# Square-free integer

10 is square-free, as its divisors are 1, 2, 5, and 10, neither of which is a perfect square (the first few perfect squares being 1, 4, 9, and 16)

In mathematics, a square-free integer is an integer which is divisible by no perfect square other than 1. For example, 10 is square-free but 18 is not, as 18 is divisible by 9 = 32. The smallest positive square-free numbers are

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, ... (sequence A005117 in the OEIS)

## Square-free factors of integers

The radical of an integer is its largest square-free factor. An integer is square-free if and only if it is equal to its radical.

Any arbitrary positive integer n can be represented in a unique way as the product of a powerful number (that is a integer such that is divisible by the square of every prime factor) and a square-free integer, which are coprime. The square-free factor, called the square-free part of the number, is the largest square-free divisor k of n that is coprime with n/k. The square-free part of an integer may be smaller than the largest square-free divisor.

Any arbitrary positive integer n can be represented in a unique way as the product of a square and a square-free integer :

${\displaystyle n=m^{2}k}$

In this factorization, m is the largest divisor of n such that m2 is a divisor of n.

In summary, there are three square-free factors that are naturally associated to every integer: the square-free part, the above factor k, and the largest square-free factor. Each is a factor of the next one. All are easily deduced from a prime factorization: if

${\displaystyle n=\prod _{i=1}^{h}p_{i}^{e_{i}}}$

is the prime factorization of n, where ${\displaystyle p_{1},\ldots ,p_{h}}$ are distinct prime numbers, then the square-free part is

${\displaystyle \prod _{e_{i}=1}p_{i},}$

The square-free factor such the quotient is a square is

${\displaystyle \prod _{e_{i}{\text{ odd}}}p_{i},}$

and the largest square-free factor is

${\displaystyle \prod _{i=1}^{h}p_{i}.}$

For example, if ${\displaystyle n=2^{4}\cdot 3^{3}\cdot 5^{2}\cdot 7,}$ the square-free part is 7, the square-free factor such that the quotient is a square is 3 ⋅ 7 = 21, and the largest square-free factor is 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, and no known polynomial-time algorithm for determining whether a number is square-free.[1] On the opposite, polynomial-time algorithms are known for primality testing.[2] 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 greatest common divisor of the polynomial and its formal derivative).[3]

## Equivalent characterizations

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

A positive integer n is square-free if and only if all abelian groups of order n are isomorphic, which is the case if and only if any such group is cyclic. This follows from the classification of finitely generated abelian groups.

A integer n is square-free if and only if the factor ring Z / nZ (see modular arithmetic) is a product of fields. This follows from the Chinese remainder theorem and the fact that a ring of the form Z / kZ is a field if and only if k is a prime.

For every positive integer n, the set of all positive divisors of n becomes a partially ordered set if we use divisibility as the order relation. This partially ordered set is always a distributive lattice. It is a Boolean algebra if and only if n is square-free.

A positive integer n is square-free if and only if μ(n) ≠ 0, where μ denotes the Möbius function.

## Dirichlet generating function

The Dirichlet generating function for the square-free numbers is

${\displaystyle {\frac {\zeta (s)}{\zeta (2s)}}=\sum _{n=1}^{\infty }{\frac {|\mu (n)|}{n^{s}}}}$ where ζ(s) is the Riemann zeta function.

This is easily seen from the Euler product

${\displaystyle {\frac {\zeta (s)}{\zeta (2s)}}=\prod _{p}{\frac {(1-p^{-2s})}{(1-p^{-s})}}=\prod _{p}(1+p^{-s}).}$

## Distribution

Let Q(x) denote the number of square-free integers between 1 and x. 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 events are independent, we obtain the approximation:

${\displaystyle Q(x)\approx x\prod _{p\ {\text{prime}}}\left(1-{\frac {1}{p^{2}}}\right)=x\prod _{p\ {\text{prime}}}{\frac {1}{(1-{\frac {1}{p^{2}}})^{-1}}}}$
${\displaystyle Q(x)\approx x\prod _{p\ {\text{prime}}}{\frac {1}{1+{\frac {1}{p^{2}}}+{\frac {1}{p^{4}}}+\cdots }}={\frac {x}{\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}}}={\frac {x}{\zeta (2)}}}$

