Let $e: \mathbb G_1 \times \mathbb G_1 \rightarrow \mathbb G_T$ be an efficient bilinear pairing. Note that the pairing is symmetric (i.e., Type 1).
The problem is, given $g \in \mathbb G_1$ and $e(g,g)^a \in \mathbb G_T$, for an unknown $a$, to compute $X_1, X_2 \in \mathbb G_1$ such that $e(X_1,X_2) = e(g,g)^a$
Is this a hard problem? My feeling is that it is not and there should be an easy way to solve it.
Update 1: I have found a related problem that it is said to be hard:
The Pairing Pre-Image Problem (PPIP) in $\mathbb G_1, \mathbb G_T$ such that $\mathbb G_1 = <g>, |\mathbb G_1| = |\mathbb G_T | = p$ with pairing $e : \mathbb G_1 \times \mathbb G_1 \rightarrow \mathbb G_T$ is defined as follows: On input a tuple $(g, Z) \in \mathbb G_1 \times$ $\mathbb G_T$, output $X \in \mathbb G_1$ such that $e(X, g) = Z$.
According to this paper: CDH in $\mathbb G_1 \leq$ PPIP $\leq$ DL $\in \mathbb G_T$
The problem I am asking is not harder than PPIP since I could use PPIP to solve my problem : simply run the PPIP solver on $(g,Z)$, take output $X$, and return $X_1 = X, X_2 = g$. That is, PPIP is my problem but requiring that $X_2=g$.
Still, I think that there should be an easy way to solve my problem. Any ideas?
Update 2: Thanks to the comment from @DrLecter, now I know that the original problem I stated was exactly the Generalized Pairing Inversion (GPI) problem, which seems to be hard. That makes me reformulate the question to my real problem (see this question).