I had asked this question here Geometric Progression: How to solve for $n$ in the following equation $\frac {5^n-1}4 \equiv 2 \pmod 7$ and accepted solution has conclusion
$$\frac{5^n-1}4\equiv 2\pmod 7\iff n\equiv 4\pmod 6$$
I wanted to understand how did the author reached to the conclusion. Yes I had asked him in the comment but maybe he is busy, so he did not reply back.
Can anybody help?
On
You have $5^n\equiv2(mod~ 7)$, then $\frac{5^n-2}{7}=k$. We can find by checking that $n=4$ is the smallest one. We got $\frac{5^4-2}{7}=k=89$. Now multiply this by $5^m$, so the result also divides $7$: $$5^m\frac{5^4-2}{7}=\frac{5^{m+4}-2\cdot5^m+2-2}{7}=\frac{5^{m+4}-2}{7}-\frac{2(5^m-1)}{7}$$
Now find the least $m$, when $5^m-1$ divides $7$. By checking $m=6$, so:$$\frac{5^{m+4}-2}{7}=\frac{5^{6+4}-2}{7}=k$$
Now, use congruence of powers: $$a\equiv b(mod~ m)\Rightarrow a^p\equiv b^p(mod~ m)$$ $$5^6\equiv 1(mod~ 7)\Rightarrow 5^{6p}\equiv 1(mod~ 7)$$
So, $$\frac{5^{m+4}-2}{7}=\frac{5^{6p+4}-2}{7}=k\; \; \; (p\in\mathbb{N})$$
I thought it in this way
From $5^n\equiv2(mod~ 7)$ it is clear that $ 5^n=7k+2$
Now $7k+2$ must end with $5$ , so $k$ must end with $9$ (I think I don't need to explain why)
So after checking values of $k=9,19,29,39,\cdots$ you will see that $89$ is first to give natural $n$ which is $4$
Also since $4$ is less than $7$ and the next solution will be $11$ as I found , we can conclude $n=7l+4$ or $n\equiv 4(mod ~7)$
Not a very great answer but hope it will help....!!!