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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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[*]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
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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
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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
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Urgent. Need them quick
sealight2107   0
8 minutes ago
With $a,b,c>1$ and $a+b+c=2abc$. Prove that:
$\sqrt[3]{ab-1}+\sqrt[3]{bc-1}+\sqrt[3]{ca-1} \le \sqrt[3]{(a+b+c)^2}$
0 replies
sealight2107
8 minutes ago
0 replies
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m^4+3^m is a perfect square number
Havu   1
N 15 minutes ago by Soupboy0
Find a positive integer m such that $m^4+3^m$ is a perfect square number.
1 reply
Havu
17 minutes ago
Soupboy0
15 minutes ago
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Divisibility..
Sadigly   3
N 23 minutes ago by Jackson0423
Source: Azerbaijan Junior MO 2025 P2
Find all $4$ consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
3 replies
Sadigly
Today at 7:37 AM
Jackson0423
23 minutes ago
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Divisibilty...
Sadigly   9
N 32 minutes ago by Jackson0423
Source: My (fake) translation error
Find all $4$ consecutive even numbers, such that the square of their product divides the sum of their squares.
9 replies
Sadigly
an hour ago
Jackson0423
32 minutes ago
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Inspired by Kosovo 2010
sqing   1
N 33 minutes ago by ytChen
Source: Own
Let $ a,b>0 , a+b\leq k $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq\left(1+\frac{4}{k(k+2)}\right)^2$$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq\left(1+\frac{2}{k+2}\right)^2$$Let $ a,b>0 , a+b\leq 2 $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq \frac{9}{4} $$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq \frac{9}{4} $$
1 reply
sqing
Today at 3:56 AM
ytChen
33 minutes ago
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how can I solve this FE
Jackson0423   5
N 33 minutes ago by Jackson0423

Let \( f : \mathbb{R} \to \mathbb{R} \) be a function that satisfies the following equation for all real numbers \( x \):
\[ f(x^2 + x + 3) + 2f(x^2 - 3x + 5) = 6x^2 - 10x + 17. \]Find the value of \( f(100) \).
5 replies
Jackson0423
an hour ago
Jackson0423
33 minutes ago
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Number Theory
VicKmath7   5
N 39 minutes ago by Adywastaken
Source: Archimedes Junior 2014
Let $p$ prime and $m$ a positive integer. Determine all pairs $( p,m)$ satisfying the equation: $ p(p+m)+p=(m+1)^3$
5 replies
VicKmath7
Mar 17, 2020
Adywastaken
39 minutes ago
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x+y in B iff x,y in A
fattypiggy123   5
N 43 minutes ago by Math2030
Source: China Mathematical Olympiad 2015 Q3
Let $n \geq 5$ be a positive integer and let $A$ and $B$ be sets of integers satisfying the following conditions:

i) $|A| = n$, $|B| = m$ and $A$ is a subset of $B$
ii) For any distinct $x,y \in B$, $x+y \in B$ iff $x,y \in A$

