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Questions tagged [squarefree-numbers]

For questions about integers divisible by no square number other than $1$ (squarefree-numbers).

Squarefree numbers are integers divisible by no square number other than $1$ aka an integer which prime decomposition contains no repeated factors. An integer which prime decomposition contains at least $1$ square number is called a "powerful number" or a "squareful number".

The first smallest few squarefree numbers are $1$, $2$, $3$, $5$, $6$, $7$, $10$, $11$, $13$, $14$, ... ($\text{OEIS A}005117$).

Read more about squarefree numbers and their properties here.

15 questions
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Is $\frac{(p-1)^p+1}{p^2}$ square-free?

Is $\frac{(p-1)^p+1}{p^2}$ squarefree for all primes $p \geq 7$? I did some small testing and it seems to hold up to $p \leq 47$. Also, note that the above expression is indeed an integer by binomial expansion.
FAN WENDI HCI
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6
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How many number square-free integer from 1 to 2013

Question: Let $Q(x)$ denote the number of square-free (quadratfrei) integers between $1$ and $x$ find $Q(2013)=?$ My try: I know $ 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33, 34, 35, 37, 38, 39,\cdots $ Square-free…
math110
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Square-free numbers sum Inequality

Let $n$ be a positive integer, and let $k$ be a fixed positive integer. Consider all integers $N_1, N_2, \dots, N_r$ less than or equal to $n$ that are square-free and have exactly $2k$ distinct prime factors. For each $N_j$, choose a factorization…
Juan Moreno
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A tight positivity conjecture about sums over divisors of square-free integers.

Let $p_n$ be the $n$th prime number and all variables, unless otherwise specified, are natural numbers. Conjecture: For all square-free $n \geq 2$, the following function evaluates to a positive integer: $$F(n) = -1 + \sum_{d \mid n} \mu(d)…
Daniel Donnelly
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3
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Partial sum of square-free integers

I have been unable to find anywhere a sharp bounding for the partial sum of the first $n$ square-free integers $\sum_{k=1}^{n} k$, where $k$ runs over the square-free integers. I guess it should be around $\frac{\pi^2}{6}…
Juan Moreno
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Conjecture about non-squarefree numbers.

Conjecture : Let $k$ be a non-negative integer. Then , there is a positive integer $j$ , such that $10^j+k$ is NOT squarefree. The conjecture is true upto $k=10^6$. We can always find a positive integer $j\le 421$ as desired. If we demand that…
Peter
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Which integers $N$ satisfy $N\mid m^2\Rightarrow N\mid m,\,$ for all integers $m$?

I'm looking for a set of conditions and maybe a proof of said conditions for the thought proposed in the title. It seems to me that what was stated in the title always is true when N is not a perfect square and N < m, but I can find instances that…
ifangy
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$\frac{a}{p}-\frac{b}{p^2}=-1$, $b-a$ is multiplication of prime numbers with exponent $1$

I was calculating following for some numbers and I found this is true for a lot of examples. Let $p\geq 3$ be a prime number such that $p|a$ and $p^2|b$ and following equation is true: $$\frac{a}{p}-\frac{b}{p^2}=-1$$ Now we write…
MAN-MADE
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Squarefreeness Mersenne Numbers

In this page of Caldwell, after the proof of the link with Wieferich primes and pointing out that the two known ones can't be divisors of any Mersenne numbers, in the Comment he asserts «... so $M_q$ is square-free for all primes less than $4 \cdot…
Falcon
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There exist two square-free numbers, each with a given amount of factors, that differ by a given amount.

Let $m,n,d$ all be positive integers with $m+n>2.$ Proposition: There exist prime numbers, $p_1,p_2,\ldots,p_m;\ \ q_1, q_2,\ldots,q_n,$ all distinct, such that $\displaystyle\prod_{i=1}^{m} p_i = \prod_{i=1}^{n} q_i + d. $ Other than obvious…
Adam Rubinson
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Existence of square free factorisation.

A polynomial $f(x)\in \mathbb{Z}_q[x]$ is said to be square-free if it has no repeated roots. The square-free factorisation of $f$ is $$f(x)=\prod_{i=1}^k f_i(x)^i,$$ where each $f_i(x)$ is square free polynomial and $\gcd(f_i(x), f_j(x))$ is $1$…
PAMG
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Question about Carmichael-factorials

Maybe , this conept is known , maybe not : If we define $C_n$ to be the $n$-th Carmichael-number and $$F(n):=\prod_{j=1}^n C_j$$ could be called "Carmichael-factorial". For large $n$ , $F(n)$ has many small factors , so the numbers $F(n)-2$ and…
Peter
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Form of prime, $p$ when $p+1$ is square-free

Today, to the stack community, I wish to share gist of observations I took in a past few days in specific problems from elementary number theory. These observations might be trivial to prove, and it is likely that I might be missing something to…
Aziz
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When is $\frac{5^n - 1}{4}$ squarefree?

While researching the topic of odd perfect numbers, I came across the following related subproblem: PROBLEM: Determine congruence conditions (on $n > 1$) such that $$\frac{5^n - 1}{4}$$ is squarefree. MY ATTEMPT Set $$m_n := \frac{5^n - 1}{4}.$$ I…
Jose Arnaldo Bebita Dris
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Estimating the number of square-free integers

Let $Q(x)$ denote the number of square-free integers not exceeding $x$. I tried to write $Q(x)$ as $$ Q(x) = \sum_{n \leq x} \mu(n)^2 $$ but I don't know how to manipulate this in any useful way. I know that $\sum_{d | n} \mu(d)^2 = 2^{\omega(n)}$,…
MAZ
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