Give an element of $ \mathbb{Z}[\sqrt{-17}] $ that is a product of two irreducibles and also a product of three irreducibles Give an element of $ \mathbb{Z}[\sqrt{-17}] $ that is a product of two irreducibles and also a product of three irreducibles.
My thoughts so far:
Using the multiplicative norm $ N(a + b\sqrt{-17}) = a^2 + 17 b^2 $, we see that the 
units are precisely 1, -1. I can also see that there are no elements of norm $ 2,3,5,6,7,8,10,11,12,13,14,15... $. So if an element has norm 4 or 9 for example, then it is irreducible.
I don't really know where to go from here. 
Any help appreciated. Thanks
 A: Hint. How much is $(1+\sqrt{-17})(1-\sqrt{-17})$? Can you express it as a product in a different way? Are all the factors you have in either factorization irreducible?
Added. Why consider this product? If $\mathbb{Z}[\sqrt{-d}]$, with $d$ an odd squarefree integer greater than $1$, is not a UFD, then $1+\sqrt{-d}$ will be part of a witness to this fact. You have $(1+\sqrt{-d})(1-\sqrt{-d}) = d^2+1$ is divisible by $2$, but neither $1+\sqrt{-d}$ nor $1-\sqrt{-d}$ are divisible by $2$ in $\mathbb{Z}[\sqrt{-d}]$. Also, $2$ is irreducible, because $a^2+db^2 = 2$ has no solutions when $d\gt 2$, so that shows that $2$ is an irreducible that is not a prime (since it divides a product but neither of the factors). So $1+\sqrt{-d}$ and $1-\sqrt{-d}$ are usually good sources of examples of things going wrong with factorizations into irreducibles in $\mathbb{Z}[\sqrt{-d}]$, when such things do indeed go wrong.
Coda. Bill Dubuque will no doubt give you a general way to approach this kind of problem once he gets around to it. As I noted in the comments, the above was not meant to be a "method", or an "algorithm", or a "solution", but merely the thought process that led me to consider that product before expending too much effort dissecting this particular problem. Since it immediately gave a solution to the desired problem, that was all she wrote.
