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:
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
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
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:
The representation m = a2b3 calculated in this way has the property that b is squarefree, and is uniquely defined by this property.
The sum of reciprocals of powerful numbers converges to
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,
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.
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:
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
and McDaniel showed that every integer has infinitely many such representations (McDaniel, 1982).
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.
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
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:
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
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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