Let $p$ be an odd prime and $d$ be a natural number such that $d \mid (p-1)$. For $a \in Z$ coprime with p, show that $x^d \equiv a$ mod $p$ if and only if $a^\frac{p-1}{d} \equiv 1$ mod $p$. I have no idea about this $x^d \equiv a$ mod p.
Asked
Active
Viewed 175 times
3
-
Also, what is $x$? I think you're missing a quantifier. Did you mean "for some $x$"? – quasi Mar 08 '17 at 06:20
-
My search with buzzwords "d" and "p-1" came up with this hit among others. Keep looking and learning! – Jyrki Lahtonen Mar 08 '17 at 06:52
-
This is even better! – Jyrki Lahtonen Mar 08 '17 at 06:53
-
@quasi updated again. i combined 2 questions. lel. Sry for the inconvenience – user3591471 Mar 08 '17 at 07:05
-
Ok, it's better, but you still need to specify a quantifier for $x$. Presumably, it was intended as "For some $x$ ...", or equivalently, "there exists $x$ such that ..." – quasi Mar 08 '17 at 07:18
-
@quasi i think it is "there exists some x such that..". thats the whole question. – user3591471 Mar 08 '17 at 07:22
-
Looks better. Enjoy the site! Deleting some of my older comments as they have served their purpose :-) – Jyrki Lahtonen Mar 08 '17 at 08:45
1 Answers
3
Assumptions:
- $p$ is prime.
- $d$ is a positive integer such that $d \mid (p-1)$.
- $a$ is an integer, $a$ not a multiple of $p$.
To be shown:
- $x^d \equiv a \pmod{p}$, for some integer $x$, if and only if $a^\frac{p-1}{d} \equiv 1 \pmod{p}$.
First we show $x^d \equiv a \pmod{p}$, for some integer $x$, implies $a^\frac{p-1}{d} \equiv 1 \pmod{p}$.
Thus, suppose there is an integer $x$ such that $x^d \equiv a \pmod{p}$.
Since $a,p$ are relatively prime, $x^d \equiv a \pmod{p}$ implies $x,p$ are relatively prime.
Then
\begin{align*} a^\frac{p-1}{d} &\equiv (x^d)^\frac{p-1}{d} \pmod{p}\\[6pt] &\equiv x^{p-1} \pmod{p}\\[6pt] &\equiv 1 \pmod{p}\\[6pt] \end{align*}
as was to be shown.
Next we show $a^\frac{p-1}{d} \equiv 1 \pmod{p}$ implies $x^d \equiv a \pmod{p}$, for some integer $x$.
Thus, suppose $a^\frac{p-1}{d} \equiv 1 \pmod{p}$.
Since $p$ is prime, there exists a primitive element, mod $p$. Let $y$ be such an element.
Since $y$ is a primitive element, mod $p$, the multiplicative order, mod $p$, of $y$ is $p-1$.
It follows that the $p-1$ elements $1,y,y^2,...y^{p-2}$ represent, in some order, all nonzero residues, mod $p$. In particular, there is an integer $k \in \{0,1,...,p-2\}$ such that $y^k \equiv a \pmod{p}$.
Then \begin{align*} &a^\frac{p-1}{d} \equiv 1 \pmod{p}\\[6pt] \implies\; &(y^k)^\frac{p-1}{d} \equiv 1 \pmod{p}\\[6pt] \implies\; &(p-1)\; \text{ divides} \left(k{\small{\left(\frac{p-1}{d}\right)}}\right)\\[6pt] \implies\; & {\frac {k{\large{\left(\frac{p-1}{d}\right)}}} {p-1}} \in \mathbb{Z}^+ \\[6pt] \implies\; &{\frac{k}{d}} \in \mathbb{Z}^+\\[6pt] \implies\; &k = jd,\text{ for some }j \in \mathbb{Z}^+\\[6pt] \end{align*}
Let $x = y^j$. Then
\begin{align*} x^d &\equiv (y^j)^d \pmod{p}\\[6pt] &\equiv y^{jd} \pmod{p}\\[6pt] &\equiv y^k \pmod{p}\\[6pt] &\equiv a \pmod{p}\\[6pt] \end{align*}
thus, we have an integer $x$ such that $x^d \equiv a \pmod{p}$, as was to be shown.
This completes the proof.
quasi
- 58,772