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 $x^{21}\equiv x\pmod{11}$.
There are simple techniques to check these, and you will find that

$$x^5\equiv2\pmod{13}\implies x=13n+6$$

$$x^3\equiv2\pmod{11}\implies x=11k+7$$
For the first, we assumed that $x=13n+a$ for each $a=0,\pm1,\pm2,\dots,\pm6$. For the second, we assumed that $x=11k+b\mod11$ for each $b=0,\pm1,\pm2,\dots.,\pm5$. As these numbers form a complete residue system, it is sufficient to consider them.