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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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[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
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[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
Inequality
MathsII-enjoy   7
N 4 minutes ago by sqing
A interesting problem generalized :-D
7 replies
+1 w
MathsII-enjoy
Saturday at 1:59 PM
sqing
4 minutes ago
F.E....can you solve it?
Jackson0423   0
25 minutes ago
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that
\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]for all real numbers \( x \) satisfying \( f(x) \neq 1 \).
0 replies
Jackson0423
25 minutes ago
0 replies
Functional equation
Math-wiz   25
N an hour ago by Adywastaken
Source: IMOC SL A1
Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x,y\in\mathbb{R}$,
$$f(xy+f(x))=f(xf(y))+x$$
25 replies
Math-wiz
Dec 15, 2019
Adywastaken
an hour ago
Nice numer theory
GeoArt   5
N an hour ago by Primeniyazidayi
$p$ is a prime number, $m, x, y$ are natural numbers ($m, x, y > 1$). It is known that $\frac{x^p + y^p}{2}$ $=$ $(\frac{x+y}{2} )^m$. Prove that $p = m$.
5 replies
GeoArt
Jan 7, 2021
Primeniyazidayi
an hour ago
IOQM 2022-23 P-7
lifeismathematics   2
N 2 hours ago by Adywastaken
Find the number of ordered pairs $(a,b)$ such that $a,b \in \{10,11,\cdots,29,30\}$ and
$\hspace{1cm}$ $GCD(a,b)+LCM(a,b)=a+b$.
2 replies
lifeismathematics
Oct 30, 2022
Adywastaken
2 hours ago
Inequalities
sqing   7
N 3 hours ago by sqing
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
7 replies
sqing
Yesterday at 12:46 PM
sqing
3 hours ago
Inequalities
sqing   10
N 4 hours ago by sqing
Let $a,b,c> 0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1.$ Prove that
$$  (1-abc) (1-a)(1-b)(1-c)  \ge 208 $$$$ (1+abc) (1-a)(1-b)(1-c)  \le -224 $$$$(1+a^2b^2c^2) (1-a)(1-b)(1-c)  \le -5840 $$
10 replies
sqing
Jul 12, 2024
sqing
4 hours ago
China MO 1996 p1
math_gold_medalist28   1
N 4 hours ago by MathsII-enjoy
Let ABC be a triangle with orthocentre H. The tangent lines from A to the circle with diameter BC touch this circle at P and Q. Prove that H, P and Q are collinear.
1 reply
math_gold_medalist28
May 2, 2025
MathsII-enjoy
4 hours ago
If it is an integer then perfect square
Ecrin_eren   1
N 4 hours ago by Pal702004


"Let a, b, c, d be non-zero digits, and let abcd and dcba represent four-digit numbers.

Show that if the number abcd / dcba is an integer, then that integer is a perfect square."



1 reply
Ecrin_eren
May 1, 2025
Pal702004
4 hours ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   6
N 5 hours ago by SomeonecoolLovesMaths
This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3

