prelim1_sol - u 2 = 1 = u 2-1 = 0 = ( u-1)( u + 1) = 0 =...

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Brief solutions to Prelim 1 1. Yes, because (11 , 41) = 1. Using the Euclidean algorithm, we find that the inverse is [15]. 2. Using the cancellation laws, we have 9 x 9 (mod 69) ⇐⇒ 3 x 3 (mod 23) ⇐⇒ x 1 (mod 23) . So the general solution is x = 1 + 23 k , where k is an arbitrary integer. 3. Since 385 = 5 · 7 · 11, it suffices to show that a 60 1 (mod p ) for p = 5 , 7 , 11. This follows from Fermat’s theorem since a 6≡ 0 (mod p ) and 60 is divisible by p - 1. 4. (a) The units are represented by ± 1 , ± 3. The squares of these integers are 1 and 9, both 1 (mod 8). (b) You could just quote the result proved in the text that a quadratic equation has at most two solutions in an integral domain. Or argue directly:
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Unformatted text preview: u 2 = 1 = u 2-1 = 0 = ( u-1)( u + 1) = 0 = u-1 = 0 or u + 1 = 0. (The last step uses the fact that R has no 0-divisors.) So u = 1 or u =-1. (c) Z 8 has 0-divisors. 5. (a) Routine calculation. (b) R is closed under addition, multiplication, and negation, and it contains 0 and 1. All the ring axioms now follow from the fact that they hold for complex numbers. 6. No. If one existed, it would have to satisfy f ([0] 2 ) = 0 and f ([1] 2 ) = 1. But that function does not preserve sums: f ([1])+ f ([1]) = 1+1 = 2, but f ([1]+[1]) = f ([0]) = 0 6 = 2....
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