Geometric Progression: How to solve for $n$ in the following equation $\frac {5^n-1}4 \equiv 2 \pmod 7$

Question

Solve $$\frac {5^n-1}4 \equiv 2 \pmod 7$$

How to find the minimum value of $n$ that can satisfy above equation?

Answer 1

Multiply by $4$: $$5^n-1\equiv 1\pmod 7. $$ Add $1$: $$5^n\equiv 2\pmod 7. $$ The first few powers of $5$, modulo $7$, are: $1,5,4,6,2,3,1$. Hence the period is $6$ (as might be expected) and we conclude $$\frac{5^n-1}4\equiv 2\pmod 7\iff n\equiv 4\pmod 6. $$

Answer 2

Just define the values of this expression for every $n$: $$5^n=\begin{cases}1&\text{if }\;n\equiv 0\mod 6\\ 5&\text{if }\;n\equiv 1\mod 6\\4&\text{if }\;n\equiv 2\mod 6\\6&\text{if }\;n\equiv 3\mod 6\\ 2&\text{if }\;n\equiv 4\mod 6\\ 3&\text{if }\;n\equiv 5\mod 6 \end{cases} \pmod 7$$ hence $\;5^n-1\equiv 0,\:4,\:3,\:5,\:1,\:2$ respectively. On the other hand $$\frac{5^n-1}4\equiv 2(5^n-1)\equiv\begin{cases}2&\text{if }\;n\equiv 0\mod 6\\ 3&\text{if }\;n\equiv 1\mod 6\\4&\text{if }\;n\equiv 2\mod 6\\5&\text{if }\;n\equiv 3\mod 6\\ 4&\text{if }\;n\equiv 4\mod 6\\ 6&\text{if }\;n\equiv 5\mod 6\end{cases} \pmod 7$$