View source for Square-free integer

{{Short description|Number without repeated prime factors}}
[[File:Composite number Cuisenaire rods 10.svg|thumb|10 is square-free, as its divisors greater than 1 are 2, 5, and 10, none of which is square (the first few squares being 1, 4, 9, and 16)]]
[[File:squarefree_numbers_sieve.svg|thumb|Square-free integers up to 120 remain after eliminating multiples of squares of primes up to √120]]
In [[Mathematics|mathematics]], a '''square-free integer''' (or '''squarefree integer''') is an [[Integer|integer]] which is [[Divisor|divisible]] by no [[Square number|square number]] other than 1. That is, its prime factorization has exactly one factor for each prime that appears in it. For example, {{nowrap|1=10 = 2 ⋅ 5}} is square-free, but {{nowrap|1=18 = 2 ⋅ 3 ⋅ 3}} is not, because 18 is divisible by {{nowrap|1=9 = 3<sup>2</sup>}}. The smallest positive square-free numbers are
{{bi|left=1.6|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, ... {{OEIS|id=A005117}}}}

==Square-free factorization==
Every positive integer <math>n</math> can be factored in a unique way as
<math display=block>n=\prod_{i=1}^k q_i^i,</math>
where the <math>q_i</math> different from one are square-free integers that are pairwise coprime.
This is called the ''square-free factorization'' of {{mvar|n}}.

To construct the square-free factorization, let
<math display=block>n=\prod_{j=1}^h p_j^{e_j}</math> 
be the prime factorization of <math>n</math>, where the <math>p_j</math> are distinct [[Prime number|prime number]]s. Then the factors of the square-free factorization are defined as 
<math display=block>q_i=\prod_{j: e_j=i}p_j.</math>

An integer is square-free if and only if <math>q_i=1</math> for all <math>i > 1</math>. An integer greater than one is the <math>k</math>th power of another integer if and only if <math>k</math> is a divisor of all <math>i</math> such that <math>q_i\neq 1.</math>

The use of the square-free factorization of integers is limited by the fact that its computation is as difficult as the computation of the prime factorization. More precisely every known [[Algorithm|algorithm]] for computing a square-free factorization computes also the prime factorization. This is a notable difference with the case of [[Polynomial|polynomial]]s for which the same definitions can be given, but, in this case, the square-free factorization is not only easier to compute than the complete factorization, but it is the first step of all standard factorization algorithms.

==Square-free factors of integers==
The radical of an integer is its largest square-free factor, that is <math>\textstyle \prod_{i=1}^k q_i</math> with notation of the preceding section. An integer is square-free [[If and only if|if and only if]] it is equal to its radical.

Every positive integer <math>n</math> can be represented in a unique way as the product of a [[Powerful number|powerful number]] (that is an integer such that is divisible by the square of every prime factor) and a square-free integer, which are coprime. In this factorization, the square-free factor is <math>q_1,</math> and the powerful number is <math>\textstyle \prod_{i=2}^k q_i^i.</math>

The ''square-free part'' of <math>n</math> is <math>q_1,</math> which is the largest square-free divisor <math>k</math> of <math>n</math> that is coprime with <math>n/k</math>. The square-free part of an integer may be smaller than the largest square-free divisor, which is <math>\textstyle \prod_{i=1}^k q_i.</math>

Any arbitrary positive integer <math>n</math> can be represented in a unique way as the product of a [[Square|square]] and a square-free integer:
<math display=block> n=m^2 k</math>
In this factorization, <math>m</math> is the largest divisor of <math>n</math> such that <math>m^2</math> is a divisor of <math>n</math>.

In summary, there are three square-free factors that are naturally associated to every integer: the square-free part, the above factor <math>k</math>, and the largest square-free factor. Each is a factor of the next one. All are easily deduced from the prime factorization or the square-free factorization: if
<math display=block>n=\prod_{i=1}^h p_i^{e_i}=\prod_{i=1}^k q_i^i</math>
are the prime factorization and the square-free factorization of <math>n</math>, where <math>p_1, \ldots, p_h</math> are distinct prime numbers, then the square-free part is
<math display=block>\prod_{e_i=1} p_i =q_1,</math>
The square-free factor such the quotient is a square is 
<math display=block>\prod_{e_i \text{ odd}} p_i=\prod_{i \text{ odd}} q_i,</math>
and the largest square-free factor is 
<math display=block>\prod_{i=1}^h p_i=\prod_{i=1}^k q_i.</math>

For example, if <math>n=75600=2^4\cdot 3^3\cdot 5^2\cdot 7,</math> one has <math>q_1=7,\; q_2=5,\;q_3=3,\;q_4=2.</math> The square-free part is {{math|7}}, the square-free factor such that the quotient is a square is {{math|1=3 ⋅ 7 = 21}}, and the largest square-free factor is {{math|1=2 ⋅ 3 ⋅ 5 ⋅ 7 = 210}}.

