If $a^3=e$, then $a$ has a square root

Hint: Try $x$ as a power of $a$ (since these are the only things being assuredly in $G$).

Hint: Multiply by $a$ to find $$a^4 = a$$

Hint $\,\ a^{\large 3}= 1\,\Rightarrow\,a^{\large 3n}=1\,\overset{\times\, a}\Rightarrow\,a^{\large 1+3n} = a\, $ is a square if $\ 2\mid 1\!+\!3n,\,$ e.g. $\, n = \,\ldots\ $ QED

Remark $ $ Here is the intuition. $\ a^{\large 3}= 1\ $ implies that exponents on $\,a\,$ can be considered mod $\,3,\,$

$$\color{#c00}{a^{\large 3}= 1}\ \Rightarrow\ a^{\large k+3n} = a^{\large k} (\color{#c00}{a^{\large 3}})^{\large n} = a^{\large k}\ \ \ {\rm so}\ \ \ a^{\large j}\! = a^{\large j\ {\rm mod}\ 3} $$

Therefore we can replace the exponent $1$ in $\,a^1\,$ by any $\,j\equiv 1\pmod 3,\,$ which includes $ $ even $\,j,\,$ i.e. $\,{\rm mod}\ 3,\,$ we have that $1$ is "even", i.e. $\ 2\mid 1,\,$ i.e. $2$ is invertible. This generalizes as follows.

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If $\,\color{#c00}{a^k = 1}\,$ and $\,\gcd(n,k)=1\,$ then $\,a\,$ is an $n$'th power. Indeed, by above it suffices to find a multiple $\,jn\,$ of $\,n\,$ that is $\,\equiv 1\pmod k,\,$ i.e. to invert $\,n\,$ mod $\,k,\,$ which is easy:

$\qquad\qquad$ by Bezout, there are $\,i,j\in\Bbb Z\,$ with $\ jn = 1 + ik\ $ so $\ (a^j)^n = a(\color{#c00}{a^k})^i = a$

Note how the problem reduces to the problem of division mod $\,k.\,$ The structure underlying this reduction will become clearer when one studies cyclic groups and modules (over $\,\Bbb Z/k)$.