Powerful number

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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 OEIS).

Equivalence of the two definitions[edit]

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

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 = αi - γi. Then, all values βi are nonnegative even integers, and all values γi are either zero or three, so

m = (\prod p_i^{\beta_i})(\prod p_i^{\gamma_i}) = (\prod p_i^{\beta_i/2})^2(\prod p_i^{\gamma_i/3})^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:

m = 21600 = 2^5 \times 3^3 \times 5^2 \, ,
b = 2 \times 3 = 6 \, ,
a = \sqrt{\frac{m}{b^3}} = \sqrt{2^2 \times 5^2} = 10 \, ,
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[edit]

The Dirichlet series generating function for powerful numbers is

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

and so the sum of reciprocals of powerful numbers converges to

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

where p runs over all primes, ζ(s) denotes the Riemann zeta function, and ζ(3) is Apéry's constant (Golomb, 1970).

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,

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

(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 x2 − ny2 = ±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. Jaroslaw Wroblewski showed that there are indeed infinitely many such pairs by showing that 33c2+1=73d2 has infinitely many solutions. It is a conjecture of Erdős, Mollin, and Walsh that there are no three consecutive powerful numbers.

Sums and differences of powerful numbers[edit]

Any odd number is a difference of two consecutive squares: (k + 1)2 = k2 + 2k +12, 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 = 32(33 − 52).

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).


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.

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