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Does the following hold in bilinear pairings? $$e(g^{a_1x_1}g^{a_2x_2},g^{c_1}g^{c_2})=e(g^{x_1+x_2},g^{a_1a_2(c_1+c_2)})$$

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    $\begingroup$ Only for $x_1=0 \wedge x_2=0$. $\endgroup$ – Artjom B. Oct 16 '14 at 7:25
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Yes, sure, in some cases. In the following I assume that we speak about pairings as used in cryptography.

Lets look at the trivial case, i.e., when your equation reduces to $1_{G_T}=1_{G_T}$ where $1_{G_T}$ is the identity in your target group $G_T$. Then any assignment of your variables that makes at least one input element of each pairing the element $g^0=1_G$ ($1_G$ the identity in your group $G$) makes the equality hold. Note if you have $e(a,g^0)=e(a,1_G)=1_{G_T}$ (Observe that for any $a,b \in G$ you have $e(a, b) = e(a, b\cdot 1_G) = e(a, b) \cdot e(a, 1)$ and thus $e(a, 1)=1_{G_T}$). For instance, as in @Artjom B.s example for the case $x_1=x_2=0$.

For the non-trivial case it is quite easy to see that $a_1=a_2=1$ makes the equation hold.

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  • $\begingroup$ Maybe it was not clearly defined but the setup is for any randomly chosen $a_1,a_2$ and $x_1,x_2$ $\endgroup$ – curious Oct 16 '14 at 9:26
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    $\begingroup$ @curious. Then, only with negligible probability (meaning in practice NO). $\endgroup$ – DrLecter Oct 16 '14 at 9:28

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