Primes of the Form 4n+3Date: 11/09/1999 at 11:14:45 From: Jessica Troise Subject: Abstract Math I I need to prove that there are infinitely many primes of the form 4n+3 where n is an element of the natural numbers. I'm having real problems knowing how to start this problem. Date: 11/09/1999 at 11:56:35 From: Doctor Wilkinson Subject: Re: Abstract Math I Hi, Jessica, Take the proof that there are an infinite number of primes as a model. In that proof, you suppose that p_1, p_2, ..., p_n are all the primes. You then multiply them all together and add 1. The result is not divisible by any of p_1, p_2, ..., p_n because when you divide it by any of those you get a remainder of 1. Therefore it must be divisible by some other prime. Suppose we try to do something similar. So suppose that p_1, p_2, ..., p_k are all the primes of the form 4n+3. Now all these primes are odd, so we probably won't get anywhere taking their product and adding 1, since that will give us an even number. But we probably want to make something out of p_1, p_2, ..., p_k and try to copy the argument from the other proof. Now the next part of the argument should say that whatever we get has to be divisible by a prime of the form 4n+3. That's not true of just any number, but it is true of numbers of the form 4n+3, because such a number first of all doesn't have 2 as a prime factor, and if you multiply two numbers of the form 4n+1 together, you always get another number of the form 4n + 1. Just multiply out (4m+1)(4n+1) and see what happens. Now suppose we multiply all the numbers p_1, p_2, ..., p_k together. Then you can work out that you'll get something of the form 4n+1 if n is even, and something of the form 4n+3 if n is odd. In the first case, if you add 2 you'll get something of the form 4n+3: p_1*p_2*...*p_k + 2 = 4x + 3 Now you can easily show that this number is not divisible by any of p_1, ..., p_k, since if you divide you get a remainder of 2. So you should be able to push through the argument in this case. Can you see what to do in the other case? - Doctor Wilkinson, The Math Forum http://mathforum.org/dr.math/ |
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