# Irreducible elements in $\mathbb{Z}[\sqrt{5}]$

In $$\mathbb{Z}[\sqrt{5}]$$, we defnie the norma $$N(a+b\sqrt{5})=|a^2-5b^2|$$. I'm trying show that if $$4\leq N(x)<16$$, then $$x$$ is irreducible.

We know that an element is a unity is equivalent to tha its norm is $$1$$. I have try and I think is enougt prove that for all $$x$$, $$N(x)\notin \{2,3\}$$. Because if there exists an $$x$$ such that $$N(x)\in \{2,3\}$$, then $$x$$ is not a unity and $$N(x^2)\in \{4,9\}$$. So $$4\leq N(x^2)<16$$ and $$x^2$$ is not irreducible. Some have a suggestion for me? Thanks.

• It is false that being a unit is equivalent to having norm $1$, because your function $N$ can take negative values. Specifically, $N(2+\sqrt{5}) = 4-5 =-1$, and $2+\sqrt{5}$ is a unit, since $(2+\sqrt{5})(-2+\sqrt{5}) = -4+5 = 1$. I cannot really understand the rest of your argument. Commented May 20 at 19:53
• I already corrected the definition of the norm. What I know is enought for show the result is that $N(x)\notin \{2,3\}$ for all $x$. Do you know if that is thrue? Commented May 20 at 20:13

Throughout this argument, we will use the fact that the norm is multiplicative, that is, $$N(x)N(y)=N(xy)$$ (I implore you to check this).

A unit is an element $$x$$ with $$N(x)=1$$.

An element $$x$$ is irreducible if every factorisation $$x=ab$$ forces $$a$$ or $$b$$ to be a unit.

As you mentioned, there are no elements in $$\mathbb{Z}[\sqrt5]$$ of norm $$2$$ or $$3$$. This follows from the fact that we cannot solve $$x^2=2$$ or $$3$$ (mod$$5$$) (check this too!).

Any element of prime norm is irreducible, by the multiplicativity of the norm, so that does $$N(x)=5,7,11,13$$.

If $$N(x)=4,6,8,10,14$$, then the factorisation of $$x$$ must include an element of norm $$2$$, which is impossible, so these are irreducible.

If $$N(x)=9,15$$, then the factorisation of $$x$$ must include an element of norm $$3$$, which is again impossible.

Finally, if $$N(x)=12$$, the factorisation of $$x$$ must include an element of norm $$2$$ or $$3$$, which is impossible.

• What about in general? How one can find all irreducible elements of that ring? Commented May 20 at 21:42
• @nozalp10 With difficulty. This is a particularly tricky ring to work with for several reasons. It is not a UFD and it is not a ring of integers (the ring of integers of $\mathbb{Q}(\sqrt 5)$ is much larger). Also this ring has infinitely many units. Commented May 21 at 6:28