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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

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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
Need help with barycentric
Sadigly   0
13 minutes ago
Hi,is there a good handout/book that explains barycentric,other than EGMO?
0 replies
Sadigly
13 minutes ago
0 replies
Combinatorics
P162008   3
N 36 minutes ago by P162008
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
3 replies
P162008
4 hours ago
P162008
36 minutes ago
Find min and max
lgx57   0
an hour ago
Source: Own
$x_1,x_2, \cdots ,x_n\ge 0$,$\displaystyle\sum_{i=1}^n x_i=m$. $k_1,k_2,\cdots,k_n >0$. Find min and max of
$$\sum_{i=1}^n(k_i\prod_{j=1}^i x_j)$$
0 replies
lgx57
an hour ago
0 replies
Find min
lgx57   0
an hour ago
Source: Own
$a,b>0$, $a^4+ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
$a,b>0$, $a^4-ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
0 replies
lgx57
an hour ago
0 replies
No more topics!
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
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#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
169 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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#5
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Thank you so much
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GioOrnikapa
76 posts
#6
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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
113 posts
#8
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solution
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Safal
169 posts
#9
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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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