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Unformatted text preview: 18.781, Fall 2007 Problem Set 1 Solutions of Selected Problems Problem 1.2.2 We apply the Euclidean algorithm, as Example 1. 3587 = 1 1819 + 1768 1819 = 1 1768 + 51 1768 = 34 51 + 34 51 = 1 34 + 17 34 = 2 17 This gives (3587 , 1819) = 17 . Lets find x, y which satisfy 1819 x + 3587 y = 17. Starting with 1819 0 + 3587 1 = 3587 . and 1819 1 + 3587 0 = 1819 . We multiply the second of these equations by 1, and subtract the result from the first equation, to obtain 1819 ( 1) + 3587 1 = 1768 . We multiply this equation by 1, and subtract from the preceding equation to find that 1819 2 + 3587 ( 1) = 51 . We multiply this equation by 34, and subtract from the preceding equation to find that 1819 ( 69) + 3587 35 = 34 . We multiply this equation by 1, and subtract from the preceding equation to find that 1819 71 + 3587 ( 36) = 17 . Hence, we may take x = 71 , y = 36. 2 Problem 1.2.13 First, recall the fact that If a 1  c, , a k  c and ( a 1 , , a k ) = 1, then a 1 a k  c . 1 This can be proved directly from theorem 1.12. 1) n 2 n n 2 n = n ( n 1). Since n 1 , n are consecutive integers, one of them is even. Hence, their product is clearly divisible by 2. 2) n 3 n n 3 n = ( n 1) n ( n +1). Since n 1 , n, n +1 are three consecutive integers, one of them is a multiple of 3. This implies 3  n 3 n . Also since 2  n ( n 1) and n ( n 1)  n 3 n , 2  n 3 n . Because (2 , 3) = 1, the above fact tells that 6  n 3 n , as desired. 3) n 5 n n 5 n = n ( n 4 1) = ( n 1) n ( n + 1)( n 2 + 1). Then n 3 n  n 5 n , so n 5 n is divisible by 6. Also, notice that every integer can be represented by 5 k + r for proper r = 0 , , 4. If r = 0 , 1 , 4, then n, n 1 , n + 1 is divisible by 5, respectively. If r = 2 , 3, then n 2 + 1 = (25 k 2 + 20 k + 5) , (25 k 2 + 30 k + 10), respectively, and both are divisible by 5. Therefore, 5  n 5 n . Since (5....
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 Spring '09
 BRUBAKER
 Number Theory

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