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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
Nice "if and only if" function problem
ICE_CNME_4   0
19 minutes ago
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!
0 replies
ICE_CNME_4
19 minutes ago
0 replies
4 variables
Nguyenhuyen_AG   11
N 23 minutes ago by arqady
Let $a,\,b,\,c,\,d$ are non-negative real numbers and $0 \leqslant k \leqslant \frac{2}{\sqrt{3}}.$ Prove that
$$a^2+b^2+c^2+d^2+kabcd \geqslant k+4+(k+2)(a+b+c+d-4).$$hide
11 replies
Nguyenhuyen_AG
Dec 21, 2020
arqady
23 minutes ago
Hard Number Theory
MuradSafarli   0
37 minutes ago
Find all odd natural numbers \( m \) such that \( m^2 - 1 \mid 3^m + 5^m \).
0 replies
MuradSafarli
37 minutes ago
0 replies
A and B take stones from a pile
WakeUp   4
N an hour ago by reni_wee
Source: CentroAmerican 2003
Two players $A$ and $B$ take turns playing the following game: There is a pile of $2003$ stones. In his first turn, $A$ selects a divisor of $2003$ and removes this number of stones from the pile. $B$ then chooses a divisor of the number of remaining stones, and removes that number of stones from the new pile, and so on. The player who has to remove the last stone loses. Show that one of the two players has a winning strategy and describe the strategy.
4 replies
WakeUp
Nov 29, 2010
reni_wee
an hour ago
Inequalities
sqing   3
N 6 hours ago by bogpt
Let $ a,b,c\geq 0 $ and $ab+bc+ca =1.$ Prove that
$$(a^2+b^2+c^2)(a+b+c-2)\ge 8abc(1-a-b-c) $$$$(a^2+b^2+c^2)(a+b+c-\frac{5}{2})\ge 2abc(1-a-b-c) $$
3 replies
sqing
Yesterday at 2:26 PM
bogpt
6 hours ago
Inequalities
sqing   1
N Today at 1:35 PM by sqing
Let $ a,b,c $ be real numbers . Prove that
$$\frac{(a-1)(b-1)(c-1)}{(a^2+3)(b^2+1)(c^2+3)} \ge -\frac{1+\sqrt 2}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+1)(b^2+3)(c^2+1)} \ge -\frac{3+2\sqrt{2}}{8}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+3)(b^2+2)(c^2+3)} \ge -\frac{1+\sqrt3}{16}$$$$\frac{(a-1)(b-1)(c-1)}{(a^2+2)(b^2+3)(c^2+2)} \ge -\frac{2+\sqrt{3}}{16}$$
1 reply
sqing
Today at 1:11 PM
sqing
Today at 1:35 PM
[PMO25 Qualifying II.8] A Square Can't Be A Floor
kae_3   2
N Today at 12:36 PM by tapilyoca
Determine the largest perfect square less than $1000$ that cannot be expressed as $\lfloor x\rfloor + \lfloor 2x\rfloor + \lfloor 3x\rfloor + \lfloor 6x\rfloor$ for some positive real number $x$.

Answer Confirmation
2 replies
kae_3
Feb 21, 2025
tapilyoca
Today at 12:36 PM
solve in R
zolfmark   1
N Today at 12:10 PM by Mathzeus1024
x+1/y=9/2 and y+1/z=11/4 and z+1/x=12/5
1 reply
zolfmark
Feb 16, 2016
Mathzeus1024
Today at 12:10 PM
not so hard problem (own)
BinariouslyRandom   1
N Today at 11:38 AM by BinariouslyRandom
An open rectangular box is made from exactly $900 m^2$ of cardboard, with 5 rectangular faces instead of 6 and there is no leftover cardboard. What is the largest volume this box can hold given that $100 m^2$ of cardboard is needed to cover the box?
1 reply
BinariouslyRandom
Today at 11:37 AM
BinariouslyRandom
Today at 11:38 AM
Minimize z.
Entrepreneur   3
N Today at 11:17 AM by no_room_for_error
Minimize $z = 6x + 3y.$ Subject to the constraints:
$$\begin{cases}
4x+y\ge80,\\
x+5y \ge115,\\
3x+2y\le150,\\
x,y\ge0.
\end{cases}$$
3 replies
Entrepreneur
May 21, 2025
no_room_for_error
Today at 11:17 AM
Inequalities
sqing   18
N Today at 7:48 AM by sqing
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc +  abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{2}{5}abc  \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc  + \frac{7}{20}abc  \leq \frac{2381411}{26460}$$
18 replies
sqing
May 21, 2025
sqing
Today at 7:48 AM
Inequalities
sqing   18
N Today at 7:20 AM by sqing
Let $ a,b>0   $ . Prove that
$$ \frac{a}{a^2+a +2b+1}+ \frac{b}{b^2+2a +b+1}  \leq  \frac{2}{5} $$$$ \frac{a}{a^2+2a +b+1}+ \frac{b}{b^2+a +2b+1}  \leq  \frac{2}{5} $$
18 replies
sqing
May 13, 2025
sqing
Today at 7:20 AM
helpppppppp me
stupid_boiii   1
N Today at 6:56 AM by vanstraelen
Given triangle ABC. The tangent at ? to the circumcircle(ABC) intersects line BC at point T. Points D,E satisfy AD=BD, AE=CE, and ∠CBD=∠BCE<90 ∘ . Prove that D,E,T are collinear.
