<s>[INST] If $2^{12^{7}+3}\equiv x \pmod{36}$, then what is the value of $x$? 
If $2^{12^{7} + 3} \equiv x \pmod{36}$, then what is the value of $x$?

We have:
$$
\begin{align}
2^5 & \equiv - 4 \pmod{36} \\
2^{10} & \equiv 16 \pmod{36} \\
2^{12} & \equiv - 8 \pmod{36} \\
(2^{12})^{12} & \equiv 8^{12} \pmod{36} \\
2^{12^2} & \equiv - 8 \pmod{36} \text{, as } 8^{12} \equiv - 8 \pmod{36} \\
(2^{12^2})^{12^5} & \equiv - 8 \pmod{36} \\
2^{12^7} & \equiv - 8 \pmod{36} \\
2^{{12^7} + 3} & \equiv - 64 \pmod{36} \\
2^{{12^7} + 3} & \equiv 8 \pmod{36} \\
\end{align}
$$
Am I right or wrong? Is there any simpler way to find it?
 [/INST] It's probably easier to work modulo $4$ and modulo $9$, and use CRT.
First we have
$$x \equiv 2^{12^{7}+3}\equiv 0 \pmod{4}$$
To compute
$$x \equiv 2^{12^{7}+3} \pmod{9},$$
it suffices to compute $12^{7} + 3 \equiv 3 \pmod{6}$, as $6 = \varphi(9)$. (Or, if you prefer, as $2^{6} \equiv 1 \pmod{9}$.) So
$$2^{12^{7}+3}\equiv 2^{3} \equiv 8 \pmod{9}.$$
Clearly, then, the solution is indeed $x = 8$ as you found it, as $8 \equiv 0 \pmod{4}$.
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