I realize this question was asked before, but I did not find the answers satisfying. Here is my attempt:
Since G is cyclic, any element can be written as $g^m$ for $ 0 \leq m \leq 11 $, so the equation reads $ (g^k)^2 = g^j $. These two elements are equal if $ 2k \equiv j \mod 12 $ which implies $j = 2(m - 6 z)$, so if we want to find a $g$, which does not then it has to be an element with $j$ odd.
However, the solution says that $ g^6 = (g^2)^6 = e $, which is a contradiction.
On
Put another way, let $g$ be an element that generates $C$; as $C$ is cyclic there indeed exists such a $g$.
Note that in a cyclic group $C$, for any $d$ that divides the order $n$ of $C$, the set of $d$-th powers is a subgroup of $C$ of order $n/d$. So putting $d=2$ and $n=12$, the set of squares $S$ of $C$ is a subgroup of $C$ of order 12/2 = 6.
So suppose $g$ is in $S$. Then by the fact that $S$ is a group, every power of $g$ is also in $S$, which (as $g$ generates the whole group $C)$ inpies that $|S| = |C| = 12$. This contradicts with what we observed in the paragraph above that $|S| = 6$.
So $g$ is not in $S$, and thus there is no element that squares to $g$.
Using additive notations, you are looking for $n\in\mathbb Z_{12}$ such that for all $k\in\mathbb Z_{12}$, $n\neq 2k$. If $1$ is not such a number, then there exists $k$ such that $2k=1$, so that $2$ is invertible in the ring $\mathbb Z_{12}$, but this is not the case because $2$ and $12$ are not coprime.