1 reply
stupid_boiii
Yesterday at 4:22 AM
vanstraelen
Today at 6:56 AM
Algebra Polynomials
Foxellar   2
N Today at 5:43 AM by Foxellar
The real root of the polynomial \( p(x) = 8x^3 - 3x^2 - 3x - 1 \) can be written in the form
\[
\frac{\sqrt[3]{a} + \sqrt[3]{b} + 1}{c},
\]where \( a, b, \) and \( c \) are positive integers. Find the value of \( a + b + c \).
2 replies
Foxellar
Today at 4:52 AM
Foxellar
Today at 5:43 AM
Problem 5
SlovEcience   6
N Apr 11, 2025 by Safal
Let \( n > 3 \) be an odd integer. Prove that there exists a prime number \( p \) such that
\[
p \mid 2^{\varphi(n)} - 1 \quad \text{but} \quad p \nmid n.
\]
6 replies
SlovEcience
Apr 10, 2025
Safal
Apr 11, 2025
Problem 5
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SlovEcience
14 posts
#1 • 1 Y
Y by PikaPika999
Let \( n > 3 \) be an odd integer. Prove that there exists a prime number \( p \) such that
\[
p \mid 2^{\varphi(n)} - 1 \quad \text{but} \quad p \nmid n.
\]
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Safal
170 posts
#2 • 1 Y
Y by PikaPika999
Solution
This post has been edited 4 times. Last edited by Safal, Apr 11, 2025, 11:03 AM
Reason: Typo
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SlovEcience
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Thank you so much
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GioOrnikapa
77 posts
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Theorem bash. Assume the contrary. Let $t$ be the minimal positive integer for which \[ n \mid 2^t - 1. \]Assume that $t < \varphi(n)$. Then, using Zsigmondy's theorem, there exists a prime number $p$ which divides $2^{\varphi(n)}-1$ and does not divide $2^t-1$ (checking exceptions manually leads to $\varphi(n) = 6$ case which itself implies $n=7,9,14,18$ and $p=7,3$ works). If \[ p \mid n \implies p \mid n \mid 2^t-1, \]contradiction.
We are left with the case when $t = \varphi(n)$. It means that $2$ is a primitive root mod $n$ and since $n$ has a primitive root and is odd, we have $n = p^k$ for some prime $p$ and a positive integer $k$. We are down to solving \[ 2^{\varphi(p^k)}-1=p^{\ell} \]which we can just finish off using Mihailescu's theorem and thus we are done. $\square$
This post has been edited 1 time. Last edited by GioOrnikapa, Apr 11, 2025, 8:44 AM
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habcy12345
44 posts
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We divide the problem into two cases:
1. $\varphi(n)\ne 6$: In this case, by Zsigmondy's Theorem, there exists a prime divisor $p_0$ of $2^{\varphi(n)} - 1$ such that ord$_{p_0}(2) = \varphi(n)$. Then, by Fermat's Little Theorem, we have $\varphi(n) \mid p_0 - 1$.
Consider the case where $p_0$ is a divisor of $n$. Then clearly we also have $(p_0 - 1) \mid \varphi(n)$, so $p_0 - 1 = \varphi(n)$. Therefore, $n = p_0$, because if $n \ne p_0$, set $n = p_0^x \cdot A$ with gcd$(A, p_0) = 1$, then we immediately get $p_0 - 1 = p_0^{x-1}(p_0 - 1)\varphi(A) > p_0 - 1$ since $A \ne 2$, which is a contradiction.
From that, we see that 3 is a prime number satisfying the problem since $3 \mid 2^{\varphi(n)} - 1 = 2^{p_0 - 1} - 1$ (as $p_0$ is odd because $p_0 = n > 3$) and $3 \nmid n = p_0$. If $p_0$ is not a divisor of $n$, then just choose $p = p_0$, which satisfies the requirement.
2. $\varphi(n) = 6$: It is easy to check that the numbers satisfying $\varphi(n) = 6$ are $\{7, 9, 14, 18\}$. For $n = 7, 14$, choose $p = 3$; for $n = 9, 18$, choose $p = 7$, which satisfy the problem.
Therefore, the problem is proven in all cases. QED.
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SimplisticFormulas
121 posts
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solution
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Safal
170 posts
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SimplisticFormulas wrote:
solution

Claim is not true take $m=2$ then $2m+1=5$ and $\varphi(2m+1)=4$ Thus $2^{\frac{\varphi(2m+1)}{2}}+1=2^2+1=5$ thus the $\gcd(2m+1,2^{\frac{\varphi(2m+1)}{2}}+1)=\gcd(5,5)=5$
This post has been edited 1 time. Last edited by Safal, Apr 11, 2025, 12:53 PM
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