# Congruence classes and prime divisors

i am very sorry i have an attempted proof but the website will not let me post it. i can't seem to figure this proof out in a way i am happy with. if someone could help i would much appreciate that.

Let p be a prime number. Prove that in ${\bf Z}_p$, if $[a] = [a]^{-1}$, for some $[a]$ does not equal [0], then $[a] = [1]$ or $[a] = [p - 1]$.

Seems it may let me add it now that its already posted :)

So what i did is i said all the class's can be labeled by defining them an [p+k] then a set k to be an element of the interval $(-p,p)$ that gives me every congruence class then i said that since a^-1 = a that $a*a$= $[1]$

using that i took $(p+k)^2$ to be $p^2$ + $k^2$ + 2kp next $k^2$ = -2kp + $p^2$ next p divides Right hand side so p divides left hand side put gcd $(p,k^2)$ = 1 thus $k^2$ = +-p or +-1 well i defined all congruence class with so k is bigger then -p and so k less than p and an integer so k = +-1 are the only solutions therefor $[p+1]$ and $[p-1]$ = $[1]$ and $[p+1]$=$[1]$ that's all i got any advice?

-
I did a little editing to improve the formatting, but I left some for you to do if you'd like. See what I did, you'll get the idea. But please don't undo what I've done, unless you can do it better. –  Gerry Myerson Jan 31 '13 at 4:48
Do they have those little dots people put at the end of a sentence where you live? –  Gerry Myerson Jan 31 '13 at 11:06

Suppose $[ab]=[a][b]=[0]$ in $\mathbb{Z}_p$. This means $p\mid ab$, so since $p$ is prime, either $p\mid a$ or $p\mid b$, in which case $[a]=[0]$ or $[b]=0$. Hence $[a][b]=[0]$ if and only if either $[a]=[0]$ or $[b]=[0]$.

You noticed $[a]^2=[1]$. This is a good start, it follows that $[a]$ is a solution to $x^2-[1]=0$ in $\mathbb{Z}_p$. But $x^2-[1]=(x-[1])(x+[1])$. So $x^2-[1]=[0]$ if and only if $x-[1]=[0]$ or $x+[1]=[0]$. What does that tell you about your $[a]$?

-
that $[a]$ must be $[p+1]$ or $[p-1]$ Im sorry i feel like im missing something obvious here ( i have been doing some form of math or anther for 16 hours or so already today) i understand that and i believe its true but i cant seem to actually prove it that its true –  Faust7 Jan 31 '13 at 5:09
@Faust7 So since $[a]^2-[1]=[0]$, that implies $([a]-[1])([a]+[1])=[0]$. By the first paragraph, either $[a]-[1]=[0]$ or $[a]+[1]=[0]$. So either $[a]=[1]$, or $[a]=[-1]=[p-1]$. And you're done. –  BDub Jan 31 '13 at 5:11
Omg that's so obvious now that you write it that way. been working on this for hours! Thank you so much for your help! –  Faust7 Jan 31 '13 at 5:13
No problem. ${}$ –  BDub Jan 31 '13 at 5:16

Hint: Knowing $[a] = [a]^{-1}$, and $[a]\neq [0]$, compute $[1] + [p-1]$ and $[p-1] + [1]$?

What is the only elements in $\mathbb{Z}_p$ such that $[a] = [a]^{-1}$?, i.e

Are there any other elements $[b] \neq [a] \in \mathbb{Z}_p$ for which [b] = [b]^{-1}? Why not?

-
well no im fairly confident of that as $p^2$ + $k^2$ + 2kp mod p is just $k^2$ hence we have the gcd = ($k^2$,p) = 1 so $k^2$ must be +-1 the problem being is i have no way to prove it :) –  Faust7 Jan 31 '13 at 4:58
+1. Is it the floor function? –  Babak S. Jan 31 '13 at 6:36
sorry those are all meant to be congruence class's –  Faust7 Jan 31 '13 at 16:54