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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.

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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
Combinatorial
|nSan|ty   7
N 3 minutes ago by SomeonecoolLovesMaths
Source: RMO 2007 problem
How many 6-digit numbers are there such that-:
a)The digits of each number are all from the set $ \{1,2,3,4,5\}$
b)any digit that appears in the number appears at least twice ?
(Example: $ 225252$ is valid while $ 222133$ is not)
[weightage 17/100]
7 replies
|nSan|ty
Oct 10, 2007
SomeonecoolLovesMaths
3 minutes ago
pairs (m, n) such that a fractional expression is an integer
cielblue   0
22 minutes ago
Find all pairs $(m,\ n)$ of positive integers such that $\frac{m^3-mn+1}{m^2+mn+2}$ is an integer.
0 replies
cielblue
22 minutes ago
0 replies
the same prime factors
andria   6
N 44 minutes ago by MathLuis
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
6 replies
andria
Sep 6, 2015
MathLuis
44 minutes ago
Inspired by RMO 2006
sqing   1
N an hour ago by SomeonecoolLovesMaths
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 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
1 reply
sqing
6 hours ago
SomeonecoolLovesMaths
an hour ago
Problem 4 of RMO 2006 (Regional Mathematical Olympiad-India)
makar   7
N an hour ago by SomeonecoolLovesMaths
Source: Combinatorics (Box Principle)
A $ 6\times 6$ square is dissected in to 9 rectangles by lines parallel to its sides such that all these rectangles have integer sides. Prove that there are always two congruent rectangles.
7 replies
makar
Sep 13, 2009
SomeonecoolLovesMaths
an hour ago
Simple FE
oVlad   52
N an hour ago by Sadigly
Source: BMO Shortlist 2022, A1
Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
\[f(x(x + f(y))) = (x + y)f(x),\]for all $x, y \in\mathbb{R}$.
52 replies
oVlad
May 13, 2023
Sadigly
an hour ago
Cool Functional Equation
Warideeb   1
N an hour ago by maromex
Find all functions real to real such that
$f(xy+f(x))=xf(y)+f(x)$
for all reals $x,y$.
1 reply
Warideeb
3 hours ago
maromex
an hour ago
Functional equation
socrates   10
N an hour ago by MathLuis
Source: Inspired by another
Determine all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that \[ \forall x, y \in \mathbb{R}^+ \ , \  \ f(x+f(xy))=f(x)+xf(y).\]
10 replies
socrates
Oct 28, 2014
MathLuis
an hour ago
Find f
Redriver   7
N 2 hours ago by aaravdodhia
Find all $: R \to R : \ \ f(x^2+f(y))=y+f^2(x)$
7 replies
Redriver
Jun 25, 2006
aaravdodhia
2 hours ago
IMO Shortlist 2011, G5
WakeUp   72
N 2 hours ago by ItsBesi
Source: IMO Shortlist 2011, G5
Let $ABC$ be a triangle with incentre $I$ and circumcircle $\omega$. Let $D$ and $E$ be the second intersection points of $\omega$ with $AI$ and $BI$, respectively. The chord $DE$ meets $AC$ at a point $F$, and $BC$ at a point $G$. Let $P$ be the intersection point of the line through $F$ parallel to $AD$ and the line through $G$ parallel to $BE$. Suppose that the tangents to $\omega$ at $A$ and $B$ meet at a point $K$. Prove that the three lines $AE,BD$ and $KP$ are either parallel or concurrent.

Proposed by Irena Majcen and Kris Stopar, Slovenia
72 replies
WakeUp
Jul 13, 2012
ItsBesi
2 hours ago
Hard Functional Equation in the Complex Numbers
yaybanana   5
N 2 hours ago by aaravdodhia
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 $x,y \in \mathbb{C}$
5 replies
yaybanana
Apr 9, 2025
aaravdodhia
2 hours ago
Power tower sum
Rijul saini   9
N 2 hours ago by ihategeo_1969
Source: India IMOTC 2024 Day 3 Problem 2
Let $a$ and $n$ be positive integers such that:
1. $a^{2^n}-a$ is divisible by $n$,
2. $\sum\limits_{k=1}^{n} k^{2024}a^{2^k}$ is not divisible by $n$.

Prove that $n$ has a prime factor smaller than $2024$.