Problem 4
6 replies
SomeonecoolLovesMaths
Yesterday at 8:16 AM
SomeonecoolLovesMaths
5 hours ago
parallelogram in a tetrahedron
vanstraelen   0
Today at 6:43 AM
Given a tetrahedron $ABCD$ and a plane $\mu$, parallel with the edges $AC$ and $BD$.
$AB \cap \mu=P$.
a) Prove: the intersection of the tetrahedron with the plane is a parallelogram.
b) If $\left|AC\right|=14,\left|BD\right|=7$ and $\frac{\left|PA\right|}{\left|PB\right|}=\frac{3}{4}$,
calculates the lenghts of the sides of this parallelogram.
0 replies
vanstraelen
Today at 6:43 AM
0 replies
Arithmetic Series and Common Differences
4everwise   6
N Today at 2:12 AM by epl1
For each positive integer $k$, let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$. For example, $S_3$ is the sequence $1,4,7,10,...$. For how many values of $k$ does $S_k$ contain the term $2005$?
6 replies
4everwise
Nov 10, 2005
epl1
Today at 2:12 AM
find number of elements in H
Darealzolt   0
Today at 1:50 AM
If \( H \) is the set of positive real solutions to the system
\[
x^3 + y^3 + z^3 = x + y + z
\]\[
x^2 + y^2 + z^2 = xyz
\]then find the number of elements in \( H \).
0 replies
Darealzolt
Today at 1:50 AM
0 replies
old problem from an open contest
Darealzolt   0
Today at 1:41 AM
Given that $a, b \in \mathbb{R}$ satisfy
\[
a + \frac{1}{a + 2015} = b - 4030 + \frac{1}{b - 2015}
\]and $|a - b| > 5000$. Determine the value of
\[
\frac{ab}{2015} - a + b.
\]
0 replies
Darealzolt
Today at 1:41 AM
0 replies
GCD of a sequence
oVlad   7
N Apr 21, 2025 by grupyorum
Source: Romania EGMO TST 2017 Day 1 P2
Determine all pairs $(a,b)$ of positive integers with the following property: all of the terms of the sequence $(a^n+b^n+1)_{n\geqslant 1}$ have a greatest common divisor $d>1.$
7 replies
oVlad
Apr 21, 2025
grupyorum
Apr 21, 2025
GCD of a sequence
G H J
G H BBookmark kLocked kLocked NReply
Source: Romania EGMO TST 2017 Day 1 P2
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oVlad
1742 posts
#1
Y by
Determine all pairs $(a,b)$ of positive integers with the following property: all of the terms of the sequence $(a^n+b^n+1)_{n\geqslant 1}$ have a greatest common divisor $d>1.$
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kokcio
69 posts
#2
Y by
Assume that $d$ divides $a+b+1, a^2+b^2+1, a^3+b^3+1$. Then we know that $ab=\frac{(a+b)^2-(a^2+b^2)}{2}\equiv1\mod d$, so $a^3+b^3=(a+b)(a^2-ab+b^2)\equiv2\mod d$, so $d$ divides $2+1=3$, which means $d=3$, because we want $d>1$. But if $3$ divides $a+b+1$ and $a^2+b^2+1$, then we have to have that $a,b\equiv1\mod3$, and then obciosuly $3$ divides $a^n+b^n+1$ for all $n$. Hence, the answer is $(a,b)=(3k+1,3l+1)$ for some non-negative integers $k,l$.
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Rohit-2006
237 posts
#3
Y by
No you are wrong.....first of all if the parity of $k$ and $l$ are same then obviously your answer is wrong. Actually the answer would be $(a,b)=(1,1),(2k+1,2l),(2k,2l+1)$ I guess. Trying to prove.....
Z K Y
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Rohit-2006
237 posts
#4
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Now it can be solved easily.....$a=b$ case is trivial which gives us $a=b=1$....and the rest as follows
Attachments:
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kokcio
69 posts
#5
Y by
I forgot about $2$.
If $a+b$ is odd then one of this numbers has to be even and one odd and then this pair works in this problem ($d=2$ or $d=6$).
Therefore, pairs which satisfy conditions of this problem are $(2k,2l+1),(2k+1,2l),(3k+1,3l+1)$.
We will have in this cases $d\in\{2,3,6\}$.
Z K Y
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kokcio
69 posts
#6
Y by
Rohit-2006 wrote:
Now it can be solved easily.....$a=b$ case is trivial which gives us $a=b=1$....and the rest as follows

This is incorrect. If $a=b$, then we can have $a=b=3k+1$ and $d=3$.
Z K Y
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Rohit-2006
237 posts
#7
Y by
Oh yeah right.....wait trying tomorrow....though instead of the equality case....you can easily see that $d$ doesn't necessarily divide that part....
This post has been edited 1 time. Last edited by Rohit-2006, Apr 21, 2025, 7:11 PM
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grupyorum
1417 posts
#8
Y by
Let $p$ be a prime such that $p\mid a^n+b^n+1$ for $n\ge 1$. Using
\[
a^3+b^3+1 - 3ab = (a+b+1)(a^2+b^2+1-ab-a-b)
\]we get $p\mid 3ab$. Suppose first that $p\mid a$ (the case $p\mid b$ is symmetric). Then, $p\mid a+b+1$ forces $b\equiv -1\pmod{p}$ which together with $p\mid a^2+b^2+1\equiv 2\pmod{p}$ forces $p=2$. So, $(a,b)=(2k,2\ell-1),(2\ell-1,2k)$, where $k,\ell\ge 1$ are arbitrary are both solution families.

Assume now $p\nmid ab$. Then $p=3$ and $a+b\equiv 2\pmod{3}$. Since $3\nmid ab$, we must have $a\equiv b\equiv 1\pmod{3}$, giving the family $(3k+1,3\ell+1)$ where $k,\ell\ge 0$ are arbitrary.
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