I just do not have a clue how to solve this exercise:
Find every natural number $m$ such that $998^m-1$ divides number $1994^m$.
A friend of mine solved it this way:
$998^m - 1 = (998 - 1) \cdot (1 + 998 + \cdots + 998^{m - 1}) = 997 \cdot (1 + 998 + \cdots + 998^{m - 1})$,
So $997$ is a divisor of $988^m - 1$
And $997$ is a divisor of $1994^m$
Is his solution correct?
Thank you for your time.
First notice that $1994=2\cdot 997$. It must happen that if $998^m-1$ divides $1994^m=997^m\cdot 2^m$, then
$$998^m-1= 997^k*2^q$$
for $k,q\leq m$. See that $998^m-1$ is always odd, so it won't have factors $2$. So
$$998^m-1=997^k$$
Then, if $m>1$, look that by the Zsigmondy theorem, it must happen that exists a prime $p$ such that $p\mid 998^m-1$, but doesn't divide $998^1-1$, so $998^m-1$ can't be a power of $997$, because it has another prime factor different than $997$. So the only answer will be $m=1$.