I tried to solve this 2 questions but without a success:

1. Is $13$ a sixth power modulo $289$?
2. Find all the solutions of $x^{8}\equiv 3\mod 13$

In question 1, I tried to see if $13$ is a quadratic residue modulo $289$, but we didn't learn how to do it (I know how to do it only modulo primes), I would like to get help and ideas of how to solve this question.

• $289 = 17^2$. If $13$ is a quadratic residue modulo $289$, then it also is a quadratic residue modulo $17$ ($x^2 \equiv 13 \pmod{289} \implies x^2 \equiv 13 \pmod{17}$). If you have never heard of Hensel lifting, that might still give you an idea how to check whether $13$ is a quadratic residue modulo $289$. – Daniel Fischer May 15 '15 at 21:47
• The answer to (1) is yes, but you're not going to find it by accident. $66^6\equiv 223^6\equiv 13 \bmod 289$. It seems odd to choose $289$ for this exercise when $6 \nmid (17-1)$, so the proof is more difficult (although that judgement depends what tools you have available for modular exponentiation). – Joffan May 15 '15 at 22:30

For (1), see Hensel's Lemma. You can also do it explicitly: first find a solution $x$ mod $17$, then try $y$ where $y = x + 17 t$.
2) $x^8\equiv 3 \implies x^4\equiv \pm 4 \equiv \{4,9\} \implies x^2 \equiv \{\pm 2^*, \pm 3 \} \\ \implies x \equiv \{\pm 4,\pm 6 \} \equiv \{4,6,7,9 \} \bmod 13$
You only need to check around the first three multiples of $13$ to get squares, by symmetry. $\pm 2$ are not quadratic residues.