Proposed by Shantanu Nene
9 replies
Rijul saini
May 31, 2024
ihategeo_1969
2 hours ago
Nice "if and only if" function problem
ICE_CNME_4   7
N 2 hours ago by ICE_CNME_4
Let $f : [0, \infty) \to [0, \infty)$, $f(x) = \dfrac{ax + b}{cx + d}$, with $a, d \in (0, \infty)$, $b, c \in [0, \infty)$. Prove that there exists $n \in \mathbb{N}^*$ such that for every $x \geq 0$
\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\](For $n \in \mathbb{N}^*$ and $x \geq 0$, the notation $f_n(x)$ represents $\underbrace{(f \circ f \circ \dots \circ f)}_{n \text{ times}}(x)$. )

Please do it at 9th grade level. Thank you!
7 replies
ICE_CNME_4
Yesterday at 7:23 PM
ICE_CNME_4
2 hours ago
Cauchy-Schwarz 2
prtoi   8
N 2 hours ago by mrtheory
Source: Handout by Samin Riasat
if $a^2+b^2+c^2+d^2=4$, prove that:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\ge4$
8 replies
prtoi
Mar 26, 2025
mrtheory
2 hours ago
Iran second round 2025-q1
mohsen   8
N May 18, 2025 by Autistic_Turk
Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.
8 replies
mohsen
Apr 19, 2025
Autistic_Turk
May 18, 2025
Iran second round 2025-q1
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G H BBookmark kLocked kLocked NReply
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mohsen
16 posts
#1 • 1 Y
Y by sami1618
Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.
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sami1618
913 posts
#2 • 1 Y
Y by Sedro
Answer: No such number exists.

Solution. We start by proving the following two claims:

Claim 1. We have that $n$ is squarefree.
Proof. If there exists a prime $p$ for which $p^2|n$, then $n+p$ must be a perfect square. But $v_p(n+p)=1$, a contradiction.

Claim 2. We have that $n$ has at most two distinct prime divisors.
Proof. Assume otherwise that $p$, $q$, and $r$ are three distinct prime divisors of $n$. Without loss of generality, assume that $p<q<r$. It is easy to see that both $p$ and $q$ must be strictly smaller than $\sqrt{n}$. On the other hand, $$\sqrt{n+q}-\sqrt{n+p}\geq 1\Rightarrow q-p\geq \sqrt{n+q}+\sqrt{n+p}\geq 2\sqrt{n},$$which is a contradiction.

By the Claims, we must have that either $n=p$ for some prime $p>2$ or $n=pq$ for some distinct primes $p$ and $q$. In the first case, $2p$ must be a perfect square, which is absurd. In the second case, both $p(q+1)$ and $q(p+1)$ must be perfect squares. This implies that $p|q+1$ and that $q|p+1$. Assume that $p<q$. Then it must be that $p=q-1$ which implies that $p=2$ and $q=3$, which fails to work.
This post has been edited 1 time. Last edited by sami1618, Apr 19, 2025, 5:33 PM
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missionjoshi.65
2 posts
#3 • 2 Y
Y by Calc_1, sami1618
Posting though very similar as above!
Claim 1: $n$ can't be a power of prime.

Suppose, $n = p^k$ for some prime $p$, then we have $v_p(p^k+p) =1+v_p(p^{k-1}+1),$ which can't be a square when $k \geq 2$. And, $k = 1$ gives $n = 2$ which is a contradiction.

Claim 2: $n$ cannot have three or more prime factors.

Suppose $p,q,r | n$ and $p<q<r$, then

$$n +p = a^2 \quad \text{and} \quad n+q = b^2$$
This implies:
$$ q-p = (b-a)(b+a)$$
Similarly,
$$ q-p \geq b+a = \sqrt{n+p} + \sqrt{n+q} > \sqrt{n}+\sqrt{n} = 2 \sqrt{n}$$
which is a contradiction since both $p, q < \sqrt{n}.$