No algorithm is known for computing any of these square-free factors which is faster than computing the complete prime factorization. In particular, there is no known polynomial-time algorithm for computing the square-free part of an integer, or even for [[Decision problem|determining]] whether an integer is square-free.<ref>{{cite conference
 | last1 = Adleman | first1 = Leonard M.
 | last2 = McCurley | first2 = Kevin S.
 | editor1-last = Adleman | editor1-first = Leonard M.
 | editor2-last = Huang | editor2-first = Ming-Deh A.
 | contribution = Open problems in number theoretic complexity, II
 | doi = 10.1007/3-540-58691-1_70
 | pages = 291–322
 | publisher = Springer
 | series = Lecture Notes in Computer Science
 | title = Algorithmic Number Theory, First International Symposium, ANTS-I, Ithaca, NY, USA, May 6–9, 1994, Proceedings
 | volume = 877
 | year = 1994}}</ref>  In contrast, polynomial-time algorithms are known for primality testing.<ref>{{Cite journal|last1=Agrawal|first1=Manindra|last2=Kayal|first2=Neeraj|last3=Saxena|first3=Nitin|date=1 September 2004|title=PRIMES is in P|journal=Annals of Mathematics|language=en-US|volume=160|issue=2|pages=781–793|doi=10.4007/annals.2004.160.781|doi-access=free|issn=0003-486X|mr=2123939|zbl=1071.11070|url=https://www.cse.iitk.ac.in/users/manindra/algebra/primality_original.pdf}}</ref> This is a major difference between the arithmetic of the integers, and the arithmetic of the univariate polynomials, as polynomial-time algorithms are known for square-free factorization of polynomials (in short, the largest square-free factor of a polynomial is its quotient by the [[Polynomial greatest common divisor|greatest common divisor]] of the polynomial and its [[Formal derivative|formal derivative]]).<ref>{{cite thesis |last=Richards |first=Chelsea |title=Algorithms for factoring square-free polynomials over finite fields |type=Master's thesis |publisher=Simon Fraser University |location=Canada |year=2009 |url=http://www.cecm.sfu.ca/CAG/theses/chelsea.pdf}}</ref>

==Equivalent characterizations==
A positive integer <math>n</math> is square-free if and only if in the prime factorization of <math>n</math>, no prime factor occurs with an exponent larger than one. Another way of stating the same is that for every prime [[Divisor|factor]] <math>p</math> of <math>n</math>, the prime <math>p</math> does not evenly divide&nbsp;<math>n/p</math>. Also <math>n</math> is square-free if and only if in every factorization <math>n=ab</math>, the factors <math>a</math> and <math>b</math> are coprime.  An immediate result of this definition is that all prime numbers are square-free.

A positive integer <math>n</math> is square-free if and only if all [[Abelian group|abelian group]]s of [[Order (group theory)|order]] <math>n</math> are [[Group isomorphism|isomorphic]], which is the case if and only if any such group is [[Cyclic group|cyclic]]. This follows from the classification of [[Finitely generated abelian group|finitely generated abelian group]]s.

A integer <math>n</math> is square-free if and only if the factor ring <math>\mathbb{Z}/n\mathbb{Z}</math> (see [[Modular arithmetic|modular arithmetic]]) is a [[Product of rings|product]] of [[Field (mathematics)|field]]s. This follows from the [[Chinese remainder theorem]] and the fact that a ring of the form <math>\mathbb{Z}/k\mathbb{Z}</math> is a field if and only if <math>k</math> is prime.

For every positive integer <math>n</math>, the set of all positive divisors of <math>n</math> becomes a [[Partially ordered set|partially ordered set]] if we use [[Divisor|divisibility]] as the order relation. This partially ordered set is always a [[Distributive lattice|distributive lattice]]. It is a [[Boolean algebra (structure)|Boolean algebra]] if and only if <math>n</math> is square-free.