This argument can be made rigorous, and a very elementary estimate yields

${\displaystyle Q(x)={\frac {x}{\zeta (2)}}+O\left({\sqrt {x}}\right)={\frac {6x}{\pi ^{2}}}+O\left({\sqrt {x}}\right)}$

(see pi and big O notation) since we use the above characterization to obtain

${\displaystyle Q(x)=\sum _{n\leq x}\sum _{d^{2}\mid n}\mu (d)=\sum _{d\leq x}\mu (d)\sum _{n\leq x,d^{2}\mid n}1=\sum _{d\leq x}\mu (d)\left\lfloor {\frac {x}{d^{2}}}\right\rfloor }$

and, observing that the last summand is zero for ${\displaystyle d>{\sqrt {x}}}$, we have

${\displaystyle Q(x)=\sum _{d\leq {\sqrt {x}}}\mu (d)\left\lfloor {\frac {x}{d^{2}}}\right\rfloor =\sum _{d\leq {\sqrt {x}}}{\frac {x\mu (d)}{d^{2}}}+O(\sum _{d\leq {\sqrt {x}}}1)=x\sum _{d\leq {\sqrt {x}}}{\frac {\mu (d)}{d^{2}}}+O({\sqrt {x}})=x\sum _{d}{\frac {\mu (d)}{d^{2}}}+O\left(x\sum _{d>{\sqrt {x}}}{\frac {1}{d^{2}}}+{\sqrt {x}}\right)={\frac {x}{\zeta (2)}}+O({\sqrt {x}}).}$

By exploiting the largest known zero-free region of the Riemann zeta function, due to Ivan Matveyevich Vinogradov, N.M. Korobov (ru) and Hans-Egon Richert, the maximal size of the error term has been reduced by Arnold Walfisz[4] and we have

${\displaystyle Q(x)={\frac {6x}{\pi ^{2}}}+O\left(x^{1/2}\exp \left(-c{\frac {(\log x)^{3/5}}{(\log \log x)^{1/5}}}\right)\right).}$

for some positive constant c. Under the Riemann hypothesis, the error term can be further reduced[5] to yield

${\displaystyle Q(x)={\frac {x}{\zeta (2)}}+O\left(x^{17/54+\varepsilon }\right)={\frac {6x}{\pi ^{2}}}+O\left(x^{17/54+\varepsilon }\right).}$

The asymptotic/natural density of square-free numbers is therefore

${\displaystyle \lim _{x\to \infty }{\frac {Q(x)}{x}}={\frac {6}{\pi ^{2}}}={\frac {1}{\zeta (2)}}}$

where ζ is the Riemann zeta function and 1/ζ(2) is approximately 0.6079. 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

${\displaystyle Q(x,n)={\frac {x}{\sum _{k=1}^{\infty }{\frac {1}{k^{n}}}}}+O\left({\sqrt[{n}]{x}}\right)={\frac {x}{\zeta (n)}}+O\left({\sqrt[{n}]{x}}\right).}$

Since a multiple of 4 must have a square factor 4=22, it cannot occur that four consecutive integers are all square-free. On the other hand, there exist infinitely many integers n for which 4n +1, 4n +2, 4n +3 are all square-free. Otherwise, observing that 4n and at least one of 4n +1, 4n +2, 4n +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

${\displaystyle Q(x)\leq {\frac {x}{2}}+C}$ for some constant C,

contrary to the above asymptotic estimate for ${\displaystyle Q(x)}$.

