If $x^3$ is a square, is $x$ a square? Very simple question here, which I feel like I should be able to answer but am struggling with. Let $k$ be a finite field, and let $x\in k^\times$. Is it true that
$$x^3\in\left(k^\times\right)^2 \Longleftrightarrow x\in\left(k^\times\right)^2?$$
 A: $x=x^3/x^2$, so if $x^3=a^2$ then $x=(a/x)^2$.
A: Let us use proof by contradiction to show that $x$ is square, too. 
We can write $x$ into its prime factors: 
$$x = a_1^{p_1} \times ... \times a_n^{p_n}$$. 
Let us assume that the conclusion is false i.e. $x$ is not square. Then at least one of the factors has an odd power: 
$$p_j = 2k+1 \hspace{0.5in} j = 1,...,n$$. 
In $x^3$ this power will be $6k+3$ which is $3(2k+1)$ (which is the product of two odd numbers and thus is odd), this means that $x^3$ has a factor with odd power which means that $x^3$ is not square. 
This contradicts with the argument's premise and therefore the initial conclusion is true. 
A: It is worth highlighting the underlying idea from an additive perspective.
$$\begin{align}  x^{\large n+1}\! = y^{\large n }\!\! &\overset{\Large (\ \ )/x^{\Large n}\!\!\!\!}\iff x = (y/x)^{\large n}\ \ \ \ \ \, \text{OP is case }\ n=2\\[.3em]
(n\!+\!1) x = n y \!\!&\ \iff x =  n(y\!-\!x)\ \ \ \ \text{in additive form}\\[.3em]
n\,\mid\, (n\!+\!1)x\!\! &\ \iff n\,\mid\, x\qquad\ \ \ \ \  \ \ \  \text{in divisibility form}\\[.3em]
\bmod n\!:\,\ 0\equiv (n\!+\!1)x\!\! &\ \iff 0\equiv x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{in congruence form}
\end{align}\qquad\qquad$$
Probably everyone can immediately see at least one of the last three, but sometimes we miss the first, because our additive intuition is often stronger than our multiplicative intuition (but compare my most popular post).  To work around that it often helps to first translate multiplicative statements into additive form to spark intuition.
