Question: I have the following equations :

$$x^5=2 \pmod {13}$$ $$x^3=2 \pmod{11}$$

I wonder how to solve such equations is there a method to get rid off the powers in order to use the Chinese Remainder Theorem without checking each $x$ value?

Any ideas?

Thank you.

Answer: 
**Hint:** If $x^3\equiv 2\pmod{11}$ then $ x^{21}\equiv 2^7\pmod{11}$.

By Fermat, we have $x^{21}\equiv x\pmod{11}$.
You can simplify the computations writing the elements in $\mathbf F\_{13}$ and $\mathbf F\_{11}$ as
$$\{0,\pm 1,\pm 2,\pm3,\pm4,\pm 5,\pm6\}\quad\text{and}\quad\{0,\pm 1,\pm 2,\pm3,\pm4,\pm 5\}$$
respectively. As $0$ and $1$ can't be solutions, we only have this computation for the first equation:
$$\begin{array}{c\*{7}{c}}
x&\pm2&\pm3&\pm4&\pm5&\pm6\\
\hline
x^2&4&-4&3&-1&-3\\
x^4&3&3&-4&1&-4\\
\hline
x^5&\pm6&\mp4&\mp3&\pm5&\pm2
\end{array}$$
Hence there is only one solution: $\;x=6$.