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Let $n>1$ a natural number for which there does not exist an $m \in \mathbb{N}$ such that $m^2|n$.

Prove that the only invertible elements of the integral domain $\mathbb{Z}[i\sqrt{n}]$ are $-1$ and $1$.

$\mathbb{Z}[i\sqrt{n}]=\{a+bi \sqrt{n}|a,b \in \mathbb{Z}\}$

Can someone help me with this?

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    Consider the product of two arbitrary elements $a + bi \sqrt{n}, c + di \sqrt{n}$ and assume it is equal to $1$ and find conditions on $a,b,c,d$. – 2017-02-15
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    Now, you're aware that for $n \equiv 3 \pmod 4$ you also have to consider numbers like $$\frac{1 + \sqrt{-3}}{2}?$$ If you intend to consider such cases, you'd only need to revise to $n > 3$, however. – 2017-02-15

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If $a+bi\sqrt{n}$ is an invertible element of $\mathbb{Z}[i\sqrt{n}]$, then its norm must be an invertible element in $\mathbb{Z}$, so must be $\pm 1$.

And the norm of $a+bi\sqrt{n}$ is $(a+bi\sqrt{n})(a-bi\sqrt{n})=a^2+b^2n$. If this is equal to $\pm 1$, what are the possible values of $a$ and $b$?

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    why the norm must be invertible? – 2017-02-15
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    Assume $x$ is invertible, with inverse $y$. Then as the norm is multiplicative, we have $1=N(1)=N(xy) = N(x)N(y)$, so $N(x)$ is invertible – 2017-02-15
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In general :To determine invertible elements in $\mathbb Z[\sqrt n]$ where $n$ free of square::

Let be $u=a+b\sqrt n $ an invertible element in $\mathbb Z[\sqrt n]$, then exists $v=c+d\sqrt n \in \mathbb Z[\sqrt n] $ where $uv=1$ , then $N(uv)=N(u)N(v)=1\Rightarrow N(u)=1$. we have three cases:

$1)$ $n=-1 \Rightarrow N(u)=a^2+b^2=1\Rightarrow a=\pm 1, b=0$ or $a=0, b=\pm 1$, then $u=\pm 1$ or $u=\pm i$

$2)$ $n<-1 \Rightarrow N(u)=a^2-nb^2=1 \Rightarrow a=\pm 1, b=0$ ,then $u=\pm 1$

$3)$ $n>0 \Rightarrow N(u)=a^2-nb^2=1$ has infinitely many of solutions , then invertible element is $\{-1,+1 \}$ and infinite many of elements.

for example for case $(3)$ in $\mathbb Z[\sqrt3]$ we have $u=2-\sqrt3$ is invertible element because $(2-\sqrt3)(2+\sqrt3)=1$

Hint: if $n<0$ ,then $\mathbb Z[\sqrt n]$ is PID if and only if $n=-1$ or $n=-2$

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    How can $\mathbb{Z}[\sqrt{n}]$ not be an integral domain ? It is a subring of a field, it must be integral. You must've meant something else – 2017-02-15
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    @Max I think he refers to *euclidean* domains. – 2017-02-15
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    soory, I edit my answer – 2017-02-15
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    I mean $\mathbb Z[\sqrt n]$ is PID when $n<0$ and $n=-1$ or $n=-2$ – 2017-02-15
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    @Xam ,thank you very much – 2017-02-15