# Write __ as a product of primes in the ring of Gaussian integers?

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:

$$5\times5\times3\times2\times2=300$$

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'.

-
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

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).
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.