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Geometrically speaking, a positive integer m is a perfect cube if and only if one can arrange m solid unit cubes into a larger, solid cube. For example, 27 small cubes can be arranged into one larger one with the appearance of a Rubik's Cube, since 3 × 3 × 3 = 27. The difference between the cubes of consecutive integers can be expressed as ...
Legendre's conjecture: Does there always exist at least one prime between consecutive perfect squares? Are there infinitely many primes p such that p − 1 is a perfect square? In other words: Are there infinitely many primes of the form n 2 + 1? As of 2025, all four problems are unresolved.
[1] [2] The integers 2 3 and 3 2 are two perfect powers (that is, powers of exponent higher than one) of natural numbers whose values (8 and 9, respectively) are consecutive. The theorem states that this is the only case of two consecutive perfect powers. That is to say, that
It is known that the prime number theorem gives an accurate count of the primes within short intervals, either unconditionally [5] or based on the Riemann hypothesis, [6] but the lengths of the intervals for which this has been proven are longer than the intervals between consecutive squares, too long to prove Legendre's conjecture.
The sum of four cubes problem [1] asks whether every integer is the sum of four cubes of integers. It is conjectured the answer is affirmative, but this conjecture has been neither proven nor disproven. [2] Some of the cubes may be negative numbers, in contrast to Waring's problem on sums of cubes, where they are required to be positive.
G(3) is at least 4 (since cubes are congruent to 0, 1 or −1 mod 9); for numbers less than 1.3 × 10 9, 1 290 740 is the last to require 6 cubes, and the number of numbers between N and 2N requiring 5 cubes drops off with increasing N at sufficient speed to have people believe that G(3) = 4; [17] the largest number now known not to be a sum of ...
The sum within each gmonon is a cube, so the sum of the whole table is a sum of cubes. [7] Visual demonstration that the square of a triangular number equals a sum of cubes. In the more recent mathematical literature, Edmonds (1957) provides a proof using summation by parts. [8]
The Riemann hypothesis implies that there exists a prime between any two consecutive cubes, allowing the sufficiently large condition to be removed, and allowing the sequence of Mills primes to begin at a 1 = 2. For all a > , there is at least one prime between and (+). [2]