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# Adv Alegbra HW Solutions 122 - F p x x 4 1 = x 2 ax 1 x 2...

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122 (ii) If x 4 + 1 = ( x 2 + ax + b )( x 2 + cx + d ) F p [ x ] , where p is an odd prime, prove that c =− a and d + b a 2 = 0 a ( d b ) = 0 bd = 1 . Solution. After expanding, we obtain the equations a + c = 0 d + ac + b = 0 ad + bc = 0 bd = 1 . Substituting c =− a yields the desired equations. (iii) Prove that x 4 + 1 factors in F p [ x ] , where p is an odd prime, if any of the following congruences are solvable: b 2 ≡− 1mod p , a 2 ≡± 2mod p . Solution. Using the hint, x 4 + 1 = ( x 2 + b )( x 2 b ) in F p [ x ] if a 0mod p .I f a ²≡ 0mod p , then d b mod p and b 2 bd 1mod p . Hence, b ≡± 1mod p , by Lemma 3.82 and Exercise 3.100. From part (i), we have a 2 ≡± 2mod p .As - suming that one of these congruences can be solved, we have fac-
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Unformatted text preview: F p [ x ] x 4 + 1 = ( x 2 + ax + 1 )( x 2 − ax + 1 ) or x 4 + 1 = ( x 2 + ax − 1 )( x 2 − ax − 1 ). (iv) Prove that x 4 + 1 factors in F p [ x ] for all primes p . Solution. In part (i), we saw that x 4 + 1 factors in F 2 [ x ] , and in part (iii), we saw that it factors in F p [ x ] for every odd prime p if certain congruences can be solved. As every odd prime (indeed, every odd integer) is congruent to 1 or 3 mod 4, Exercise 3.100 completes the proof....
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