math111_midterm_solutions

math111_midterm_solutions - Math 111 Midterm Solutions...

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Math 111 Midterm Solutions Prof. Yehuda Shalom TA: Darren Creutz Date: 4 May 2009 Problem 1. Compute d = gcd (29 + 4 i, 1 + 2 i ) in Z [ i ], the Gaussian integers ( i = - 1). Find Gaussian integers n and m such that (29 + 4 i ) n + (1 + 2 i ) m = d . Solution. First observe that 29 + 4 i 1 + 2 i = (29 + 4 i )(1 - 2 i ) (1 + 2 i )(1 - 2 i ) = 29 + 8 - 58 i + 4 i 5 = 7 - 10 i + 2 - 4 i 5 = 7 - 10 i + 2 1 + 2 i Therefore gcd (29 + 4 i, 1 + 2 i ) = gcd (1 + 2 i, 2) = gcd (2 , 1 + 2 i - 2 i ) = gcd (2 , 1) = 1 Now tracing what we did, set: 29 + 4 i = (1 + 2 i )(7 - 10 i ) + 2 so (29 + 4 i ) - (7 - 10 i )(1 + 2 i ) = 2 and 1 + 2 i = i * 2 + 1 = i * [(29 + 4 i ) - (7 - 10 i )(1 + 2 i )] + 1 so - i (29 + 4 i ) + [ i (7 - 10 i ) + 1](1 + 2 i ) = 1 and so n = - i and m = 11 + 7 i will work: - i (29 + 4 i ) + (11 + 7 i )(1 + 2 i ) = - 29 i + 4 + 11 + 22 i + 7 i - 14 = 1 Problem 2. Find all integer solutions to the following system of congruences: x = 2 mod 7 3 x = 1 mod 5 x = 7 mod 10 Solution. First since 3 - 1 mod 5 = 2 replace the second equation by x = 2 mod 5. Since 7 is relatively prime to 5 and 10 and since gcd (5 , 10) = 5 divides 7 - 2 = 5, the Chinese Remainder Theorem applies. There is then a unique solution
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math111_midterm_solutions - Math 111 Midterm Solutions...

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