# Powerful number

Short description: Numbers whose prime factors all divide the number more than once
\displaystyle{ \begin{align}144000&=2^7\times 3^2\times 5^3\\ &=2^3\times 2^4\times 3^2 \times 5^3\\ &=(2\times5)^3 \times (2^2\times 3)^2\end{align} } 144000 is a powerful number.
Every exponent in its prime factorization
is larger than 1.
It is the product of a square and a cube.

A powerful number is a positive integer m such that for every prime number p dividing m, p2 also divides m. Equivalently, a powerful number is the product of a square and a cube, that is, a number m of the form m = a2b3, where a and b are positive integers. Powerful numbers are also known as squareful, square-full, or 2-full. Paul Erdős and George Szekeres studied such numbers and Solomon W. Golomb named such numbers powerful.

The following is a list of all powerful numbers between 1 and 1000:

1, 4, 8, 9, 16, 25, 27, 32, 36, 49, 64, 72, 81, 100, 108, 121, 125, 128, 144, 169, 196, 200, 216, 225, 243, 256, 288, 289, 324, 343, 361, 392, 400, 432, 441, 484, 500, 512, 529, 576, 625, 648, 675, 676, 729, 784, 800, 841, 864, 900, 961, 968, 972, 1000, ... (sequence A001694 in the OEIS).

## Equivalence of the two definitions

If m = a2b3, then every prime in the prime factorization of a appears in the prime factorization of m with an exponent of at least two, and every prime in the prime factorization of b appears in the prime factorization of m with an exponent of at least three; therefore, m is powerful.

In the other direction, suppose that m is powerful, with prime factorization

$\displaystyle{ m = \prod p_i^{\alpha_i}, }$

where each αi ≥ 2. Define γi to be three if αi is odd, and zero otherwise, and define βi = αiγi. Then, all values βi are nonnegative even integers, and all values γi are either zero or three, so

$\displaystyle{ m = \left(\prod p_i^{\beta_i}\right)\left(\prod p_i^{\gamma_i}\right) = \left(\prod p_i^{\beta_i/2}\right)^2\left(\prod p_i^{\gamma_i/3}\right)^3 }$

supplies the desired representation of m as a product of a square and a cube.

Informally, given the prime factorization of m, take b to be the product of the prime factors of m that have an odd exponent (if there are none, then take b to be 1). Because m is powerful, each prime factor with an odd exponent has an exponent that is at least 3, so m/b3 is an integer. In addition, each prime factor of m/b3 has an even exponent, so m/b3 is a perfect square, so call this a2; then m = a2b3. For example:

$\displaystyle{ m = 21600 = 2^5 \times 3^3 \times 5^2 \, , }$
$\displaystyle{ b = 2 \times 3 = 6 \, , }$
$\displaystyle{ a = \sqrt{\frac{m}{b^3}} = \sqrt{2^2 \times 5^2} = 10 \, , }$
$\displaystyle{ m = a^2b^3 = 10^2 \times 6^3 \, . }$

The representation m = a2b3 calculated in this way has the property that b is squarefree, and is uniquely defined by this property.

## Mathematical properties

The sum of the reciprocals of the powerful numbers converges. The value of this sum may be written in several other ways, including as the infinite product

$\displaystyle{ \prod_p\left(1+\frac{1}{p(p-1)}\right)=\frac{\zeta(2)\zeta(3)}{\zeta(6)} = \frac{315}{2\pi^4}\zeta(3)=1.9435964368..., }$

where p runs over all primes, ζ(s) denotes the Riemann zeta function, and ζ(3) is Apéry's constant.[1] (sequence A082695 in the OEIS) More generally, the sum of the reciprocals of the sth powers of the powerful numbers (a Dirichlet series generating function) is equal to

$\displaystyle{ \frac{\zeta(2s)\zeta(3s)}{\zeta(6s)} }$

whenever it converges.

Let k(x) denote the number of powerful numbers in the interval [1,x]. Then k(x) is proportional to the square root of x. More precisely,

$\displaystyle{ cx^{1/2}-3x^{1/3}\le k(x) \le cx^{1/2}, c = \zeta(3/2)/\zeta(3) = 2.173 \ldots }$

(Golomb, 1970).

The two smallest consecutive powerful numbers are 8 and 9. Since Pell's equation x2 − 8y2 = 1 has infinitely many integral solutions, there are infinitely many pairs of consecutive powerful numbers (Golomb, 1970); more generally, one can find consecutive powerful numbers by solving a similar Pell equation x2ny2 = ±1 for any perfect cube n. However, one of the two powerful numbers in a pair formed in this way must be a square. According to Guy, Erdős has asked whether there are infinitely many pairs of consecutive powerful numbers such as (233, 2332132) in which neither number in the pair is a square. (Walker 1976) showed that there are indeed infinitely many such pairs by showing that 33c2 + 1 = 73d2 has infinitely many solutions. Walker's solutions to this equation are generated, for any odd integer k, by considering the number

$\displaystyle{ (2\sqrt{7}+3\sqrt{3})^{7k}=a\sqrt{7}+b\sqrt{3}, }$

for integers a divisible by 7 and b divisible by 3, and constructing from a and b the consecutive powerful numbers 7a2 and 3b2 with 7a2 = 1 + 3b2. The smallest consecutive pair in this family is generated for k = 1, a = 2637362, and b = 4028637 as