A positive integer <math>n</math> is square-free [[If and only if|if and only if]] <math>\mu(n)\ne 0</math>, where <math>\mu</math> denotes the Möbius function.

==Dirichlet series==
The absolute value of the Möbius function is the [[Indicator function|indicator function]] for the square-free integers &ndash; that is, {{math|{{mabs|''&mu;''(''n'')}}}} is equal to 1 if {{mvar|n}} is square-free, and 0 if it is not.  The [[Dirichlet series]] of this indicator function is
:<math>\sum_{n=1}^{\infty}\frac{|\mu(n)|}{n^{s}} = \frac{\zeta(s)}{\zeta(2s)},</math>
where {{math|''ζ''(''s'')}} is the [[Riemann zeta function]].  This follows from the [[Euler product]]
:<math>  \frac{\zeta(s)}{\zeta(2s) } =\prod_p \frac{(1-p^{-2s})}{(1-p^{-s})}=\prod_p (1+p^{-s}), </math>
where the products are taken over the prime numbers.

==Distribution==
Let ''Q''(''x'') denote the number of square-free integers between 1 and ''x'' ({{OEIS2C|A013928}} shifting index by 1). For large ''n'', 3/4 of the positive integers less than ''n'' are not divisible by 4, 8/9 of these numbers are not divisible by 9, and so on. Because these ratios satisfy the multiplicative property (this follows from [[Chinese remainder theorem]]), we obtain the approximation:

:<math>\begin{align}Q(x) &\approx x\prod_{p\ \text{prime}} \left(1-\frac{1}{p^2}\right) = x\prod_{p\ \text{prime}} \frac{1}{(1-\frac{1}{p^2})^{-1}}\\
&=x\prod_{p\ \text{prime}} \frac{1}{1+\frac{1}{p^2}+\frac{1}{p^4}+\cdots} = \frac{x}{\sum_{k=1}^\infty \frac{1}{k^2}} = \frac{x}{\zeta(2)} = \frac{6x}{\pi^2}.\end{align}</math>

This argument can be made rigorous for getting the estimate (using [[Big O notation|big O notation]])

:<math>Q(x) =  \frac{6x}{\pi^2} + O\left(\sqrt{x}\right).</math>

''Sketch of a proof:'' the above characterization gives
:<math>Q(x)=\sum_{n\leq x} \sum_{d^2\mid n} \mu(d)=\sum_{d\leq x} \mu(d)\sum_{n\leq x, d^2\mid n}1=\sum_{d\leq x} \mu(d)\left\lfloor\frac{x}{d^2}\right\rfloor;</math>
observing that the last summand is zero for <math>d>\sqrt{x}</math>, it results that
:<math>\begin{align}Q(x)&=\sum_{d\leq\sqrt{x}} \mu(d)\left\lfloor\frac{x}{d^2}\right\rfloor
=\sum_{d\leq\sqrt{x}} \frac{x\mu(d)}{d^2}+O\left(\sum_{d\leq\sqrt{x}} 1\right)
=x\sum_{d\leq\sqrt{x}} \frac{\mu(d)}{d^2}+O(\sqrt{x})\\
&=x\sum_{d} \frac{\mu(d)}{d^2}+O\left(x\sum_{d>\sqrt{x}}\frac{1}{d^2}+\sqrt{x}\right)
=\frac{x}{\zeta(2)}+O(\sqrt{x}).
\end{align}</math>

By exploiting the largest known zero-free region of the Riemann zeta function [[Biography:Arnold Walfisz|Arnold Walfisz]] improved the approximation to<ref>{{cite book |last1=Walfisz |first1=A. |title=Weylsche Exponentialsummen in der neueren Zahlentheorie |date=1963 |publisher=VEB Deutscher Verlag der Wissenschaften |location=Berlin}}</ref>
:<math>Q(x) = \frac{6x}{\pi^2} + O\left(x^{1/2}\exp\left(-c\frac{(\log x)^{3/5}}{(\log\log x)^{1/5}}\right)\right),</math>
for some positive constant {{math|''c''}}.

