Solutions to a'th root of 1 mod p if gcd(a, p-1)=1

Question

Let $p$ be prime, and $a$ be an integer such that $gcd(a,p) = gcd(a, p-1)=1$. Show that the equation $x^a \equiv 1$ (mod $p$) has exactly one solution.

Answer 1

Since, $GCD(a, p-1) = 1$, by the Extended Euclid's algorithm we can write $ta + s(p-1) = 1$ for some integers $s$ and $t$.

If $x^a \equiv 1 \text{ (mod p)}$ and by Fermat's Little Theorem, $x^{p-1} \equiv 1 \text{ (mod p)}$. Then $x \equiv \left(x^{p-1}\right)^s \cdot \left(x^{a}\right)^t \equiv 1 \cdot 1 \equiv 1 \text{ (mod p)}$. This means the only solution to the equation is the trivial solution, where $x \equiv 1 \text{ (mod p)}$.

Answer 2

Claim 1: If $x$ is a multiple of $p$ then there is not solution.

Reason: If $p|x$ then $x \equiv 0 \pmod p$ and $x^a\equiv 0 \pmod p$.

Claim 1: Fermat's little theorem says that if $x$ and $p$ are relatively prime (in other words if $p\not \mid x$. then $x^{p-1}\equiv 1 \pmod p$.

Reason: It's in a book.

As a corollary the order of $x$ (the smallest positive integer so that $a^k\equiv 1 \pmod p$) must divide $p-1$.

Reason: Let $k$ be the order of $x$. $x^{p-1}\equiv 1$ so $k \le p-1$. If $k\not \mid p-1$ then there are $m$ and $r; 0 < r < k$ so that $p-1 = mk + r$ and $1 \equiv x^{p-1} \equiv x^{mk + r} \equiv (x^k)^mx^r\equiv 1^mx^4 \equiv x^r\pmod p$. But $k$ was the smallest positive integer that was possible so that's a contradiction.

And $x^m \equiv 1 \pmod p$ if and only if $m$ is a multiple of the order of $x$.

Reason: same.

So if $x^a \equiv 1 \pmod p$ and $x^{p-1}\equiv 1 \pmod p$ and if the order of $x$ is $k$ then $a$ and $p-1$ are both multiples of $k$. But $\gcd(a,p-1) =1$ so $k = 1$ and $x^1 \equiv 1 \pmod p$ and... that's it.... $x\equiv 1\pmod p$ and that's the only solution.