Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\def\ord{\operatorname{ord}}$I could not prove the following statement. Could you please help me?

Let $\ord_p a = p-1$. Show that for every $c∈\mathbb Z$, $\gcd(c,p)=1$, there exists $1≤i≤p-1$ such that $c≡a^i \pmod p$

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

Consider the numbers $a^1, a^2, a^3,\dots, a^{p-1}$. None of them is congruent to $0$ modulo $p$. For if $a^i$ is divisible by $p$, then so is $a$, contradicting the fact that $a^{p-1}\equiv 1\pmod{p}$.

Suppose there are two numbers $i$ and $j$, with $1\le i\lt j\le p-1$, such that $a^i\equiv a^j \pmod{p}$. Then $a^{j-i}\equiv 1\pmod{p}$. (This can be proved by cancellation, or more precisely by multiplying both sides by $b^i$, where $b$ is the inverse of $a$ modulo $p$.) Since $1\le j-i\lt p-1$, this contradicts the fact that $a$ has order $p-1$.

So all of our $a^i$ are distinct modulo $p$. That means that their remainders on division by $p$ take on $p-1$ distinct values. That means that the remainders travel, in some order, through all numbers from $1$ to $p-1$. It follows that for any $c\not\equiv 0\pmod{p}$, there is an $i$ with $1\le i\le p-1$ such that $a^i\equiv c\pmod{p}$.

share|improve this answer
    
Oops. Thanks for clarifying :P –  user1296727 Nov 13 '12 at 15:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.