Plan ahead for the next school year. Schedule your class today!

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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
and train with the best! Please note that early bird pricing ends August 19th!
Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
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0 replies
jwelsh
Jul 1, 2025
0 replies
Integer sequences with a two-step recurrence
Assassino9931   1
N 10 minutes ago by Assassino9931
Source: Bulgaria Autumn Tournament 2009 Grade 12
Fix an integer $m$. Determine the number of sequences $(a_n)_{n\geq 1}$ of integers such that $a_na_{n+2} = n^2 + m$ for all $n\geq 1$.
1 reply
+1 w
Assassino9931
Today at 1:02 AM
Assassino9931
10 minutes ago
old product!
teomihai   2
N 12 minutes ago by teomihai
It is posible to prove ,without induction :
$(\frac{1}{2}\frac{3}{4}...\frac{2n-1}{2n})^2\leq{\frac{1}{3n+1}} $ for any positiv integer number $n$.?
2 replies
teomihai
Yesterday at 3:51 PM
teomihai
12 minutes ago
2025 IMO Results
ilikemath247365   5
N 12 minutes ago by cheltstudent
Source: https://www.imo-official.org/year_info.aspx?year=2025
Congrats to China for getting 1st place! Congrats to USA for getting 2nd and congrats to South Korea for getting 3rd!
5 replies
ilikemath247365
Yesterday at 4:48 PM
cheltstudent
12 minutes ago
A feasible refinement of GMA 567
Rhapsodies_pro   2
N an hour ago by Rhapsodies_pro
Source: Own?
Let \(a_1,a_2,\dotsc,a_n\) (\(n>3\)) be non-negative real numbers fulfilling \[\sum_{k=1}^na_k^2+{\left(n^2-3n+1\right)}\prod_{k=1}^na_k\geqslant{\left(n-1\right)}^2\text.\]Prove or disprove: \[\frac1{n-1}\sum_{1\leqslant i<j\leqslant n}{\left(a_i-a_j\right)}^2\geqslant{\left({\left(n^2-2n-1\right)}\sum_{k=1}^na_k-n{\left(n-1\right)}{\left(n-3\right)}\right)}{\left(n-\sum_{k=1}^na_k\right)}\textnormal.\]
2 replies
1 viewing
Rhapsodies_pro
Jul 21, 2025
Rhapsodies_pro
an hour ago
Nothing but a game
AlexCenteno2007   2
N an hour ago by vanstraelen
Let ABCD be a trapeze with AD ∥ BC. M and N are the midpoints of CD and BC
respectively, and P is the common point of the lines AM and DN. If PM/AP = 4, show that
ABCD is a parallelogram.
2 replies
AlexCenteno2007
Yesterday at 5:11 PM
vanstraelen
an hour ago
evenually periodic sequence
Tofa7a._.36   3
N 2 hours ago by Jupiterballs
Let $N$ be a positive integer and let $a_1,a_2,\dots,a_N$ be positive integers. Define for all $n > N$, $a_n$ to be the least occurring integer in $a_1,a_2,\dots ,a_{n-1}$(if the smallest number of occurrences is shared between two distinct integers $a_k$ , let $a_n$ be the smaller of the two).
Prove that the sequence $(a_n)_{n\in\mathbb{N^*}}$ is eventually periodic.
Example: For $N=5$ , for
$$a_1 = 1, a_2 = 6, a_3 = 8, a_4 = 3, a_5 = 1$$We get: $a_6 = 3, a_7 = 6, a_8 = 8,...$
Example of an eventually periodic sequence : the following sequence is periodic starting from the fourth term and its period is 2:
$$5,4,3,1,2,1,2,1,2,...$$
3 replies
Tofa7a._.36
Jul 15, 2025
Jupiterballs
2 hours ago
Function
Ecrin_eren   1
N 3 hours ago by alexheinis

Is there a function from non-negative real numbers to non-negative real numbers satisfying

f(f(x)) = |x - 1| for all x ≥ 0?











