Prove that if there exists elements $z$ and $x$ of $\textbf{Z}[\sqrt{3}]$ such that $N(z)=ab$ and $N(x)=a$ where $a$ and $b$ are integers, then there exists a $y \in \textbf{Z}[\sqrt{3}]$ such that $N(y)=b$.

Background statements:

The norm of an element of $\textbf{Z}[\sqrt{3}]$ referred to as $z = a+b\sqrt{3}$ where $a$ and $b$ are integers is defined to be $N(z) = a^2 - 3b^2$.

We already proved that the norm is always an integer.

  • 1
    $\begingroup$ What if you could divide elements in $\mathbb{Z}[\sqrt{3}]$? What would you do then? Even if you couldn't divide an arbitrary element by another arbitrary element, maybe you can do it for the particular elements you might be interested in. $\endgroup$ Jun 30, 2017 at 2:31
  • $\begingroup$ I mean in the sense of dividing $z$ by $x$ under the hypothesis that $N(z)$ is divisible by $N(x)$. $\endgroup$ Jun 30, 2017 at 3:09
  • $\begingroup$ I claim that if $N(x)$ divides $N(z)$, then $z/x$ (which, as you noted, is a priori in $\mathbb{Q}[\sqrt{3}]$) is also in $\mathbb{Z}[\sqrt{3}]$. $\endgroup$ Jun 30, 2017 at 3:51
  • $\begingroup$ Is $\Bbb Z[\sqrt3]$ a unique factorization domain? Then look at the prime factorizations of $x$ and $z$. (Incidentally, the conjecture in your last comment is false; $\frac{19}{13}+\frac8{13}\sqrt3$ has norm $1$. That's probably not the smallest counterexample.) $\endgroup$ Jun 30, 2017 at 4:08
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    $\begingroup$ @Alqatrkapa $N(19+8\sqrt3)$ divides $N(13)$ (indeed, they are equal), but $(19+8\sqrt3)/13\notin\Bbb Z[\sqrt3]$. (However, Typhon should notice that the prime factorizations of these numbers are $(4+\sqrt3)(4+\sqrt3)$ and $(4+\sqrt3)(4-\sqrt3)$, respectively.) $\endgroup$ Jun 30, 2017 at 4:17

2 Answers 2


There is nothing wrong with Akiva Weinberger's post. However, since this post was made in the context of a somewhat more rigorous and formal setting, I'm posting a formal proof that I believe would most likely be considered acceptable in that setting.

Let $z$ and $y$ be extended integers such that for integers $a > 1$ and $b > 1$, $N(z) = ab$ and $N(x) = a$. Then there exists an extended integer $y$ such that $N(y) = b$.


First some background information. Since $N(z) > 1$ we can write $z = p_1 p_2 p_3 \cdots p_n$ where all $p$'s are prime extended integers and $n$ is the unique number of prime factors. By the properties of the conjugate $\overline{z} = \overline{p_1} \overline{p_2} \overline{p_3} \cdots \overline{p_n}$.

Since the number of prime factors in an extended integer is unique we will perform complete induction along the number of factors in $x$.

For the base case we assume that $x$ is a prime extended integer. Therefore $\overline{x}$ is also a prime extended integer. Since $N(x) | N(z)$ we can state that there is a $j \in Z$ such that $x | p_j$ and $0 < j \leq n$. Therefore since they are both prime numbers we can say that $N(x) = N(p_j)$. We can assume without loss of generality that $j = 1$. Therefore $\frac {N(z)}{N(x)} = N(p_2) N(p_3) \cdots N(p_n)$. By the properties of the norm, there exists an extended integer $y$ such that $N(y) = N(p_2) N(p_3) \cdots N(p_n) = b$. This completes the base case.

For the inductive step, we assume that the statement is true whenever $x$ has $m$ prime factors and seek to prove it when $x$ has $m+1$ extended integer prime factors. First, we can write $x = q_1 q_2 q_3 \cdots q_{m+1}$. Due to factoring we know that $N(q_{m+1}) | N(z)$. Therefore there exists an extended integer $w$ such that $N(w) = N(q_1) N(q_2) N(q_3) \cdots N(q_{m}) b$. Since $q_1 q_2 q_3 \cdots q_{m}$ has $m$ extended integer prime factors than by induction there exists an extended integer $y$ such that $y = \frac {N(z)}{N(x)} = b$. This completes the inductive step.

Therefore by induction the proof is complete for all extended integers.


It is quite possible that I could do it without induction, but that is the method which felt natural to me. I think it also makes the proof easier to read, which is by far the second most important component (the first is rigorous justification). Credit to Akiva Weinberger for thinking of the splendid method of factorization. I had been trying to do it algebraically through division for some time with complete failure. I see now that factoring was both the intuition that led to me thinking of the conjecture as well as the only simplistic argument that can sufficiently prove it.

For those who may be curious, I am accepting this answer as it is the one that best fits the style of proofs the rest were constructed in.


$x\bar x$'s prime factorization is a product of conjugate pairs; $z\bar z$, whose prime factorization is a subset of $x\bar x$'s, is also a product of conjugate pairs. The quotient $\frac{x\bar x}{z\bar z}$ is thus the product of conjugate pairs as well (inductively by canceling out one conjugate pair at a time from the numerator and denominator), which means it's the norm of something (choose one of each conjugate pair).


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