If $a$ is not a multiple of a prime $p$ , then prove that there is an integer $b$ such that $p^b-1$ is a multiple of $a$
I have no idea where to start I would be grateful if anyone can give me hint.
NOTE: Please try to prove it using only basic properties of GCD and LCM .Please avoid fancy theorems because I am not that advanced
On
If $a$ is not a multiple of $p$ , we can write :
$$p\equiv r \mod a \quad\quad \text{such that }r \ne 0$$
We use Euler's theorem , to say that :
$$p^{\phi(a)} \equiv 1 \mod a$$
This follows from the fact that $p$ is a prime and $a$ is not a multiple of $p.$Hence it is a necessary condition for the question to remain true. Now taking $\phi (n) = b$ , we get :
$$p^b \equiv1\mod a \implies \boxed{p^b-1 \equiv \mod a}$$
Hence there always exist a multiple $b$ , for which $p^b-1$ is a multiple of $a.$
The set of numbers $\{p^x\}$ for $x$ a natural number, is infinite. If you divide then all by $a$ and look at the remainders, there are only $a$ possibilities. Therefore at least two of the numbers have the same remainder when you divide by $a$. (Pigeonhole principle.)
That leads to $p^x - p^y$ being a multiple of $a$ for some $x$ and $y$. Now you have a clue.