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Proof by exhaustion can be used to prove that if an integer is a perfect cube, then it must be either a multiple of 9, 1 more than a multiple of 9, or 1 less than a multiple of 9. [3] Proof: Each perfect cube is the cube of some integer n, where n is either a multiple of 3, 1 more than a multiple of 3, or 1 less than a multiple of 3. So these ...
As a result, there is a one-to-one correspondence between Mersenne primes and even perfect numbers, so a list of one can be converted into a list of the other. [ 1 ] [ 5 ] [ 6 ] It is currently an open problem whether there are infinitely many Mersenne primes and even perfect numbers.
Illustration of the perfect number status of the number 6. In number theory, a perfect number is a positive integer that is equal to the sum of its positive proper divisors, that is, divisors excluding the number itself. [1] For instance, 6 has proper divisors 1, 2 and 3, and 1 + 2 + 3 = 6, so 6 is a perfect number.
A perfect parallelepiped is a parallelepiped with integer-length edges, face diagonals, and body diagonals, but not necessarily with all right angles; a perfect cuboid is a special case of a perfect parallelepiped. In 2009, dozens of perfect parallelepipeds were shown to exist, [19] answering an open question of Richard Guy. Some of these ...
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.
1729 can be expressed as a sum of two positive cubes in two ways, illustrated geometrically. 1729 is also known as Ramanujan number or Hardy–Ramanujan number, named after an anecdote of the British mathematician G. H. Hardy when he visited Indian mathematician Srinivasa Ramanujan who was ill in a hospital.
Note that if n 2 is the closest perfect square to the desired square x and d = x - n 2 is their difference, it is more convenient to express this approximation in the form of mixed fraction as . Thus, in the previous example, the square root of 15 is 4 − 1 8 . {\displaystyle 4{\tfrac {-1}{8}}.}
If all P of the power cubes are perfect, the P-multimagic cube is said to be perfect. The first known example of a bimagic cube was given by John Hendricks in 2000; it is a semiperfect cube of order 25 and magic constant 195325. In 2003, C. Bower discovered two semi-perfect bimagic cubes of order 16, and a perfect bimagic cube of order 32. [2]