Under the [[Riemann hypothesis]], the error term can be reduced to<ref>Jia, Chao Hua. "The distribution of square-free numbers", ''Science in China Series A: Mathematics'' '''36''':2 (1993), pp. 154–169. Cited in Pappalardi 2003, [http://www.mat.uniroma3.it/users/pappa/papers/allahabad2003.pdf A Survey on ''k''-freeness]; also see Kaneenika Sinha, "[http://www.math.ualberta.ca/~kansinha/maxnrevfinal.pdf Average orders of certain arithmetical functions]", ''Journal of the Ramanujan Mathematical Society'' '''21''':3 (2006), pp. 267–277.</ref>
:<math>Q(x) = \frac{x}{\zeta(2)} + O\left(x^{17/54+\varepsilon}\right) = \frac{6}{\pi^2}x + O\left(x^{17/54+\varepsilon}\right).</math>

In 2015 the error term was further reduced to<ref>{{cite journal |last1=Liu |first1=H.-Q. |title=On the distribution of squarefree numbers |journal=Journal of Number Theory |date=2016 |volume=159 |pages=202–222|doi=10.1016/j.jnt.2015.07.013 |doi-access=free }}</ref>

:<math>Q(x) =  \frac{6}{\pi^2}x + O\left(x^{11/35+\varepsilon}\right).</math>

The asymptotic/[[Natural density|natural density]] of square-free numbers is therefore

:<math>\lim_{x\to\infty} \frac{Q(x)}{x} = \frac{6}{\pi^2}\approx 0.6079</math>

Therefore over 3/5 of the integers are square-free.

Likewise, if ''Q''(''x'',''n'') denotes the number of ''n''-free integers (e.g. 3-free integers being cube-free integers) between 1 and ''x'', one can show<ref>{{cite journal | first1 = E. H. | last1 = Linfoot |  first2 = C. J. A. | last2 = Evelyn | title = On a Problem in the Additive Theory of Numbers | journal = Mathematische Zeitschrift | date = 1929 | volume = 30 | pages = 443–448 | doi = 10.1007/BF01187781 | s2cid = 120604049 | url = https://gdz.sub.uni-goettingen.de/id/PPN266833020_0030?tify={%22pages%22:[437],%22view%22:%22info%22} }}</ref>
:<math>Q(x,n) = \frac{x}{\sum_{k=1}^\infty \frac{1}{k^n}} + O\left(\sqrt[n]{x}\right) = \frac{x}{\zeta(n)} + O\left(\sqrt[n]{x}\right).</math>

Since a multiple of 4 must have a square factor 4=2<sup>2</sup>, it cannot occur that four consecutive integers are all square-free. On the other hand, there exist infinitely many integers ''n'' for which 4''n'' +1, 4''n'' +2, 4''n'' +3 are all square-free. Otherwise, observing that 4''n'' and at least one of 4''n'' +1, 4''n'' +2, 4''n'' +3 among four could be non-square-free for sufficiently large ''n'', half of all positive integers minus finitely many must be non-square-free and therefore
:<math>Q(x) \leq \frac{x}{2}+C</math> for some constant ''C'',
contrary to the above asymptotic estimate for <math>Q(x)</math>.

There exist sequences of consecutive non-square-free integers of arbitrary length.  Indeed, if ''n'' satisfies a simultaneous congruence
:<math>n\equiv -i\pmod{p_i^2} \qquad (i=1, 2, \ldots, l)</math>
for distinct primes <math>p_1, p_2, \ldots, p_l</math>, then each <math>n+i</math> is divisible by ''p<sub>i </sub>''<sup>2</sup>.<ref>{{cite book | last1= Parent | first1=D. P. |title=Exercises in Number Theory | publisher=[[Physics:Springer-Verlag|Springer-Verlag]] New York | isbn=978-1-4757-5194-9 | url=https://www.springer.com/book/9780387960630 | doi=10.1007/978-1-4757-5194-9 | year=1984| series=Problem Books in Mathematics }}</ref> On the other hand, the above-mentioned estimate <math>Q(x) = 6x/\pi^2 + O\left(\sqrt{x}\right)</math> implies that, for some constant ''c'', there always exists a square-free integer between ''x'' and <math>x+c\sqrt{x}</math> for positive ''x''. Moreover, an elementary argument allows us to replace <math>x+c\sqrt{x}</math> by <math>x+cx^{1/5}\log x.</math><ref>{{cite journal
 | last1 = Filaseta | first1 = Michael
 | last2 = Trifonov | first2 = Ognian
 | doi = 10.1112/jlms/s2-45.2.215
 | issue = 2
 | journal = Journal of the London Mathematical Society
 | mr = 1171549
 | pages = 215–221
 | series = Second Series
 | title = On gaps between squarefree numbers. II
 | volume = 45
 | year = 1992}}</ref> The ABC conjecture would allow <math>x+x^{o(1)}</math>.<ref>{{cite journal | last1 = Andrew | first1 = Granville | year = 1998 | title = ABC allows us to count squarefrees | journal = Int. Math. Res. Not. | volume = 1998 | issue = 19 | pages = 991–1009 | doi = 10.1155/S1073792898000592 | doi-access =  free}}</ref>

