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Unformatted text preview: 18.781, Fall 2007 Problem Set 4 Solutions to Selected Problems Problem 2.7.2 You may want to solve this problem by taking x as 0 through 6 and find the value of x which makes the given equation true. It might be easier, but here we will use the Theorem 2.29. Since (4 , 7) = 1, by multiplying 4, the given equation has same solution with x 3 + 6 x 2 + 3 x + 4 ≡ 0(mod 7) . Since degree of this equation is 3, if we show that x 3 +6 x 2 +3 x +4 is a factor of x 7 x modulo 7, we can conclude that x 3 + 6 x 2 + 3 x + 4 ≡ 0(mod 7) has three solutions by Theorem 2.29. Keeping the fact that every coefficient is in modulo 7 in your mind, divide x 7 x by x 3 + 6 x 2 + 3 x + 4. Then we can calculate like following : ( x 7 x ) ( x 3 + 6 x 2 + 3 x + 4)( x 4 ) ≡ ( x 6 + 4 x 5 + 3 x 4 x ) ( x 6 + 4 x 5 + 3 x 4 x ) ( x 3 + 6 x 2 + 3 x + 4)( x 3 ) ≡ (5 x 5 + 3 x 3 x ) (5 x 5 + 3 x 3 x ) ( x 3 + 6 x 2 + 3 x + 4)(5 x 2 ) ≡ 5 x 4 + 2 x 3 + x 2 x (5 x 4 + 2 x 3 + x 2 x ) ( x 3 + 6 x 2 + 3 x + 4)(5 x ) ≡ This implies that x 3 + 6 x 2 + 3 x + 4 is a factor of x 7 x modulo 7, so we’ve done. 2 Problem 2.7.3 We can find that x 14 + 12 x 2 ≡ x 14 x 2 ≡ x ( x 13 x ) (mod 13) . Since ( x 13 x ) ≡ 0(mod 13) for all integer x by Fermat’s theorem, x 14 + 12 x 2 ≡ 0(mod 13) has 13 solutions. 2 Problem 2.7.4 First of all, if the degree of f is strictly less than 1, f ( x ) ≡ 0 (mod p ) has a solution if and only if f ( x ) is identically zero. Then if we let q ( x ) = 0, we get a desired conclusion. Now assume that degree of f > 0. We will use an induction on j . Before proceeding, we prove the following claim : Suppose that f ( x ) ≡ 0 (mod p ) has a solution x ≡ a (mod p ). Then there is a polynomial q ( x ) such that f ( x ) ≡ ( x a ) q ( x ) (mod p ). 1 Dividing f ( x ) by ( x a ), we have f ( x ) ≡ ( x a ) q ( x ) + r ( x ) (mod p ) where deg( r ) < 1, that is, r ( x ) is constant in modulo p . Since f ( a ) ≡ 0 (mod p ), r ( a ) ≡ 0 (mod p ). Hence r ( x ) ≡ in modulo p , so we can find that f ( x ) ≡ ( x a ) q ( x ) (mod p ). Now we prove the statement of problem by induction. The case of j = 1 is just proved by the claim. Suppose that the statement is true for j = k , and consider the case of j = k +1. Because that f ( x ) ≡ 0 (mod p ) has k solutions, we can say that f ( x ) ≡ ( x a 1 )( x a 2 ) ··· ( x a k ) q ( x ) (mod p ). Applying x = a k +1 , we have ≡ f ( a k +1 ) ≡ ( a k +1 a 1 )( a k +1 a 2 ) ··· ( a k +1 a k ) q ( a k +1 ) (mod p ) Since a k +1 is different from a 1 , ··· , a k in modulo p , ( a k +1 a i ) is not 0 for i = 1 , ··· , k ....
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 Spring '09
 BRUBAKER
 Number Theory, Prime number, Carmichael, 6x2

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