There exist sequences of consecutive non-square-free integers of arbitrary length. Indeed, if n satisfies a simultaneous congruence

${\displaystyle n\equiv -i{\pmod {p_{i}^{2}}}\qquad (i=1,2,\ldots ,l)}$

for distinct primes ${\displaystyle p_{1},p_{2},\ldots ,p_{l}}$, then each ${\displaystyle n+i}$ is divisible by pi 2.[6] On the other hand, the above-mentioned estimate ${\displaystyle Q(x)=6x/\pi ^{2}+O\left({\sqrt {x}}\right)}$ implies that, for some constant c, there always exists a square-free integer between x and ${\displaystyle x+c{\sqrt {x}}}$ for positive x. Moreover, an elementary argument allows us to replace ${\displaystyle x+c{\sqrt {x}}}$ by ${\displaystyle x+cx^{1/5}\log x}$.[7] The ABC conjecture would allow ${\displaystyle x+x^{o(1)}}$.[8]

## Encoding as binary numbers

If we represent a square-free number as the infinite product

${\displaystyle \prod _{n=0}^{\infty }(p_{n+1})^{a_{n}},a_{n}\in \lbrace 0,1\rbrace ,{\text{ and }}p_{n}{\text{ is the }}n{\text{th prime}},}$

then we may take those ${\displaystyle a_{n}}$ and use them as bits in a binary number with the encoding

${\displaystyle \sum _{n=0}^{\infty }{a_{n}}\cdot 2^{n}.}$

The square-free number 42 has factorization 2 × 3 × 7, or as an infinite product 21 · 31  · 50 · 71 · 110 · 130 ··· Thus the number 42 may be encoded as the binary sequence ...001011 or 11 decimal. (Note that 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 101010. This decodes to 20 · 31 · 50 · 71 · 110 · 131 = 3 × 7 × 13 = 273.

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

(See sequences A019565, A048672 and A064273 in the OEIS.)

## Erdős squarefree conjecture

${\displaystyle {2n \choose n}}$

is never squarefree for n > 4. This was proven in 1985 for all sufficiently large integers by András Sárközy,[9] and for all integers > 4 in 1996 by Olivier Ramaré and Andrew Granville.[10]

## Squarefree core

The multiplicative function ${\displaystyle \mathrm {core} _{t}(n)}$ is defined to map positive integers n to t-free numbers by reducing the exponents in the prime power representation modulo t:

${\displaystyle \mathrm {core} _{t}(p^{e})=p^{e{\bmod {t}}}.}$

The value set of ${\displaystyle \mathrm {core} _{2}}$, in particular, are the square-free integers. Their Dirichlet generating functions are

${\displaystyle \sum _{n\geq 1}{\frac {\mathrm {core} _{t}(n)}{n^{s}}}={\frac {\zeta (ts)\zeta (s-1)}{\zeta (ts-t)}}}$.

OEIS representatives are (t=2), (t=3) and (t=4).

## Notes

1. ^ Adleman, Leonard M.; Mccurley, Kevin S. "Open Problems in Number Theoretic Complexity, II". Lecture Notes in Computer Science: 9.
2. ^ Agrawal, Manindra; Kayal, Neeraj; Saxena, Nitin (2004-09-01). "PRIMES is in P". Annals of Mathematics. 160 (2): 781–793. doi:10.4007/annals.2004.160.781. ISSN 0003-486X.
3. ^ Richards, C.: Algorithms for factoring square-free polynomials over finite fields. Master thesis, Simon Fraser University, Canada (2009)
4. ^ A. Walfisz. "Weylsche Exponentialsummen in der neueren Zahlentheorie" (VEB Deutscher Verlag der Wissenschaften, Berlin 1963.
5. ^ 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, A Survey on k-freeness; also see Kaneenika Sinha, "Average orders of certain arithmetical functions", Journal of the Ramanujan Mathematical Society 21:3 (2006), pp. 267–277.
6. ^ Parent, D. P. (1984). Exercises in Number Theory. Springer-Verlag New York. doi:10.1007/978-1-4757-5194-9. ISBN 978-1-4757-5194-9.
7. ^ Michael, Filaseta; Ognian, Trifonov (1992). "On gaps between squarefree numbers II". J. London Math. Soc. (2) 45: 215–221.
8. ^ Andrew, Granville (1998). "ABC allows us to count squarefrees". Int. Math. Res. Notices. 1998 (19): 991–1009.
9. ^ András Sárközy. On divisors of binomial coefficients, I. J. Number Theory 20 (1985), no. 1, 70–80.
10. ^ Olivier Ramaré and Andrew Granville. Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73–107