781hw3sol

781hw3sol - 18.781, Fall 2007 Problem Set 3 Solutions to...

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18.781, Fall 2007 Problem Set 3 Solutions to Selected Problems Problem 2.3.17 First of all, we can observe that 143 = 11 · 13 and x 3 - 9 x 2 + 23 x - 15 = ( x - 1)( x - 3)( x - 5) . Hence, x is a solution of given equation if and only if ( x - 1)( x - 3)( x - 5) 0(mod 11) and ( x - 1)( x - 3)( x - 5) 0(mod 13) . Clearly, this means that x 1 , 3 , 5(mod 11) and x 1 , 3 , 5(mod 13) . Using the relation 6 · 11 + ( - 5) · 13 = 1, we have x a 1 (mod 11) and x a 2 (mod 13) m x ≡ - 65 a 1 + 66 a 2 (mod 143) . (Using the Chinese Remainder theorem with m 1 = 11 , m 2 = 13 , b 1 = - 5 , b 2 = 6. ) Therefore, we can conclude that the solutions are For ( a 1 , a 2 ) = (1 , 1) , x 1(mod 143) . For ( a 1 , a 2 ) = (1 , 3) , x 133(mod 143) . For ( a 1 , a 2 ) = (1 , 5) , x 265 122(mod 143) . For ( a 1 , a 2 ) = (3 , 1) , x ≡ - 129 14(mod 143) . For ( a 1 , a 2 ) = (3 , 3) , x 3(mod 143) . For ( a 1 , a 2 ) = (3 , 5) , x 135(mod 143) . For ( a 1 , a 2 ) = (5 , 1) , x ≡ - 259 27(mod 143) . For ( a 1 , a 2 ) = (5 , 3) , x ≡ - 127 16(mod 143) . For ( a 1 , a 2 ) = (5 , 5) , x 5(mod 143) . 2 1
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Problem 2.3.21 First we prove ”if” part. This is quite trivial. Suppose that a i a r (mod p α i ) for i = 1 , 2 , ··· , r . Then x = a r is the solution of the given system. Now we prove ”only if” part. Suppose that there is a simultaneous solution x . Then for any i , x a i (mod p α i ) , hence we can express x as x = a i + t i p α i . Then for fixed i , a i + t i p α i = x = a r + t r p α r , that is, a i - a r = t r p α r - t i p α i = p α i ( t r p α r - α i - t i ) , where α r - α i 0. This gives that
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This note was uploaded on 04/28/2009 for the course MATH 18.781 taught by Professor Brubaker during the Spring '09 term at MIT.

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781hw3sol - 18.781, Fall 2007 Problem Set 3 Solutions to...

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