It's February and we'd love to help you find the right course plan!

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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Feb 2, 2025
0 replies
Polynomials OG's
vicentev   1
N 16 minutes ago by pco
Source: TNO 2008 Senior
Consider the polynomial with real coefficients:
\[ p(x) = a_{2008}x^{2008} + a_{2007}x^{2007} + \dots + a_1x + a_0 \]and it is given that its coefficients satisfy:
\[ a_i + a_{i+1} = a_{i+2}, \quad i \in \{0,1,2,\dots,2006\} \]If $p(1) = 2008$ and $p(-1) = 0$, compute $a_{2008} - a_0$.
1 reply
vicentev
3 hours ago
pco
16 minutes ago
a round-robin football tournament
nhathhuyyp5c   1
N an hour ago by nhathhuyyp5c
In a round-robin football tournament with $n>4$ teams participating, each team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. At the end of the tournament, it is observed that all teams have the same total points.
$\text{a)}$ Prove that there exist 4 teams with the same number of wins, losses, and draws.
$\text{b)}$ Find the smallest positive integer $n$ such that there do not exist $5$ teams with the same number of wins, losses, and draws.




1 reply
nhathhuyyp5c
Yesterday at 2:16 PM
nhathhuyyp5c
an hour ago
Number theory
XAN4   2
N an hour ago by XAN4
Prove that for any positive integer $n$, if prime $p$ satisfies $p^3|n^2+n-1$, then $p^2\equiv1(mod\ 40)$.
2 replies
XAN4
Yesterday at 2:13 PM
XAN4
an hour ago
Fun inequality
DottedCaculator   50
N an hour ago by Juno_34
Source: 2022 Revenge ELSMO P5
Prove that $a^3 + b^3 + c^3 + abc +a^{3}b^{2}c^{-1}+a^{3}c^{2}b^{-1}+b^{3}a^{2}c^{-1}+b^{3}c^{2}a^{-1}+c^{3}a^{2}b^{-1}+c^{3}b^{2}a^{-1}+a^{5}b^{3}c^{-3}+ abc^{14} + 
a^{5}c^{3}b^{-3}+b^{5}a^{3}c^{-3}+b^{5}c^{3}a^{-3}+c^{5}a^{3}b^{-3}+c^{5}b^{3}a^{-3}+a^{6}b^{1}c^{-1}+a^{6}c^{1}b^{-1}+b^{6}a^{1}c^{-1}+b^{6}c^{1}a^{-1}+c^{6}a^{1}b^{-1}+c^{6}b^{1}a^{-1}+  a^{6}b^{4}c^{-3}+a^{6}c^{4}b^{-3}+b^{6}a^{4}c^{-3}+b^{6}c^{4}a^{-3}+c^{6}a^{4}b^{-3}+c^{6}b^{4}a^{-3}+a^{7}b^{2}c^{-1}+a^{7}c^{2}b^{-1}+b^{7}a^{2}c^{-1}+b^{7}c^{2}a^{-1}+c^{7}a^{2}b^{-1}+ abc + a^{14}bc + c^{7}b^{2}a^{-1}+a^{4}b^{1}c^{4}+a^{4}c^{1}b^{4}+b^{4}a^{1}c^{4}+b^{4}c^{1}a^{4}+c^{4}a^{1}b^{4}+c^{4}b^{1}a^{4}+a^{6}c^{4}+a^{6}b^{4}+b^{6}c^{4}+b^{6}a^{4}+c^{6}b^{4}+c^{6}a^{4}+a^{9}b^{6}c^{-4}+a^{9}c^{6}b^{-4}+ ab^{14}c + b^{9}a^{6}c^{-4}+b^{9}c^{6}a^{-4}+c^{9}a^{6}b^{-4}+ abc + c^{9}b^{6}a^{-4}+a^{12}b^{1}c^{-1}+a^{12}c^{1}b^{-1}+b^{12}a^{1}c^{-1}+b^{12}c^{1}a^{-1}+c^{12}a^{1}b^{-1}+ c^5 b^5 a^5 - c^5 b^5 a^2 + 3 c^5 b^5 - c^5 b^2 a^5 + c^5 b^2 a^2 - 3 c^5 b^2 + 3 c^5 a^5 - 3 c^5 a^2 + 9 c^5 - c^2 b^5 a^5 + c^2 b^5 a^2 - 3 c^2 b^5 + c^2 b^2 a^5 - c^2 b^2 a^2 + 3 c^2 b^2 - 3 c^2 a^5 + 3 c^2 a^2 - 9 c^2 + 3 b^5 a^5 - 3 b^5 a^2 + 9 b^5 - 3 b^2 a^5 + 3 b^2 a^2 - 9 b^2 + 9 a^5 - 9 a^2 + 27 +  c^{12}b^{1}a^{-1}+a^{13}b^{9}c^{-9}+a^{13}c^{9}b^{-9}+b^{13}a^{9}c^{-9}+b^{13}c^{9}a^{-9}+c^{13}a^{9}b^{-9}+c^{13}b^{9}a^{-9}+a^{12}b^{11}c^{-9}+a^{12}c^{11}b^{-9}+b^{12}a^{11}c^{-9}+b^{12}c^{11}a^{-9}+c^{12}a^{11}b^{-9}+c^{12}b^{11}a^{-9}+a^{8}b^{7}+a^{8}c^{7}+b^{8}a^{7}+b^{8}c^{7}+c^{8}a^{7}+c^{8}b^{7} + a^{16} + b^{16} + c^{16} + a^{16} + b^{16} + c^{16} + a^{16} + b^{16} + c^{16}\ge c^3 + 3 c^2 a + 3 c b^2 + 6 c b a + b^3 + 3 b^2 a  + a^3 + a^{1}c^{2}+a^{1}b^{2}+4b^{1}c^{2}+4b^{1}a^{2}+c^{1}b^{2}+4c^{1}a^{2}+a^{1}c^{3}+a^{1}b^{3}+b^{1}c^{3}+b^{1}a^{3}+c^{1}b^{3}+c^{1}a^{3}+a^{3}b^{2}+a^{3}c^{2}+b^{3}a^{2}+b^{3}c^{2}+c^{3}a^{2}+c^{3}b^{2}+a^{5}c^{1}+a^{5}b^{1}+b^{5}c^{1}+b^{5}a^{1}+c^{5}b^{1}+c^{5}a^{1}+a^{2}b^{1}c^{4}+a^{2}c^{1}b^{4}+b^{2}a^{1}c^{4}+b^{2}c^{1}a^{4}+c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}+a^{1}c^{7}+a^{1}b^{7}+b^{1}c^{7}+b^{1}a^{7}+c^{1}b^{7}+c^{1}a^{7}+a^{1}c^{8}+a^{1}b^{8}+b^{1}c^{8}+b^{1}a^{8}+c^{1}b^{8}+c^{1}a^{8}+a^{5}b^{1}c^{4}+a^{5}c^{1}b^{4}+b^{5}a^{1}c^{4}+b^{5}c^{1}a^{4}+c^{5}a^{1}b^{4}+c^{5}b^{1}a^{4}+a^{2}b^{1}c^{8}+a^{2}c^{1}b^{8}+b^{2}a^{1}c^{8}+b^{2}c^{1}a^{8}+c^{2}a^{1}b^{8}+c^{2}b^{1}a^{8}+a^{1}c^{11}+a^{1}b^{11}+b^{1}c^{11}+b^{1}a^{11}+c^{1}b^{11}+c^{1}a^{11}+a^{6}b^{2}c^{5}+a^{6}c^{2}b^{5}+b^{6}a^{2}c^{5}+b^{6}c^{2}a^{5}+c^{6}a^{2}b^{5}+c^{6}b^{2}a^{5}+a^{3}b^{2}c^{9}+a^{3}c^{2}b^{9}+b^{3}a^{2}c^{9}+b^{3}c^{2}a^{9}+c^{3}a^{2}b^{9}+c^{3}b^{2}a^{9}+a^{3}b^{1}c^{11}+a^{3}c^{1}b^{11}+b^{3}a^{1}c^{11}+b^{3}c^{1}a^{11}+c^{3}a^{1}b^{11}+c^{3}b^{1}a^{11} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15}+c^{2}a^{1}b^{4}+c^{2}b^{1}a^{4}+a^{1}c^{7}+a^{1}b^{7}+b^{1}c^{7}+b^{1}a^{7}+c^{1}b^{7}+c^{1}a^{7}+a^{1}c^{8}+a^{1}b^{8}+b^{1}c^{8}+b^{1}a^{8}+c^{1}b^{8}+c^{1}a^{8}+a^{5}b^{1}c^{4}+a^{5}c^{1}b^{4}+b^{5}a^{1}c^{4}+b^{5}c^{1}a^{4}+c^{5}a^{1}b^{4}+c^{5}b^{1}a^{4}+a^{2}b^{1}c^{8}+a^{2}c^{1}b^{8}+b^{2}a^{1}c^{8}+b^{2}c^{1}a^{8}+c^{2}a^{1}b^{8}+c^{2}b^{1}a^{8}+a^{1}c^{11}+a^{1}b^{11}+b^{1}c^{11}+b^{1}a^{11}+c^{1}b^{11}+c^{1}a^{11}+a^{6}b^{2}c^{5}+a^{6}c^{2}b^{5}+b^{6}a^{2}c^{5}+b^{6}c^{2}a^{5}+c^{6}a^{2}b^{5}+c^{6}b^{2}a^{5}+a^{3}b^{2}c^{9}+a^{3}c^{2}b^{9}+b^{3}a^{2}c^{9}+b^{3}c^{2}a^{9}+c^{3}a^{2}b^{9}+c^{3}b^{2}a^{9}+a^{3}b^{1}c^{11}+a^{3}c^{1}b^{11}+b^{3}a^{1}c^{11}+b^{3}c^{1}a^{11}+c^{3}a^{1}b^{11}+c^{3}b^{1}a^{11} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15} + a^{15}b + ab^{15} + a^{15}c + ac^{15} + b^{15}c + bc^{15}$ for all $a,b,c\in\mathbb R^+$.

