0
$\begingroup$

$\newcommand{\ord}{\operatorname{ord}}$

Let $p$ is any prime and $(a,p)=1$

(i)If $\ord_pa=2,$

We know there can be $\phi(2)=1$ element that belongs to order$=2$.

(ii)If $\ord_pa=3,a^3\equiv 1\pmod p\implies a^2+a+1\equiv 0$ as $a≢-1\pmod p$

So, $-(a+1)\equiv a^{-1}$

We know, $\ord_ma=d, \ord_m(a^k)=\frac{d}{(d,k)}$ (Proof @Page#95)

So, $\ord_p(a^{-1})=\ord_pa$

We know there can be $\phi(3)=2$ elements that belongs to order$=3$.

If $a$ is one, $-(a+1)$ is the other.

(iii)If $\ord_pa=4,a^4\equiv 1\pmod p\implies a^2+1\equiv 0$ as $a^2≢1\pmod p$

So, $-a=a^{-1}$

We know there can be $\phi(4)=2$ elements that belongs to order$=4$.

If $a$ is one, $-a$ is the other.

(iii) If $\ord_pa=6$, we have proved, if $a$ is one, $1-a$ is the other.

So, in all cases, we find a linear expression of $a$ for the other element.

My question is how to generalize this for any order with $\pmod p$, then with $\pmod {p^n}$ and $\pmod m$ (where $m,n$ are positive integers).

  • 0
    Do you really mean *linear* expression? That comes about because of because of the smallness of the orders considered. And anyway $a^{-1}$ (for order $4$) does not really qualify as linear.2012-10-26
  • 0
    @AndréNicolas, by linear,I meant $Aa+B$ where $A,B$ are integer constants. For order $4,$ it's $-a$, not $a^{-1}$.2012-10-27
  • 0
    For general order $d$, there will be many elements of order $d$. So do you mean a family of integer constants? Your observations about small orders can be extended to $p^n$ or $2p^n$ for odd $p$.2012-10-27
  • 0
    @AndréNicolas, ya a set of pair of integer constants. So, there is no known standard formula for this, right? But, how to extend the observation for the other orders, whose $\phi>2$ for $p^n$ and $2p^n$?2012-10-27
  • 0
    I have not seen anything of the kind for general orders, all the standard stuff is in terms of *powers* of $a$.2012-10-27
  • 0
    @AndréNicolas, thanks for your feedback.2012-10-27
  • 0
    For example, suppose $m$ has a primitive root, and there is an element of order $4$. Then there are $\varphi(4)=2$ such elements. And in general if $a$ has order $4r$, so does $-a$. That extends what you wrote about order $4$ to everybody with a primitive root, thus to $p^n$ and $2p^n$.2012-10-27
  • 0
    +1 for your nice question. May I ask you to have a look at one of my questions here? Thanks2013-03-09
  • 0
    @BabakS., where is your question?2013-03-09
  • 0
    @labbhattacharjee: http://math.stackexchange.com/q/311273/8581. It is about multivariable functions which we encounter always in Modern Calculus. Thanks for the time.2013-03-09

0 Answers 0