===Table of ''Q''(''x''), {{Sfrac|6|π<sup>2</sup>}}''x'', and ''R''(''x'')   ===

The table shows how  <math> Q(x)</math> and <math>\frac{6}{\pi^2}x</math>  (with the latter rounded to one decimal place) compare at powers of 10.

<math> R(x) = Q(x) -\frac{6}{\pi^2}x </math> , also denoted as <math> \Delta(x) </math>.

:{| class="wikitable" style="text-align: right"
! <math> x </math>
! <math> Q(x) </math>
! <math>  \frac{6}{\pi^2}x</math>
!<math> R(x)</math>
|-
| 10
| 7
| 6.1
| 0.9
|-
| 10<sup>2</sup>
| 61
| 60.8
| 0.2
|-
| 10<sup>3</sup>
| 608
| 607.9
| 0.1
|-
| 10<sup>4</sup>
| 6,083
| 6,079.3
| 3.7
|-
| 10<sup>5</sup>
| 60,794
| 60,792.7
| 1.3
|-
| 10<sup>6</sup>
| 607,926
| 607,927.1
| {{font color|red|- 1.3}}
|-
| 10<sup>7</sup>
| 6,079,291
| 6,079,271.0
| 20.0
|-
| 10<sup>8</sup>
| 60,792,694
| 60,792,710.2
| {{font color|red|- 16.2}}
|-
| 10<sup>9</sup>
| 607,927,124
| 607,927,101.9
| 22.1
|-
| 10<sup>10</sup>
| 6,079,270,942
|  6,079,271,018.5
| {{font color|red|- 76.5}}
|-
| 10<sup>11</sup>
| 60,792,710,280
| 60,792,710,185.4
| 94.6
|-
| 10<sup>12</sup>
| 607,927,102,274
|  607,927,101,854.0
| 420.0
|-
| 10<sup>13</sup>
| 6,079,271,018,294
| 6,079,271,018,540.3
| {{font color|red|- 246.3}}
|-
| 10<sup>14</sup>
| 60,792,710,185,947
| 60,792,710,185,402.7
| 544.3
|-
| 10<sup>15</sup>
| 607,927,101,854,103
| 607,927,101,854,027.0
| 76.0
|-
|}
<math> R(x) </math> changes its sign infinitely often as <math> x </math> tends to infinity.<ref>{{cite journal |last1=Minoru |first1=Tanaka |title=Experiments concerning the distribution of squarefree numbers |journal=Proceedings of the Japan Academy, Series A, Mathematical Sciences |year=1979 |volume=55 |issue=3 |doi=10.3792/pjaa.55.101 |s2cid=121862978 |url=https://projecteuclid.org/euclid.pja/1195517398|doi-access=free }}</ref>

The absolute value of <math> R(x) </math> is astonishingly small compared with <math> x </math>.

==Encoding as binary numbers==
If we represent a square-free number as the infinite product

:<math>\prod_{n=0}^\infty (p_{n+1})^{a_n}, a_n \in \lbrace 0, 1 \rbrace,\text{ and }p_n\text{ is the }n\text{th prime}, </math>

then we may take those <math>a_n</math> and use them as bits in a binary number with the encoding

:<math>\sum_{n=0}^\infty {a_n}\cdot 2^n .</math>

The square-free number 42 has factorization {{nowrap|2 × 3 × 7}}, or as an infinite product {{nowrap|2<sup>1</sup> · 3<sup>1</sup>  · 5<sup>0</sup> · 7<sup>1</sup> · 11<sup>0</sup> · 13<sup>0</sup> ···}}  Thus the number 42 may be encoded as the binary sequence <code>...001011</code> or 11 decimal. (The binary digits are reversed from the ordering in the infinite product.)

