gcd(u,v)=1$.
This means that, $b-a=d(v^2 - u^2)$ for a particular $n$.
Since, the original condition needs to be true for infinitely many $n$, by infinite PHP we will have infinitely many $n$ for which we get the same value of $d$ for a particular value of $k$.
So, for infinitely many $n$, which in turn give infinitely many value of the pair $(u,v)$, the value $\frac{b-a}{d}$ is constant, say $t$.
From the above argument, since infinitely many pairs $(u,v)$ satisfy $t = v^2-u^2 = (v-u)(v+u)$, where $v,u$ are integers, this means that $t$ has infinitely many factors $\implies$ $t=0$, i.e., $b=a$ which contradicts that $a$ and $b$ are distinct.
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