Determine the minimum value of $m$.
5 replies
fattypiggy123
Dec 20, 2014
Math2030
43 minutes ago
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IMO Genre Predictions
ohiorizzler1434   65
N an hour ago by Oksutok
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
65 replies
ohiorizzler1434
May 3, 2025
Oksutok
an hour ago
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k^2/p for k =1 to (p-1)/2
truongphatt2668   1
N an hour ago by Double07
Let $p$ be a prime such that: $p = 4k+1$. Simplify:
$$\sum_{k=1}^{\frac{p-1}{2}}\begin{Bmatrix}\dfrac{k^2}{p}\end{Bmatrix}$$
1 reply
truongphatt2668
3 hours ago
Double07
an hour ago
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Interesting inequality
imnotgoodatmathsorry   1
N an hour ago by Bergo1305
Let $x,y,z > \frac{1}{2}$ and $x+y+z=3$.Prove that:
$\sqrt{x^3+y^3+3xy-1}+\sqrt{y^3+z^3+3yz-1}+\sqrt{z^3+x^3+3zx-1}+\frac{1}{4}(x+5)(y+5)(z+5) \le 60$
1 reply
1 viewing
imnotgoodatmathsorry
2 hours ago
Bergo1305
an hour ago
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every lucky set of values {a_1,a_2,..,a_n} satisfies a_1+a_2+...+a_n >n2^{n-1}
parmenides51   6
N an hour ago by jonh_malkovich
Source: 2020 International Olympiad of Metropolises P3
Let $n>1$ be a given integer. The Mint issues coins of $n$ different values $a_1, a_2, ..., a_n$, where each $a_i$ is a positive integer (the number of coins of each value is unlimited). A set of values $\{a_1, a_2,..., a_n\}$ is called lucky, if the sum $a_1+ a_2+...+ a_n$ can be collected in a unique way (namely, by taking one coin of each value).
(a) Prove that there exists a lucky set of values $\{a_1, a_2, ..., a_n\}$ with $$a_1+ a_2+...+ a_n < n \cdot 2^n.$$(b) Prove that every lucky set of values $\{a_1, a_2,..., a_n\}$ satisfies $$a_1+ a_2+...+ a_n >n \cdot 2^{n-1}.$$
Proposed by Ilya Bogdanov
6 replies
parmenides51
Dec 19, 2020
jonh_malkovich
an hour ago
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A strange NT problem
flower417477   0
an hour ago
Source: unknown
$p$ is a given prime number.$A=\{a_1,a_2,\cdots,a_{p-1}\}$ is a set which $\prod\limits_{i=1}^{p-1}a_i\equiv\frac{p-1}{2}\pmod p$.
Prove that there're at least $\frac{p-1}{2}$ non-empty subsets $B$ of $A$ such that $\sum\limits_{b\in B}b\equiv 1\pmod p$
0 replies
flower417477
an hour ago
0 replies
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combi/nt
blug   0
an hour ago
Prove that every positive integer $n$ can be written in the form
$$n=2^{a_1}3^{b_1}+2^{a_2}3^{b_2}+..., $$where $a_i, b_j$ are non negative integers, such that
$$2^x3^y\nmid 2^z3^t$$for every $x, y, z, t$.
0 replies
blug
an hour ago
0 replies
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Sum of divisors
DinDean   1
N Apr 27, 2025 by Tintarn
Does there exist $M>0$, such that $\forall m>M$, there exists an integer $n$ satisfying $\sigma(n)=m$?
$\sigma(n)=$ the sum of all positive divisors of $n$.
1 reply
DinDean
Apr 18, 2025
Tintarn
Apr 27, 2025
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Sum of divisors
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DinDean
16 posts
DinDean
#1 h  Apr 18, 2025, 2:47 PM
Y by
Does there exist $M>0$, such that $\forall m>M$, there exists an integer $n$ satisfying $\sigma(n)=m$?
$\sigma(n)=$ the sum of all positive divisors of $n$.
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Tintarn
9042 posts
Tintarn
#2 h  Apr 27, 2025, 10:17 PM
Y by
No. For example, take $m=q$ as a large prime. Then by multiplicativity of $\sigma(n)$, the only chance for $\sigma(n)=q$ would be that $n=p^k$ is a prime power, so $q=p^k+p^{k-1}+\dots+p+1$ for some (even) $k \ge 2$.
But it is easy to see that there are only roughly $\sqrt{X}$ numbers of the form $x^k+x^{k-1}+\dots+x+1$ up to $X$, while of course there are roughly $X/log(X)$ many choices for $q$, so "most" $q$ are not of the form $\sigma(n)$.

In fact, a slight modification of this argument shows that "most" numbers are not values of $\sigma(n)$ in the sense that the set of such values has density zero in the natural numbers.
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