Since the prime factorization of every number is unique, so also is every binary encoding of the square-free integers.

The converse is also true. Since every positive integer has a unique binary representation it is possible to reverse this encoding so that they may be decoded into a unique square-free integer.

Again, for example, if we begin with the number 42, this time as simply a positive integer, we have its binary representation <code>101010</code>. This decodes to {{nowrap|2<sup>0</sup> · 3<sup>1</sup> · 5<sup>0</sup> · 7<sup>1</sup> · 11<sup>0</sup> · 13<sup>1</sup> {{=}} 3 × 7 × 13 {{=}} 273.}}

Thus binary encoding of squarefree numbers describes a [[Bijection|bijection]] between the nonnegative integers and the set of positive squarefree integers.

(See sequences A019565, A048672 and A064273 in the [[On-Line Encyclopedia of Integer Sequences|OEIS]].)

==Erdős squarefree conjecture==
The [[Central binomial coefficient|central binomial coefficient]]

: <math>{2n \choose n}</math>

is never squarefree for ''n'' > 4. This was proven in 1985 for all sufficiently large integers by [[Biography:András Sárközy|András Sárközy]],<ref>{{cite journal
 | doi = 10.1016/0022-314X(85)90017-4 | doi-access=free
 | issue = 1
 | journal = Journal of Number Theory
 | mr = 777971
 | pages = 70–80
 | title = On divisors of binomial coefficients. I
 | volume = 20
 | year = 1985}}</ref> and for all integers > 4 in 1996 by [[Biography:Olivier Ramaré|Olivier Ramaré]] and [[Biography:Andrew Granville|Andrew Granville]].<ref>{{cite journal
| last1=Ramaré | first1=Olivier | last2=Granville | first2=Andrew
| title=Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients
| journal=[[Mathematika]]
| volume=43
| date=1996
| issue=1
| pages=73–107
| doi=10.1112/S0025579300011608}}</ref>

==Squarefree core==
Let us call "''t''-free" a positive integer that has no ''t''-th power in its divisors. In particular, the 2-free integers are the square-free integers.

The multiplicative function <math>\mathrm{core}_t(n)</math> maps every positive integer ''n'' to the quotient of ''n'' by its largest divisor that is a ''t''-th power. That is, 
: <math>\mathrm{core}_t(p^e) = p^{e\bmod t}.</math>

The integer <math>\mathrm{core}_t(n)</math> is ''t''-free, and every ''t''-free integer is mapped to itself by the function <math>\mathrm{core}_t.</math>

The Dirichlet generating function of the sequence  <math>\left(\mathrm{core}_t(n) \right)_{n\in \N}</math> is
: <math>\sum_{n\ge 1}\frac{\mathrm{core}_t(n)}{n^s}
= \frac{\zeta(ts)\zeta(s-1)}{\zeta(ts-t)}</math>.

See also {{OEIS2C|A007913}} (''t''=2), {{OEIS2C|A050985}} (''t''=3) and {{OEIS2C|A053165}} (''t''=4).

== Notes ==
<references/>

== References ==
*{{cite book | last1=Shapiro | first1=Harold N. |title=Introduction to the theory of numbers | publisher=[[Organization:Oxford University Press|Oxford University Press]] Dover Publications | isbn=978-0-486-46669-9 | year=1983}}
*{{cite journal | first1=Andrew | last1=Granville | first2=Olivier | last2=Ramaré | title=Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients | mr=1401709 | zbl=0868.11009 |  year=1996 | journal=Mathematika | volume=43 | pages=73–107 | doi=10.1112/S0025579300011608 | citeseerx=10.1.1.55.8 }}
* {{cite book |last=Guy | first=Richard K. |  title=Unsolved problems in number theory | publisher=[[Physics:Springer-Verlag|Springer-Verlag]] |edition=3rd | year=2004 |isbn=978-0-387-20860-2 | zbl=1058.11001 }}
* {{cite web
 |url= http://oeis.org/wiki/Squarefree_numbers
 |title= OEIS Wiki
 |last=
 |first=
 |date= 
 |website=
 |publisher=
 |access-date= September 24, 2021
 |quote=}}

{{Divisor classes}}


{{DEFAULTSORT:Square-Free Integer}}
[[Category:Number theory]]
[[Category:Integer sequences]]

{{Sourceattribution|Square-free integer}}