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Prove that if $w$ is an extended integer, that is $ w \in \mathbb{Z[\sqrt{3}]}$ (of the form $a+b\cdot\sqrt{3}$) with $|N(w)|>1$, then $w$ can be written as a product of primes in the extended integers.

Proof:

We proceed with complete induction.

For the base case, we take any $w$ for which $|N(w)| = 2$ and, since $2$ is prime we know it can be written as a product of primes and since $N(w) = w\cdot \bar{w}$, then $w$ is also prime and can therefore be written as a product of primes hence this case is finished.

For the inductive step, we assume that if $w$ is an extended integer such that $|N(w)| = n$, where $1< n \leq k$, then $w$ can be written as a product of extended integer primes.

We now need to show that, for $k+1$, $w$ can be written as a product of extended integer primes. We are left with two cases. In case 1, we say that $k+1$ is prime, hence $w$ can be written as a product of extended integer primes. In case 2, we say that $k+1$ is not prime. If $k+1$ is not prime, we can say that there exist integers, $s$ and $t$, such that $k+1 = s\cdot t$. We then say that $1<s<k+1$ and $1<t<k+1$ which implies that $1<s\leq k$ and $1<t\leq k$. However, both $s$ and $t$ are within the bounds that we assumed to be true in the induction hypothesis and therefore can both be written as products of primes. Since, with $k+1 = s\cdot t$, we have two products of primes being multiplied and it follows that $k+1$ is also a product of primes, hence $w$ can also be written as a product of primes and therefore the statement is proven.

I mainly just want to know if the above proof is correct/makes sense?

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  • $\begingroup$ There is no question here. $\endgroup$ – Matthew Conroy Apr 5 '17 at 0:31
  • $\begingroup$ I was under the impression that the tag proof verification meant that the proof would be checked? $\endgroup$ – Deez1133 Apr 5 '17 at 0:55
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    $\begingroup$ Well, some human interaction always helps: tagging with proof-verification does not automatically mean that your proof will be checked. Instead, it is nice to actually ask for help, perhaps saying what part(s) of the proof you are unsure about, etc. I see that you have added a question: that's great. Cheers! $\endgroup$ – Matthew Conroy Apr 5 '17 at 1:34
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    $\begingroup$ Understood, you're totally right, thank you for the insight and advice. $\endgroup$ – Deez1133 Apr 5 '17 at 2:39

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