$5^a - 5^b$ is divisible by $n$ (prove)

Question

Prove that for every n natural number exist natural numbers $a,b \leq 4n, a\not= b $, which accomplish, that number $ 5^a - 5^b $ is divisible by n. How many of these pairs exist?

Help please, I'm stuck with this problem. Any help is appreciated.

Answer 1

Let $n=5^km$ with $gcd(5,m)=1$.

Now, the pair $(a,b)$ is a solution if and only if $(b,a)$ is a solution. So it suffices to look for the solutions where $a>b$. Don't forget at the end to double the number of solutions.

Then

$$n|5^a-5^b =5^b(5^{a-b}-1)) \Leftrightarrow k \leq b \mbox{ and } m | 5^{a-b}-1$$ The existence is easy then, pick $b =k $ and $a-b =\phi(m) \leq m \leq n$.

For the number of solutions, you need

$$k \leq b \leq 4n$$ Thus you have $4n-k$ choices for $b$.

Morever, for each $b$, if $j$ is the order of $5$ modulo $m$, you must have

$$j |a-b \,.$$

Thus, $a-b$ has to be a multiple of $j$ between $1$ and $4n-b$. From here you figure out how many choices of $a$ you have for each $b$. Add them, double and you are done.