# 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

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$18=2\cdot3\cdot3=(1-\sqrt{-17})(1+\sqrt{-17})$ –  yoyo May 6 '11 at 19:25
Now that you have the reputation for it, don't forget to vote up answers you find useful and questions you find interesting (as a participant). Also, don't do it just yet (I'll explain why in a second), but don't forget to eventually accept an answer to your questions once you are satisfied. You should accept whatever answer you found most intersting/helpful, etc, by clicking on the checkmark you will see on the left margin, right under the links to vote up and down (you don't have the reputation to vote down yet). (cont...) –  Arturo Magidin May 6 '11 at 19:45
(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. –  Arturo Magidin May 6 '11 at 19:46

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.
@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). –  Arturo Magidin May 6 '11 at 20:18
@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 $\mathbb{Q}(\sqrt{-d})$. –  Arturo Magidin May 6 '11 at 20:32