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.

Question: "Show that if $p$ is prime and $\gcd(d,p-1) = 1$, then every positive integer less than p is congruent modulo $p$ to the $d$-th power of some other integer."

I understand that this is related with primitive roots but I get confused when trying to explain the theory. I was wondering if someone would be point me in the general direction? Thanks!

share|improve this question
add comment

3 Answers

Let $a$ be a positive integer less than $p$.

Since $d$ and $p-1$ are relatively prime, there exist integers $x$ and $y$ such that $dx+(p-1)y=1$. Without loss of generality we may assume that $x$ is positive. Putting $z=-y$, we get $dx=1+(p-1)z$. Since $a^{(p-1)z}\equiv 1\pmod{p}$, we have $$a\equiv a^{1+(p-1)z}\pmod{p}.$$ But $$a^{1+(p-1)z}=a^{dx}=(a^x)^{d}.$$

share|improve this answer
add comment


1) If we have a finite cyclic group $\,G=\langle x\,\rangle\,$ of order $\,n\,$, then

$$G=\langle x^k\rangle\Longleftrightarrow (n,k)=1$$

2) The multiplicative group $\,\left(\Bbb Z/p\Bbb Z\right)^*\,$ has order $\,p-1\,$

3) Your claim is trivially true for zero...

share|improve this answer
add comment

Let $r$ be a primitive root $\mod p$. Since $(d,p-1)=1 \Rightarrow r^d$ is a primitive root $\mod p$. Therefore $\{1,r^d,(r^d)^2,\ldots,(r^d)^{p-1}\}\equiv\{1,2,3,\ldots,p-1\} \pmod p$.

share|improve this answer
add comment

Your Answer


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.