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For example, 3 × 5 is an integer factorization of 15, and (x – 2)(x + 2) is a polynomial factorization of x 2 – 4. Factorization is not usually considered meaningful within number systems possessing division , such as the real or complex numbers , since any x {\displaystyle x} can be trivially written as ( x y ) × ( 1 / y ) {\displaystyle ...
If one of these values is 0, we have a linear factor. If the values are nonzero, we can list the possible factorizations for each. Now, 2 can only factor as 1×2, 2×1, (−1)×(−2), or (−2)×(−1). Therefore, if a second degree integer polynomial factor exists, it must take one of the values p(0) = 1, 2, −1, or −2. and likewise for p(1).
The polynomial P = x 4 + 1 is irreducible over Q but not over any finite field. On any field extension of F 2, P = (x + 1) 4. On every other finite field, at least one of −1, 2 and −2 is a square, because the product of two non-squares is a square and so we have; If =, then = (+) ().
Squares are always congruent to 0, 1, 4, 5, 9, 16 modulo 20. The values repeat with each increase of a by 10. In this example, N is 17 mod 20, so subtracting 17 mod 20 (or adding 3), produces 3, 4, 7, 8, 12, and 19 modulo 20 for these values. It is apparent that only the 4 from this list can be a square.
As a contrasting example, if n is the product of the primes 13729, 1372933, and 18848997161, where 13729 × 1372933 = 18848997157, Fermat's factorization method will begin with ⌈ √ n ⌉ = 18848997159 which immediately yields b = √ a 2 − n = √ 4 = 2 and hence the factors a − b = 18848997157 and a + b = 18848997161.
First we compute 2P. We have s(P) = s(1,1) = 4, so the coordinates of 2P = (x ′, y ′) are x ′ = s 2 – 2x = 14 and y ′ = s(x – x ′) – y = 4(1 – 14) – 1 = –53, all numbers understood (mod n). Just to check that this 2P is indeed on the curve: (–53) 2 = 2809 = 14 3 + 5·14 – 5. Then we compute 3(2P). We have s(2P) = s(14 ...
The entry 4+2i = −i(1+i) 2 (2+i), for example, could also be written as 4+2i= (1+i) 2 (1−2i). The entries in the table resolve this ambiguity by the following convention: the factors are primes in the right complex half plane with absolute value of the real part larger than or equal to the absolute value of the imaginary part.
Consider the polynomial Q(x) = 3x 4 + 15x 2 + 10.In order for Eisenstein's criterion to apply for a prime number p it must divide both non-leading coefficients 15 and 10, which means only p = 5 could work, and indeed it does since 5 does not divide the leading coefficient 3, and its square 25 does not divide the constant coefficient 10.