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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

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    (cont) You may want to wait before accepting answers to your recent questions because questions without an accepted answer tend to attract a *bit* more attention, and you may yet receive other answers that prove to be better/more interesting/more informative/ etc. Waiting a day or so is not amiss, but you'll want to *eventually* accept an answer to each of your questions once you are satisfied.2011-05-06

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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.

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    @elgeorges: http://en.wiktionary.org/wiki/that%27s_all_she_wrote2011-05-07