1 reply
Ecrin_eren
Yesterday at 2:26 PM
alexheinis
3 hours ago
Inequalities
sqing   28
N 3 hours ago by sqing
Let $ a,b> 0 ,\frac{a}{2b+1}+\frac{b}{3}+\frac{1}{2a+1} \leq 1.$ Prove that
$$  a^2+b^2 -ab\leq 1$$$$ a^2+b^2 +ab \leq3$$Let $ a,b,c> 0 , \frac{a}{2b+1}+\frac{b}{2c+1}+\frac{c}{2a+1} \leq 1.$ Prove that
$$    a +b +c +abc \leq 4$$
28 replies
sqing
May 24, 2025
sqing
3 hours ago
Polynomials
Roots_Of_Moksha   5
N 4 hours ago by vanstraelen
The polynomial $x^3 - 3(1+\sqrt{2})x^2 + (6\sqrt{2}-55)x -(7+5\sqrt{2})$ has three distinct real roots $\alpha$, $\beta$ and $\gamma$. The polynomial $p(x)=x^3+ax^2+bx+c$ has roots $\sqrt[3]{\alpha}$, $\sqrt[3]{\beta}$, $\sqrt[3]{\gamma}$. Find the integer closest to $a^2 + b^2 + c^2$.
Answer
5 replies
Roots_Of_Moksha
Jul 20, 2025
vanstraelen
4 hours ago
Basic Inequalities Doubt
JetFire008   9
N 4 hours ago by anduran
If $x+y=1$, find the maximum value of $x^2+y^2=1$.
I saw a solution to this question where Titu's lemma was applied and the answer was $\frac{1}{2}$. But my doubt is can't we apply other inequality to get the maximum result? or did they us titu's lemma because the given information can fit only in this lemma?
9 replies
JetFire008
Jul 18, 2025
anduran
4 hours ago
Inequality from CTPCM
Gaussian0000   3
N Today at 4:14 AM by P0tat0b0y
For a fixed positive integer $n$ compute the minimum value of the sum:
$x_1 + \frac{(x_2)^2}{2} + \frac{(x_3)^3}{3} +...+\frac{(x_n)^n}{n}$ given that
$\frac{1}{x_1} + \frac{1}{x_2}+ \frac{1}{x_3}+...+ \frac{1}{x_n}= n$
3 replies
Gaussian0000
Yesterday at 2:55 PM
P0tat0b0y
Today at 4:14 AM
Inequalities
sqing   13
N Today at 2:06 AM by sqing
Let $ a,b,c,ab+bc+ca = 3. $Prove that$$\sqrt[3]{ \frac{1}{a} + 7b} + \sqrt[3]{\frac{1}{b} + 7c} + \sqrt[3]{\frac{1}{c} + 7a } \leq \frac{6}{abc}$$
13 replies
sqing
Jun 18, 2025
sqing
Today at 2:06 AM
Inequalities
sqing   18
N Today at 2:01 AM by sqing
Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $ and $ abc = -16 $. Prove that$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b} -a^2\geq 2$$$$ \frac{a^2 + b^2}{c} + \frac{b^2 + c^2}{a} + \frac{c^2 + a^2}{b}-a^2-bc\geq -2$$Equality holds when $a=-4,b=c=2.$
18 replies
sqing
Jul 9, 2025
sqing
Today at 2:01 AM
Play a lot
AlexCenteno2007   2
N Yesterday at 10:56 PM by ohiorizzler1434
In a triangle ABC, let D be the foot of the altitude from A, and M the midpoint of BC. Through M
draw straight lines parallel to AB and AC, which intersect AD at P and Q respectively. Show
that A and D are harmonic conjugates of P and Q.
2 replies
AlexCenteno2007
Tuesday at 6:32 PM
ohiorizzler1434
Yesterday at 10:56 PM
Pythagoras...
Hip1zzzil   1
N May 17, 2025 by Primeniyazidayi
Source: KMO 2025 Round 1 P20
Find the sum of all $k$s such that:
There exists two odd positive integers $a,b$ such that ${k}^{2}={a}^{2b}+{(2b)}^{4}.$
1 reply
Hip1zzzil
May 17, 2025
Primeniyazidayi
May 17, 2025
Pythagoras...
G H J
G H BBookmark kLocked kLocked NReply
Source: KMO 2025 Round 1 P20
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Hip1zzzil
24 posts
#1
Y by
Find the sum of all $k$s such that:
There exists two odd positive integers $a,b$ such that ${k}^{2}={a}^{2b}+{(2b)}^{4}.$
This post has been edited 1 time. Last edited by Hip1zzzil, May 17, 2025, 3:41 AM
Reason: E
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Primeniyazidayi
163 posts
#2
Y by
Because $2(x^4+y^4) > (x^2+y^2)^2$ where $x \neq y$,some algebraic manipulation gives $(x^2-y^2)^2 > 2xy$ for positive integers.On the other hand we clearly have $4x^2y^2 > x^2-y^2$.
By Pythagorian Triples Lemma $k^2=x^2+y^2$ and $(a^b,4b^2)$ is a permutation of $(2xy,x^2-y^2)$ where $x>y$ are positive integers.Because the stuff which is showen above one must have $4b^2 > a^b$.$a=1$ gives clearly no solution.For $a=3$ one have $b=1,3$ which gives the triples $(3,1,2)$ and $(3,3,45)$.For $a>3$ there is clearly no $b$ which satisfies the condition,done.
This post has been edited 3 times. Last edited by Primeniyazidayi, May 17, 2025, 9:11 AM
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