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I tried to write $3 = (a+bi \sqrt{5})(c+di \sqrt{5})$ and got:

  • $ac -5bd = 3$

  • $ad + bc = 0$

But I couldn't show that one of the two terms is identity.

How to proceed?

Douglas Souza
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    Are you familiar with the norm map $N \colon \mathbb{Z}[\sqrt{-5}] \to \mathbb{Z}$? – Alex Wertheim Nov 03 '19 at 23:57
  • I think that no. – Douglas Souza Nov 03 '19 at 23:58
  • With norms, maybe? – Oscar Lanzi Nov 03 '19 at 23:59
  • I will try read about this in a book. – Douglas Souza Nov 04 '19 at 00:00
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    The norm map is a valuable tool for answering this kind of question. In your case, it can be defined very explicitly: it is the map $N \colon \mathbb{Z}[\sqrt{-5}] \to \mathbb{Z}$ defined by $N(a+b\sqrt{-5}) = a^{2}+5b^{2}$. (It is just the restriction of the usual complex norm to $\mathbb{Z}[\sqrt{-5}]$.) A start towards your question would be to show that $N$ is multiplicative, i.e. $N(xy) = N(x)N(y)$ for any $x, y \in \mathbb{Z}[\sqrt{-5}]$. The next step would be characterizing elements $x \in \mathbb{Z}[\sqrt{-5}]$ such that $N(x) = 1$. – Alex Wertheim Nov 04 '19 at 00:02
  • To make the introduction of the norm maybe seem less opaque: the main obstruction to this question is that factorization in $R := \mathbb{Z}[\sqrt{-5}]$ is not so nice. In particular, $R$ is not a UFD; your question is a step on the way to showing this, by observing that $6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$. Introducing the norm $N$ allows us to transport factorization questions in $R$ to related factorization questions in $\mathbb{Z}$, where we have a very powerful theory of factorization. – Alex Wertheim Nov 04 '19 at 00:05

2 Answers2

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The equations you have are a linear system for $(c,d)$. The solution is $$ c = \frac{3a}{a^2+5b^2}, \quad d = -\frac{3b}{a^2+5b^2} $$ We may assume that $a\ge0$ and so $c\ge0 $.

If $a=0$ then $c=0$. But then $3=-5bd$, which cannot happen in $\mathbb Z$.

Thus $c\ge 1$ and so $3a-a^2 \ge 5b^2$. The maximum value of $3x-x^2$ is $2.25$ and so $b^2 \le 0.45$. Since $b$ is an integer, we must have $b=0$ and so $d=0$. But then $ac=3$. Thus, $a+bi \sqrt{5}=1$ or $c+di \sqrt{5}=1$, that is, one of the factors is a unit.

lhf
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In disguise, the ring is actually $ R = \mathbb{Z}[\sqrt{-5}]$. To show that 3 is irreducible in $R$, we need to show that if $3=\alpha \beta$, then either $\alpha$ or $\beta$ is a unit. Define a norm function $$N: R \to \mathbb{Z}$$ This is equivalent to showing that either $N(\alpha) = 1$ or $N(\beta) = 1$. A beautiful discussion on how we come to this can be found here. Unless given a different norm, it is acceptable to use the field norm given by $$N: \mathbb{Q}[\sqrt{-5}] \to \mathbb{Q}\quad \text{ defined by } \quad N(a+b\sqrt{-5}) = a^2 +5b^2$$

Notice that the field norm is multiplicative, that is $N(AB) = N(A)N(B)$. So the work comes down to solving the following equation: $$9 = N(\alpha) = N(\alpha)N(\beta)$$

Write $\alpha = a+ b\sqrt{-5}$ and $\beta = c + d\sqrt{-5}$, then $N(\alpha) = a^2+5d^2$ and $N(\beta) = c^2+5d^2$.

Doing the algebra comes down to: $$9 = (a^2+ 5b^2)(c^2+5d^2) $$

WLOG, the possible options for $(\alpha, \beta)$ are: $(1,9), (3,3)$. It cannot be $(3,3)$. So, that leaves us with one option: $N(\alpha) = 1$ (or $N(\beta)= 1$).

Sajid Bin Mahamud
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