Gaussian IntegersDate: 05/12/99 at 10:07:03 From: Marie Sandstrom Subject: Prime numbers and complex numbers Hi! I hope you can help me with a math problem I got stuck on. It was originally in Swedish, but I've done my best translating it. Gaussian prime numbers. Some definitions: 1. A Gaussian integer is a complex number z = a + bi, where a and b are real integers. 2. The absolute value of z = a + bi is the square root of (a^2 + b^2). The square of the absolute value is called the number's complex norm. norm (z) = a^2 + b^2 3. A gaussian prime number is a Gaussian integer with an absolute value greater than 1 that is not the product of Gaussian integers of smaller absolute values. Examine and prove if true: - Is norm (c times d) = norm (c) times norm (d) if c and d are Gaussian integers? - Are all real numbers also Gaussian prime numbers? - Are all Gaussian integers whose norm is a real prime number also Gaussian prime numbers? If so, is the opposite also true? There are three questions and I didn't have any problems with the first one. I've already proved that it is true, so I don't need help with that part of the problem. I started on the second question but couldn't solve it, and I haven't been able to solve the last question either. I would be very grateful for some help. Thanks! Marie Date: 05/12/99 at 11:49:53 From: Doctor Floor Subject: Re: Prime numbers and complex numbers Hi, Marie, Thanks for your questions! I think your second question should be: are all real prime numbers also Gaussian prime numbers? The answer is no. We know for instance that 5 is a real prime, but we see that 5 = (2+i)(2-i), so 5 is not a Gaussian prime. The first part of your third question is not very difficult to prove using the result of the first question. So we have a Gaussian integer x, with norm(x) = p, where p is a real prime. Suppose that x is not a Gaussian prime, so x = y*z for some Gaussian integers y and z, in such a way that abs(x)>abs(y) and abs(x)>abs(z). Since norm(y*z) = norm(y) *norm(z) = p, we find that either norm(y) or norm(z) must be equal to p, say norm(y) = p. But that means norm(x) = norm(y), and thus abs(x) = abs(y). And we have a contradiction. Finally the second part of your third question: the converse is is also true. If x = a+bi is a Gaussian prime, then norm(x) = p, where p is a real prime. To see this, note that if x = a+bi is a prime, so is its complex conjugate x' = a-bi. Also note that norm(x) = x*x'. Now suppose that norm(x) is a composite number, then we can write norm(x) = PRODUCT(p_i), where p_i are 2 or more real prime numbers. If necessary these primes can be factored into Gaussian primes q_i, such that norm(x) = PRODUCT(q_i), where q_i are Gaussian primes. These primes are different from x and x'. But that would mean that norm(x) could be factored into two Gaussian primes (a+bi)(a-bi) and also in a couple of other Gaussian primes q_i. And we have a contradiction. About primes factored into Gaussian primes, you can read from the Dr. Math archives: http://mathforum.org/dr.math/problems/scomazzon8.11.98.html I hope this helped. If you need more help, just write us back! Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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