$\displaystyle{ 7\cdot 2637362^2 = 2^2\cdot 7^3\cdot 13^2\cdot 43^2\cdot 337^2=48689748233308 }$

and

$\displaystyle{ 3\cdot 4028637^2 = 3^3\cdot 139^2\cdot 9661^2 = 48689748233307. }$
 Unsolved problem in mathematics:Can three consecutive numbers be powerful?(more unsolved problems in mathematics)

It is a conjecture of Erdős, Mollin, and Walsh that there are no three consecutive powerful numbers. If a triplet of consecutive powerful numbers exists, then its smallest term must be congruent to 7, 27, or 35 modulo 36.[2]

## Sums and differences of powerful numbers

Any odd number is a difference of two consecutive squares: (k + 1)2 = k2 + 2k + 1, so (k + 1)2 − k2 = 2k + 1. Similarly, any multiple of four is a difference of the squares of two numbers that differ by two: (k + 2)2 − k2 = 4k + 4. However, a singly even number, that is, a number divisible by two but not by four, cannot be expressed as a difference of squares. This motivates the question of determining which singly even numbers can be expressed as differences of powerful numbers. Golomb exhibited some representations of this type:

2 = 33 − 52
10 = 133 − 37
18 = 192 − 73 = 35 − 152.

It had been conjectured that 6 cannot be so represented, and Golomb conjectured that there are infinitely many integers which cannot be represented as a difference between two powerful numbers. However, Narkiewicz showed that 6 can be so represented in infinitely many ways such as

6 = 5473 − 4632,

and McDaniel showed that every integer has infinitely many such representations (McDaniel, 1982).

Erdős conjectured that every sufficiently large integer is a sum of at most three powerful numbers; this was proved by Roger Heath-Brown (1987).

## Generalization

More generally, we can consider the integers all of whose prime factors have exponents at least k. Such an integer is called a k-powerful number, k-ful number, or k-full number.

(2k+1 − 1)k,  2k(2k+1 − 1)k,   (2k+1 − 1)k+1

are k-powerful numbers in an arithmetic progression. Moreover, if a1, a2, ..., as are k-powerful in an arithmetic progression with common difference d, then

a1(as + d)k,

a2(as + d)k, ..., as(as + d)k, (as + d)k+1

are s + 1 k-powerful numbers in an arithmetic progression.

We have an identity involving k-powerful numbers:

ak(al + ... + 1)k + ak + 1(al + ... + 1)k + ... + ak + l(al + ... + 1)k = ak(al + ... +1)k+1.

This gives infinitely many l+1-tuples of k-powerful numbers whose sum is also k-powerful. Nitaj shows there are infinitely many solutions of x+y=z in relatively prime 3-powerful numbers(Nitaj, 1995). Cohn constructs an infinite family of solutions of x+y=z in relatively prime non-cube 3-powerful numbers as follows: the triplet

X = 9712247684771506604963490444281, Y = 32295800804958334401937923416351, Z = 27474621855216870941749052236511

is a solution of the equation 32X3 + 49Y3 = 81Z3. We can construct another solution by setting X′ = X(49Y3 + 81Z3), Y′ = −Y(32X3 + 81Z3), Z′ = Z(32X3 − 49Y3) and omitting the common divisor.

## Notes

1. (Golomb, 1970)
2. Beckon, Edward (2019). "On Consecutive Triples of Powerful Numbers". Rose-Hulman Undergraduate Mathematics Journal 20 (2): 25–27.

## References

• Cohn, J. H. E. (1998). "A conjecture of Erdős on 3-powerful numbers". Math. Comp. 67 (221): 439–440. doi:10.1090/S0025-5718-98-00881-3.
• Erdős, Paul; Szekeres, George (1934). "Über die Anzahl der Abelschen Gruppen gegebener Ordnung und über ein verwandtes zahlentheoretisches Problem". Acta Litt. Sci. Szeged 7: 95–102.
• Golomb, Solomon W. (1970). "Powerful numbers". American Mathematical Monthly 77 (8): 848–852. doi:10.2307/2317020.
• Guy, Richard K. (2004). Unsolved Problems in Number Theory (3rd ed.). Springer-Verlag. pp. Section B16. ISBN 978-0-387-20860-2.
• Heath-Brown, Roger (1988). "Ternary quadratic forms and sums of three square-full numbers". Boston: Birkhäuser. pp. 137–163.
• Heath-Brown, Roger (1990). "Sums of three square-full numbers". Colloq. Math. Soc. János Bolyai, no. 51. pp. 163–171.
• Ivić, Aleksandar (1985). The Riemann zeta-function. The theory of the Riemann zeta-function with applications. A Wiley-Interscience Publication. New York etc.: John Wiley & Sons. pp. 33–34,407–413. ISBN 978-0-471-80634-9.
• McDaniel, Wayne L. (1982). "Representations of every integer as the difference of powerful numbers". Fibonacci Quarterly 20: 85–87.
• Nitaj, Abderrahmane (1995). "On a conjecture of Erdős on 3-powerful numbers". Bull. London Math. Soc. 27 (4): 317–318. doi:10.1112/blms/27.4.317.
• Walker, David T. (1976). "Consecutive integer pairs of powerful numbers and related Diophantine equations". The Fibonacci Quarterly 14 (2): 111–116.