amount of solutions of congruent relation

Question

$a_1 = 5, a_2 = 15, i = 1, 2$
$x^6 ≡ a_i \pmod{98}$

What is the best way to get the amount of solutions? Do I have to do something with the primitive roots?

Answer 1

By Chinese Remainder Theorem the solutions correspond to pairs $(x,y)$ of solutions of $$\begin{eqnarray}x^6 &\equiv& a_i \bmod 2 \\y^6 &\equiv& a_i \bmod 7^2\end{eqnarray}$$

For the case $a_1=5$ note that $5$ is not a quadratic residue $\bmod \;7$, so $x^6 \equiv 5 \bmod 7$ has no solution.

For the case $a_2=15$ note that $\varphi(49)=2 \cdot 3 \cdot 7$, therefore $(\Bbb Z/49\Bbb Z)^\times$ is cyclic and there is a primitive root $\bmod\; 49$. Therefore $x^6 \equiv m \not \equiv 1 \bmod 49$ has exactly $6$ = $\varphi(49)/6-1$ solutions iff $m^7 \equiv 1 \bmod 49$ or no solution otherwise. As $15^7 \equiv 1 \bmod 49$, there are $6$ solutions.

Clearly, $x^6 \equiv 15 \bmod 2$ has exactly one solution, so all in all there are exactly $6$ solutions.