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I want to know how many $2\times 2$ orthogonal matrices exist over the ring $\mathbb{Z}_n$ or the field $\mathbf{F}_p$. And how many $2\times 1$ orthogonal vectors exist over the ring $\mathbb{Z}_n$ or the field $\mathbf{F}_p$.

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    It makes no sense to ask "how many $2\times 1$ orthogonal vectors exist", because a vector is not "orthogonal" or "not orthogonal". For vectors, orthogonality is a relation *between* vectors, not a property of vectors.2011-08-03
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    The order of the orthogonal group over the field of $q$ elements can be found [in Wikipedia](http://en.wikipedia.org/wiki/Orthogonal_group#Over_finite_fields).2011-08-03
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    @Arturo, does the answer for ${\bf Z}_n$ follow from the answer for ${\bf F}_p$ by the Chinese Remainder Theorem, at least for squarefree $n$?2011-08-04
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    @Gerry: It should for squarefree $n$. Don't know off-hand for arbitrary $n$.2011-08-04
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    An anonymous user suggested an edit to this question that I rejected because it didn't make any sense to me (and was misspelled). Karan, in case this was you, please edit the question under the account that you used in creating it; then it won't require peer review.2011-08-06
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    Ha. Another anonymous suggested edit, 22 hours later.2011-08-07
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    @Karan: Again an anonymous user suggested an edit to this question, this time deleting the second sentence. I approved the edit because as Arturo pointed out that part of the question didn't make sense. Karan, again, if this is you editing your own question, please use the account that you used in creating it.2011-08-07
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    I rolled back the edit because I realized that Jyrki's answer (the only answer so far) referred to that part of the question. It would be helpful if the anonymous editor could comment here on what's going on.2011-08-07
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    @joriki: I am as much in the dark as the rest of you about what is being asked, so don't worry about my answer. I took this as an exercise in guessing what the intended question is. In view of the attempts at editing (may the OP lost his/her login credentials?) it is unlikely that it matches at all with what the OP had in mind, so I delete it for now.2011-08-07
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    @Jyrki, we don't know what OP wants, because OP hasn't been back since posting the question. We do know that what you posted was good mathematics, and may help someone whether it helps OP or not, so please consider undeleting.2011-08-07

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If by orthogonal vectors you mean a collection $S$ of non-zero vectors such that all pairs of distinct vectors $(a,b)$ and $(c,d)$ from the set $S$ satisfy the relation $ac+bd=0$, then the question is a bit subtle, because it is possible for a vector to be `perpendicular to itself'. In $F_p^2$ this happens, iff $p\equiv 1\pmod{4}$, because the equation $$a^2+b^2=0\Leftrightarrow (a/b)^2=-1 $$ has non-trivial ($ab\neq0$) solutions, iff $-1$ has a square root in $F_p$. In those cases we can include all (non-zero) multiples of such a vector to get a total of $p-1$ non-zero vectors with pairwise vanishing inner products. This is also the maximum. A way to see this is as follows: if an orthogonal set contains a (non-zero) vector $\vec{v}$ perpendicular to itself, then all the vectors $\perp\vec{v}$ must be scalar multiples of $\vec{v}$. This is because $\vec{v}$ cannot be orthogonal to the entire space, so the subspace $\langle\vec{v}\rangle^\perp$ has dimension one, and must be spanned by $\vec{v}$. If the orthogonal set has no such vectors as elements, then it is linearly *IN*dependent by the usual proof for "orthogonality $\Rightarrow$ linear independence", and has size at most 2.

If $p\equiv3 \pmod 4$, then all the non-zero vectors of $F_p^2$ have non-zero inner products with themselves, and again, the set is linearly independent.

If we start varying the (non-degenerate) bilinear form, then the answers vary, but in a 2-dimensional space these are the basic cases.

If you mean something else, please edit your question accordingly. I don't know what happens with the ring $\mathbf{Z}_n$. In line with the comments by Arturo and Gerry, I think that CRT will help in the case of a square-free $n$.

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    @user2154420: The congruence $x^2\equiv-1\pmod p$ has a solution if and only if $p\not\equiv3\pmod4$. This is a fact from elementary number theory.2015-02-10