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Unformatted text preview: 2 mod p is solvable. Solution. Since p = 3 mod 4, we have p = 4 k + 3 for some integer k , and so p 1 = ( 4 k + 3 ) 1 = 4 k + 2 = 2 ( 2 k + 1 ) ; that is, p 1 = 2 m , where m = 2 k + 1 is odd. By Theorem 2.131, F p = h 1 i H for some subgroup H of order m . Since F 2 I m = ( { 1 } H ) ( { 1 } H ) , however, either 2 H or 2 H . But Exercise 2.82 says that every element in H has a square root. If 2 has a square root, then there is an integer a with a 2 = 2 in F p . and this says that a 2 2 mod p . 3.100 (i) Prove that x 4 + 1 factors in F 2 [ x ] . Solution. x 4 + 1 factors in F 2 [ x ] because 1 is a root....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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