LTE flavoured divisibility

[mutasimmim]

Find all natural numbers $n$ such that $7^n$ divides $9^n-1$.

[Fatin Farhan]

$$7^n \mid 9^n-1 \Rightarrow 7 \mid 9^n-1$$.
But $$7 \nmid 9^n-1$$.
As $$n \geq 1$$ there is no such $$n$$.

[Nirjhor]

But $$7 \nmid 9^n-1$$.
As $$n \geq 1$$ there is no such $$n$$.

This is wrong. Smallest counterexample: \(n=3\). In fact \(9^n-1\equiv 2^n-1~(\bmod~7)\). And \(2^{3m}=8^m\equiv 1~(\bmod~7)\). So all multiples of \(3\) works (for the case of \(7\)).

Fun fact: \(7\mid 2^n-1\) is IMO 1964 Problem 1.

[mutasimmim]

Fatin, how did you guess that $7 \nmid (9^n-1)$?

[mutasimmim]

Hints:
1. Consider order of $9(mod 7)$
2.Then LTE

[mutasimmim]

Solution:

The order of $9(mod 7)=3$.Thus $n$ is divisible by $3$. Letting $n=3m$, and using LTE, we have $v_7(9^{3m}-1)=v_7(9^3-1)+v_7(m)=1+v_7(m) \ge 3m$. Which is possible only when $m=0$. Thus there is no such $n$.