Solving $4957^k \equiv 1 \pmod{7007}$ with the chinese remainder theorem

Question

Find all natural numbers $k$ that satisfy $$4957^k \equiv 1 \pmod{7007}$$ where $7007 = 7 \cdot 7 \cdot 11 \cdot 13$.

I know the chinese remainder theorem, but the $k$ is bugging me?

Answer 1

Hint: The chinese remainder theorem states that your equation is equivalent to the following system of equations:

$$\begin{align*} 4957^k &\equiv 1 \pmod{49}\\ 4957^k &\equiv 1 \pmod{11}\\ 4957^k &\equiv 1 \pmod{13} \end{align*}$$

This is in turn equivalent to $$\begin{align*} 8^k &\equiv 1 \pmod{49}\\ 7^k &\equiv 1 \pmod{11}\\ 4^k &\equiv 1 \pmod{13} \end{align*}$$

Now it's your turn to solve these equations.

Edit: The chinese remainder theorem is actually not even necessary to solve this problem. The equivalence between the original equation and the system of equations can easily be deduced without the CRT.