Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Write 300 as a product of primes 'in the ring of Gaussian integers'.

I can write 300 as a product of primes with no problem:


However, i'm not sure if this is what i'm supposed to be doing, because of what was stated in the original question, which was, 'in the ring of Gaussian integers'.

share|cite|improve this question
Do you know what the Gaussian integers are? – Chris Eagle Nov 6 '12 at 18:53
Do you know what it means to be prime in the Gaussian integers? – JSchlather Nov 6 '12 at 18:53
No, I don't know what Gaussian Integers are. The professor briefly covered it, and it is not in the book. – maroon.elephants Nov 6 '12 at 18:56
@maroon.elephants If the professor briefly covered it, you can just look it up in the notes that you have taken. – Phira Nov 6 '12 at 18:56
@maroon.elephants, gaussian integers are complex numbers $a + ib$ but $a$,$b$ are integers. – sperners lemma Nov 6 '12 at 19:02
up vote 1 down vote accepted

The Gaussian integers are the integers among the complex numbers. They are of the form $a+bi$ where $a,b\in\Bbb Z$ (i.e. $a,b$ are integers) and $i$ is the imaginarius unit: while $\Bbb R$ is drawn horizontally, $i$ is drawn to be the vertical unit (in coordinates of the complex plane: $(0,1)$ corresponds to $i$, and $(x,0)$ to a real number $x$).

Algebraically $$i^2=-1$$ is all what you have to know about it.

It follows that $(a+bi)(a-bi) = a^2-(bi)^2 = a^2+b^2$, in particular, as Sp.Lemma wrote, $5=(1+2i)(1-2i)$, and you can also check that $2=i(1-i)^2$, and that $i$ 'doesn't matter' here, because divides $1$ (as $1=(-i)i$), hence divides all Gaussian integers (so called unit in the ring $\Bbb Z[i]$ of Gaussian integers).

share|cite|improve this answer
OK, I think I got it. Would $((1+2i)(1-2i))^2((1+i)(1-i))^2(3)=300$? – maroon.elephants Nov 6 '12 at 20:01

Odd primes of the form $4k+1$ factor in the Gaussian integers, ones of the form $4k+3$ don't. So $3$ is fine but $5 = 1^2 + 2^2$ should be replaced by $(1 + 2i)(1-2i)$ (or some associate of that). $2$ also factors.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.