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Gaussian Integers

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Date: 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
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Date: 05/12/99 at 11:49:53
From: Doctor Floor
Subject: Re: Prime numbers and complex numbers

Hi, Marie,

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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Associated Topics:
High School Number Theory

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