Search results
Results From The WOW.Com Content Network
Now let m be the smallest positive integer such that mp is the sum of four squares, x 1 2 + x 2 2 + x 3 2 + x 4 2 (we have just shown that there is some m (namely n) with this property, so there is a least one m, and it is smaller than p).
The number of ways to represent n as the sum of four squares is eight times the sum of the divisors of n if n is odd and 24 times the sum of the odd divisors of n if n is even (see divisor function), i.e.
The tallest Leyland cypress documented is about 40 m (130 ft) tall and still growing. [18] However, because their roots are relatively shallow, a large leylandii tends to topple over. The shallow root structure also means that it is poorly adapted to areas with hot summers, such as the southern half of the United States.
The squared Euclidean distance between two points, equal to the sum of squares of the differences between their coordinates; Heron's formula for the area of a triangle can be re-written as using the sums of squares of a triangle's sides (and the sums of the squares of squares) The British flag theorem for rectangles equates two sums of two ...
Subset sum problem, an algorithmic problem that can be used to find the shortest representation of a given number as a sum of powers; Pollock's conjectures; Sums of three cubes, discusses what numbers are the sum of three not necessarily positive cubes; Sums of four cubes problem, discusses whether every integer is the sum of four cubes of integers
Comment: The proof of Euler's four-square identity is by simple algebraic evaluation. Quaternions derive from the four-square identity, which can be written as the product of two inner products of 4-dimensional vectors, yielding again an inner product of 4-dimensional vectors: ( a · a )( b · b ) = ( a × b )·( a × b ) .
Since 5 is odd, and has only itself as a divisor (or do we count the trivial divisor 1?) We should have 8(5) = 40 possible combinations of sums of squares. But there exists only one: 2 2 + 1 2 + 2(0 2) = 5 Assuming we allow all permutations of order, that only gives us 4!, and this is not unique to this decomposition either. Allowing squares of ...
[1] Every positive integer can be expressed as the sum of at most 19 fourth powers; every integer larger than 13792 can be expressed as the sum of at most 16 fourth powers (see Waring's problem). Fermat knew that a fourth power cannot be the sum of two other fourth powers (the n = 4 case of Fermat's Last Theorem; see Fermat's right triangle ...