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Let $X=a+\sqrt{-5} b:a, b \in \mathbb Z$. An element $x \in X$ is called special if there exists $y \in X$ such that $xy=1$. Then the question is to find out the number of special elements in $X$

I tried by letting $(a_1+\sqrt 5 ib_1)(a_2+\sqrt 5 ib_2)=1$ which led me to a equation involving four variables. I am unable to proceed further. Any help shall be highly appreciated. Thanks.

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If $(a_1+\sqrt{5}ib_1)(a_2+\sqrt{5}ib_2)=1$ then $|a_1+\sqrt{5}ib_1|^2|a_2+\sqrt{5}ib_2|^2=1$, that is $(a_1+5b_1^2)(a_2^2+5b_2^2)=1$. Now each of $a_j^2+5b_j^2$ is a nonnegative integer.

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  • $\begingroup$ Thanks for your answer. So that implies $b_j=0$ and $a_j=1, -1$ implying two special elements $1, -1$ $\endgroup$
    – Navin
    Commented Apr 27, 2017 at 4:19
  • $\begingroup$ @navinstudent Exactly! $\endgroup$
    – Angina Seng
    Commented Apr 27, 2017 at 4:19

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