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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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    $18=2\cdot3\cdot3=(1-\sqrt{-17})(1+\sqrt{-17})$2011-05-06
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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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    Ok. So $ \alpha = (1+ \sqrt{-17})(1-\sqrt{-17}) = 18 = 2.3.3 $ satisfies the condition. But what are the thought processes behind deciding to consider $ (1+\sqrt{-17})(1+\sqrt{-17}) $?2011-05-06
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    @user938272: I'll add them.2011-05-06
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    Thank you for the addition, it has helped a lot.2011-05-06
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    The "explanation" is quite misdleading. What it the problem instead asked for 3 or 4 irreducible factors? Your "explanation" worked only due to pure luck / coincidence. It has nothing to do with the underlying structure of the problem.2011-05-06
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    @Bill, Arturo is right - you should explain your way to deal with the problem instead of saying his is bad.. if his actually isn't the best way it will be clear to anyone who understands both.2011-05-06
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    @Arturo, How do we know that if it's not a UFD then $1+\sqrt{-d}$ is a witness? (I hadn't noticed this before but it's true in the examples I know)2011-05-06
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    @quanta: For odd $d\gt 1$, just as I said above: $(1+\sqrt{-d})(1-\sqrt{-d})$ is a multiple of $2$, but neither factor is a multiple of $2$, and $2$ is irreducible because no element can have norm $2$ under these circumstances. So $2$ is an irreducible that is not a prime, hence the ring is not a UFD. If $d$ is even this doesn't work, but you can consider $(2+\sqrt{-d})(2-\sqrt{-d}) = 4+d$, which is divisible by $2$, but neither factor is (and if $d\gt 2$, then no element has norm $2$ either).2011-05-06
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    @quanta: I never claimed I had a "way". It seemed like a good thing to try *first* and see if it worked, before spending too much time thinking about it. It worked. I was asked why I tried it, I said why.2011-05-06
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    @quanta: In the explanation in the prior comment, I'm assuming $d$ is squarefree for the argument. I'm not including $d=1$ and $d=2$, but in those cases $\mathbb{Z}[\sqrt{-d}]$ is a UFD; and of course, if $d\equiv 3\pmod{4}$, then $\mathbb{Z}[\sqrt{-d}]$ is not the ring of integers of $\mathbb{Q}(\sqrt{-d})$.2011-05-06
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    To answer Bill: I never claimed I had a "method"; I was asked what let me to consider that product, and I said what led me to consider that product. If Bill thinks this is not the right way to think about, I have no objection; if he thinks it's the *wrong* way to think about it, he may very well be correct; that it worked "by luck" and not by design, I am also quite happy to grant. To claim that the reply "has nothing to do with the problem" suggests that I what I wrote has nothing to do with what led me to try that product, and I find that an astounding statement, whether intended or not.2011-05-07
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    @Arturo Fair enough. I was concerned that, without explicitly pointing out the element of luck, a reader might be misled into thinking that this approach works generally. I don't consider sheer luck to be a mathematical way of thinking about problems. It's not clear what the point of this problem is, since to really understand what's going on requires deeper results than one meets in first course, e.g. google "factorization elasticity".2011-05-07
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    Arturo, the penultimate word in your answer is "she". I find this strange since from the context that pronoun can only refer to yourself or to Bill, who are the only humans mentioned in the whole coda :-)2011-05-07
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    @elgeorges: http://en.wiktionary.org/wiki/that%27s_all_she_wrote2011-05-07