Finding product of two unknown numbers each raised to a known power Let $G$ be a group, and let $a, b\in G$, and let $[3] = \{0,1,2\}$. Let $(x_i,y_i)$ for $i\in[3]$ be known constants. Assume that I know the elements:
$$c_i = a^{x_i}b^{y_i},\quad i\in[3]$$
Then, given $(x', y')$, is there a way to efficiently compute $a^{x'}b^{y'}$?
To say more about this, you need more assumptions on $(x_0, y_0),\dots, (x_2, y_2)$.
For example, if you write
$$B = \begin{pmatrix}x_0 &amp; x_1 &amp; x_2 \\ y_0 &amp; y_1 &amp; y_2\end{pmatrix}$$
to be the $2\times 3$ matrix of your known constants, and let:
$$\mathsf{span}_{\mathbb{Z}}(B) = \{(x', y')\in\mathbb{Z}^2 \mid \exists \vec{\alpha}\in\mathbb{Z}^3\text{ s.t. }(x', y') = B\vec{\alpha}\}$$
be the span of all of their integer linear combinations, then you can efficiently compute $a^{x'}b^{y'}$ for any $(x', y')\in\mathsf{span}_{\mathbb{Z}}(B)$ by writing:
\begin{align*}
\prod_{i\in[3]}c_i^{\vec{\alpha}_i} &amp;= \prod_{i\in[3]}(a^{x_i}b^{y_i})^{\vec{\alpha}_i}\\
&amp;=\prod_{i\in[3]}a^{\vec{\alpha}_i x_i}b^{\vec{\alpha}_i y_i}\\
&amp;=a^{x'}b^{y'},\quad (x', y') = B\vec{\alpha}
\end{align*}
The above technically uses the assumption that $G$ is abelian (commutative), but this is likely intended.
So if $\mathsf{span}_{\mathbb{Z}}(B) = \mathbb{Z}^2$ (which is definitely possible), you can compute your desired function for any pair of $(x', y')$. But if the span of $B$ is not all of $\mathbb{Z}^2$, it is not clear how to compute your desired function for $(x', y') \in \mathbb{Z}^2\setminus\mathbb{span}_{\mathbb{Z}}(B)$ (at least the prior technique no longer works).