Question 849360: Prove that every prime of the form 3m + 1 with m (in) N is also of the form 6n + 1
with n (in) N.
N for natural numbers
Answer by LinnW(1048)   (Show Source):
You can put this solution on YOUR website! We need to show that if p is a prime number of the form
6n + 1 , then there exists an m such that 3m + 1 = p.
We also need to show that if p is a prime number of the form
3m + 1, then there exists an n such that 6n + 1 = p.
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Taking the first case, showing that if p is a prime number of the form
6n + 1 , then there exists an m such that 3m + 1 = p.
Let p be such a prime number and n is a number such that 6n + 1 = p .
Certainly, 6n + 1 = 3(2n) + 1. So setting m = 2n , 3m + 1 = p.
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For the second case, we need to show that if p is a prime number of the form
3m + 1, then there exists an n such that 6n + 1 = p.
Let p be a prime number such that 3m + 1 = p.
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Suppose m is odd and greater than 2.
This means that 3m is odd, since an
odd number times an odd number is odd.
This means that 3m + 1 is even. But if
3m + 1 is even, 3m + 1 is not prime.
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Suppose m is even. Being even, there exists a number n
such that 2n = m. So 3m + 1 = 3(2n) + 1 = 6n + 1.
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