Now, for two primes $p<q$ we have $p|q+1$ and $q|p+1$ which implies $p=q-1,$ so only possible values are $p=2$ and $q=3$, which do not work. And, we are done!
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Parsia--
79 posts
#4 • 2 Y
Y by Mahdi_Mashayekhi, amirhsz
Here's a much different and overkill approach.
suppose $n$ has at least two different primes $q,r$ such that $q \equiv r \equiv 3 \mod 4$. Note that $n$ is also square-free.
So we get
$$q(\frac{n}{q}+1) = x^2 \Rightarrow (\frac{r}{q})=1 $$Similarly $ (\frac{q}{r})=1$.
Now note that $$ (\frac{r}{q})(\frac{q}{r})=(-1)^{\frac{r-1}{2}. \frac{q-1}{2}}=-1 \Rightarrow  1=(\frac{r}{q})=-(\frac{q}{r})=-1$$which is a contradiction.
if $n = 2p$, we must have $3p$ is a perfect square which gives $p=3$ which doesn't work.
Now $n$ odd won't work because of mod $4$. so $n$ is even. Consider a prime divisor $p$ of $n$ such $p \equiv 1 \mod 4$ we get $p(\frac{n}{p}+1) =x^2$ but $\frac{n}{p}+1 \equiv 3 \mod 4$.
Thus we conclude that no such $n$ exists.
This post has been edited 2 times. Last edited by Parsia--, May 10, 2025, 3:18 PM
Reason: Shortened the solution
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MathLuis
1553 posts
#5
Y by
Suppose FTSOC there existed such $n$, then if $n$ was not squarefree then there exists a prime $p$ with $p^2 \mid n$ however we have $n+p$ is a perfect square and thus $2 \mid \nu_p(n+p)=1$, a clear contradiction!, therefore $n$ is squarefree.
Now let $n=p_1 \cdot p_2 \cdots p_k$ where $p_1<p_2 \cdots p_k$. Clearly $k \ge 2$ as if $k=1$ then $2p_1$ is a perfect square but $p_1 \ne 2$ in this case so this isn't true, and if $k \ge 3$ then notice we must have $\sqrt{n+p_2}-\sqrt{n+p_1} \ge 1$ which implies that $n+p_2 \ge n+p_1+2\sqrt{n+p_1}+1$ which implies that $p_2^2>(p_2-p_1-1)^2 \ge 4n+4p_1>4p_2^2+4p_1$ which is clearly a contradiction.
And now if $n=p_1p_2$ then $p_2(p_1+1), p_1(p_2+1)$ are both perfect squares then $p_2 \mid p_1+1$ and $p_1 \mid p_2+1$ so $p_1p_2 \mid p_1p_2+p_1+p_2+1$ which means $p_1p_2 \mid p_1+p_2+1$ which means that $2 \le (p_1-1)(p_2-1) \le 2$ where equality is attained when $p_1=2$ and $p_2=3$ however $2 \cdot 4=8$ is not a perfect square thus contradiction either way and we are done :cool:
This post has been edited 1 time. Last edited by MathLuis, Apr 23, 2025, 12:31 AM
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math-helli
13 posts
#6
Y by
Here you can find some solutions of second round
https://t.me/matholampiad123
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Marco22
3 posts
#7
Y by
Answer:such number do not exist.
Suppose on the contrary that a number $n$ satisfying the properties exists, and let $p$ and $q$ be two prime divisors of $n$. Then by the hypothesis there exist $x,y \in \mathbb{N}$ such that $n+p=x^2$ and $n+q=y^2$. clearly we can see that $p|x$ and $p|y$. Then by a size argument it follows:
$p-q=x^2-y^2=(x-y)(x+y)=(ap-bq)(ap+bq)>p+q$ for some integers $a$ and $b$. So the answer clearly follows
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Mathgloggers
90 posts
#8
Y by
Notice that each prime sums with $n$ to a distinct perfect square.
So let $k \in N$ such that $k^2>n$
We have, $(k+3)^2-n<\frac{n}{2} $,because the max prime is $< \frac{n}{2}$
$k^2+9+6k<\frac{3n}{2} $ $\implies$ $n^2<k<\frac{\frac{n}{2}-9}{6} $,now just solve this equation for $n$ you will get $-2<n<2$ ,So you it also does not work.

So it has clearly at most 2 prime factors for $n=p.q>1$ which respectively sums to $n+p=k^2$ and $n+q=(k+1)^2$.
hence one has to be even =$2$ and the other must be $2k+3$ and $n=k^2-2$



Now just put all in that equation you will get :$k^2-4k-8=0$ no integer solution .
(I don't why my solution was so basic:(
This post has been edited 1 time. Last edited by Mathgloggers, May 18, 2025, 6:41 AM
Reason: m
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Autistic_Turk
11 posts
#9
Y by
mohsen wrote:
Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.

Here is my solution that I actually used in contest
Trivialy there exist positive intiger m and non negative integer q so that n=m^2+q and n<(m+1)^2 so q<2m+1 now there are 2 cases
Case 1: n has at least two different prime divisor
Let p_1 and p_2 be prime divisor of n and WLOG p_2>p_1 notice there exist intiger r_1 and r_2 such that n+p_1=(m+r_1)^2 and n+p_2=(m+r_2)^2 and p_2 divide n therfore p_2 divide n+p_2 so p_2 divide (m+r_2)^2 thus p_2 divide m+r_2 but we know p_2=(m+r_2)^2-n=m^2+2(r_2)m+(r_2)^2-m^2-q and we know any natural divisor of a natural number is smaller or equal to that number so we have
2(r_2)m +(r_2)^2-q =< m+r_2
But we have r_2=>2 and q<=2m+1 thus contradiction is trivial
Case 2: n has only one prime factor
Thus there exist prime number p and positive intiger a such that n=p^a therfore p^a+p is square but for any a larger then 1 we know that p divide n and p^2 do not divide n thus n can't be a square so a=1 and for a=1 we would get 2p is square thus 4 divide 2p and p is even thus p=2
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