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Unformatted text preview: 18.781, Fall 2007 Problem Set 8 Solutions to Selected Problems Problem 2 First, observe the following statement. If a prime integer (in Z ) p is of the form 3 k + 2, p is a prime element in R . This can be proved easily using the problem 6 in problem set 7. (Note that if a is a prime factor of p ,then N ( a ) should be p . ) By this, we can conclude that 2 , 5 , 11 are still primes in R . If x = r + s ( r, s Z ),by easy computation, we have x x = N ( x ). It can help us to find the factorization of the prime integer (in Z ) p . Once we find r, s such that r 2 rs + s 2 = p , then we have ( r + s )( r + s ) = ( r + s )( r s s ) = p . (Each factor has prime integer value by the function N , so they are primes in R .) By above observation, 2 2 2 1+1 2 = 3, 3 2 3 1+1 2 = 7 and 3 2 3 4+4 2 = 13 implies that 3 = (2+ )(1 ), 7 = (3+ )(2 ) and 13 = (3+4 )( 1 4 ) can be the prime factorization. In conclusion, we can have following prime factorizations. 7 = (3 + )(2 ) 21 = 3 7 = (2 + )(1 )(3 + )(2 ) 45 = 3 2 5 = (2 + ) 2 (1 ) 2 5 22 = 2 11 143 = 11 13 = 11 (3 + 4 )( 1...
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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.
 Spring '09
 BRUBAKER
 Number Theory

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