Proposed by Henry Jiang and C++
50 replies
DottedCaculator
Jul 11, 2022
Juno_34
an hour ago
No more topics!
Proth's theorem
puuhikki   2
N Nov 12, 2013 by Pirer
How to prove: Let $ n = h\cdot 2^k + 1, h < 2^k, 2\nmid h$. Then $ n$ is prime if and only if $ a^ {(n - 1)/2}\equiv - 1\pmod n$ for some integer $ a$.
2 replies
puuhikki
Jan 22, 2008
Pirer
Nov 12, 2013
Proth's theorem
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puuhikki
979 posts
#1 • 2 Y
Y by Adventure10, Mango247
How to prove: Let $ n = h\cdot 2^k + 1, h < 2^k, 2\nmid h$. Then $ n$ is prime if and only if $ a^ {(n - 1)/2}\equiv - 1\pmod n$ for some integer $ a$.
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scorpius119
1677 posts
#2 • 2 Y
Y by Adventure10, Mango247
If $ n$ is prime, then any nonsquare $ a$ will satisfy $ a^\frac{n-1}{2}\equiv -1\pmod{n}$. It remains to show the converse.

If $ a^\frac{n-1}{2}\equiv -1\pmod{n}$, then let $ p$ be a prime divisor of $ n$. We must have $ a^\frac{n-1}{2}\equiv -1\pmod{p}$ so the order of $ a$ modulo $ p$ must divide $ 2^k$ and be divisible by $ p-1$. This gives $ p\equiv 1\pmod{2^k}$.

Now $ p$ is of the form $ 1+a\cdot 2^k$. Also, $ \frac{n}{p}\equiv 1\pmod{2^k}$, so we have $ n=(1+a\cdot 2^k)(1+b\cdot 2^k)$ for some $ b$.

This means $ h=a+b+ab\cdot 2^k$. If $ b=0$, then $ n=p$, a prime. Suppose otherwise. Then $ b\geq 1$, and we cannot have $ a=b=1$ since $ h$ is odd. This means
\[ a+b\geq 3,ab\geq 2\Rightarrow h\geq 3+2^{k+1}\]
an absurd contradiction for $ h<2^k$. So $ n$ prime.
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Pirer
1 post
#3 • 1 Y
Y by Adventure10
I think that there is a mistake in the proof:

When you have $a^{\frac{n-1}{2}} \equiv -1\pmod{p}$, you can deduce that
$\text{ord}_p(a)$ divides $n-1 = h2^k$, and
$\text{ord}_p(a)$ divides $p - 1$.

Nevertheless, you can't deduce than $\text{ord}_p(a)$ is divided by $p-1$, so you can't deduce that $p-1$ divides $\text{ord}_p(a)$ and, thus, divides $h2^k$.

I think that to aviod it we can deduce $\text{ord}_p(a) = 2^\alpha$, for any $\alpha \in {0,\ldots, k}$ and then write $p = a\cdot 2^\alpha +1$. If so,

\[N/p \equiv 1/(-1) \equiv -1 \pmod{2^\alpha} \Rightarrow N/p = b\cdot 2^\alpha +1\]

Then,

\[ h\cdot 2^k = N = (N/p)\cdot p = (a\cdot 2^\alpha + 1)\cdot (b\cdot 2^\alpha + 1) \]

I suppose that I should try to use that $h < 2^k$ so $h\cdot 2^k < 2^{2k}$, but I don't know how to continue...


PD. I know that this is an old post, but I'm trying to find a correct proof of Proth's Theorem and I can't find it anywhere!
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