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I have to find the units in $\mathbb{Z}[\sqrt{-7}]$ so take $x,y\in\mathbb{Z}[\sqrt{-7}]$ then x has the form $a+b\sqrt{-7}$ and y has the form $c+d\sqrt{-7}$ for arbitrary x and y.

If $(a+b\sqrt{-7})(c+d\sqrt{-7})=1$ then $(a-b\sqrt{-7})(c-d\sqrt{-7})=1$ and so:

$1=(a+b\sqrt{-7})(c+d\sqrt{-7})(a-b\sqrt{-7})(c-d\sqrt{-7})=1$ which gives $1=(a^2+7b^2)(c^2+7d^2)$ so $b=d=0$ and $1=a^2c^2$ so either $a=c=1$ or $a=c=-1$ and these are the only units in $\mathbb{Z}[\sqrt{-7}]$

Is the above correct? (Sorry for so many of these type of questions)

Thanks very much for any help

If $(a+b\sqrt{-7})(c+d\sqrt{-7})=1$ then $(a-b\sqrt{-7})(c-d\sqrt{-7})=1$ and so:

$1=(a+b\sqrt{-7})(c+d\sqrt{-7})(a-b\sqrt{-7})(c-d\sqrt{-7})=1$ which gives $1=(a^2+7b^2)(c^2+7d^2)$ so $b=d=0$ and $1=a^2c^2$ so either $a=c=1$ or $a=c=-1$ and these are the only units in $\mathbb{Z}[\sqrt{-7}]$

Is the above correct? (Sorry for so many of these type of questions)

Thanks very much for any help

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