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[3]: 127 We call a triple (,,) admissible for K if such an identity exists. [1]: 125 Trivial cases of admissible triples include (,,). The problem is uninteresting for K of characteristic 2, since over such fields every sum of squares is a square, and we exclude this case. It is believed that otherwise admissibility is independent of the field ...
Pierre de Fermat gave a criterion for numbers of the form 8a + 1 and 8a + 3 to be sums of a square plus twice another square, but did not provide a proof. [1] N. Beguelin noticed in 1774 [2] that every positive integer which is neither of the form 8n + 7, nor of the form 4n, is the sum of three squares, but did not provide a satisfactory
On the other hand, the primes 3, 7, 11, 19, 23 and 31 are all congruent to 3 modulo 4, and none of them can be expressed as the sum of two squares. This is the easier part of the theorem, and follows immediately from the observation that all squares are congruent to 0 (if number squared is even) or 1 (if number squared is odd) modulo 4.
Legendre's three-square theorem states which numbers can be expressed as the sum of three squares; Jacobi's four-square theorem gives the number of ways that a number can be represented as the sum of four squares. For the number of representations of a positive integer as a sum of squares of k integers, see Sum of squares function.
The prime decomposition of the number 2450 is given by 2450 = 2 · 5 2 · 7 2. Of the primes occurring in this decomposition, 2, 5, and 7, only 7 is congruent to 3 modulo 4. Its exponent in the decomposition, 2, is even. Therefore, the theorem states that it is expressible as the sum of two squares. Indeed, 2450 = 7 2 + 49 2.
The number of ways to write a natural number as sum of two squares is given by r 2 (n).It is given explicitly by = (() ())where d 1 (n) is the number of divisors of n which are congruent to 1 modulo 4 and d 3 (n) is the number of divisors of n which are congruent to 3 modulo 4.
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 ...
A result of Albrecht Pfister [8] shows that a positive semidefinite form in n variables can be expressed as a sum of 2 n squares. [9] Dubois showed in 1967 that the answer is negative in general for ordered fields. [10] In this case one can say that a positive polynomial is a sum of weighted squares of rational functions with positive ...