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781hw8sol

# 781hw8sol - 18.781 Fall 2007 Problem Set 8 Solutions to...

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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 + ( 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 + )( r + s ¯ ω ) = ( r + )( r - 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+ ω )(1 - ω ), 7 = (3+ ω )(2 - ω ) and 13 = (3+4 ω )( - 1 - 4 ω ) can be the prime factorization. In conclusion, we can have following prime factorizations. 7 = (3 + ω )(2 - ω ) 21 = 3 · 7 = (2 + ω )(1 - ω )(3 + ω )(2 - ω ) 45 = 3 2 · 5 = (2 + ω ) 2 (1 - ω ) 2 · 5 22 = 2 · 11 143 = 11 · 13 = 11 · (3 + 4 ω )( - 1 - 4 ω ) Problem 3 First, prove the following claim.

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781hw8sol - 18.781 Fall 2007 Problem Set 8 Solutions to...

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