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Dixon's method replaces the condition "is the square of an integer" with the much weaker one "has only small prime factors"; for example, there are 292 squares smaller than 84923; 662 numbers smaller than 84923 whose prime factors are only 2,3,5 or 7; and 4767 whose prime factors are all less than 30. (Such numbers are called B-smooth with ...
The former are ≡ ±1 (mod 12) and the latter are all ≡ ±5 (mod 12). −3 is in rows 7, 13, 19, 31, 37, and 43 but not in rows 5, 11, 17, 23, 29, 41, or 47. The former are ≡ 1 (mod 3) and the latter ≡ 2 (mod 3). Since the only residue (mod 3) is 1, we see that −3 is a quadratic residue modulo every prime which is a residue modulo 3.
This corresponds to a set of y values whose product is a square number, i.e. one whose factorization has only even exponents. The products of x and y values together form a congruence of squares. This is a classic system of linear equations problem, and can be efficiently solved using Gaussian elimination as soon as the number of rows exceeds ...
Since a prime number has factors of only 1 and itself, and since m = 2 is the only non-zero value of m to give a factor of 1 on the right side of the equation above, it follows that 3 is the only prime number one less than a square (3 = 2 2 − 1). More generally, the difference of the squares of two numbers is the product of their sum and ...
The CRT says that this is the same as p ≡ 1 (mod 840), and Dirichlet's theorem says there are an infinite number of primes of this form. 2521 is the smallest, and indeed 1 2 ≡ 1, 1046 2 ≡ 2, 123 2 ≡ 3, 2 2 ≡ 4, 643 2 ≡ 5, 87 2 ≡ 6, 668 2 ≡ 7, 429 2 ≡ 8, 3 2 ≡ 9, and 529 2 ≡ 10 (mod 2521).
Name First elements Short description OEIS Mersenne prime exponents : 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, ... Primes p such that 2 p − 1 is prime.: A000043 ...
In mathematics, a square-free integer (or squarefree integer) is an integer which is divisible by no square number other than 1. That is, its prime factorization has exactly one factor for each prime that appears in it. For example, 10 = 2 ⋅ 5 is square-free, but 18 = 2 ⋅ 3 ⋅ 3 is not, because 18 is divisible by 9 = 3 2. The smallest ...
In another case, testing p = 13, we have 17 (13 − 1)/2 = 17 6 ≡ 1 (mod 13), therefore 17 is a quadratic residue modulo 13. As confirmation, note that 17 ≡ 4 (mod 13), and 2 2 = 4. We can do these calculations faster by using various modular arithmetic and Legendre symbol properties.