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The question is the natural generalization of the

Find all positive integers solutions such that $3^k$ divides $2^n-1$ .

Let $p$ to be a prime number and let $k$ & $a$ be fixed natural numbers.

Find all natural numbers $n$, such that $p^k \mid a^n-1$?

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At first notice that if $\gcd(a,p) > 1$, then clearly there is no such natural number $n$. **So we can assume that $\gcd(a,p) = 1$.**

Let $a$ be a natural number such that $\gcd(a,M)=1$.

By the $\text{ord}_M(a)$ we mean the minimum power $r$

for which $a^r-1$ is divisible by $M$.

**Lemma(II)**: Let $a$ be a natural number such that $\gcd(a,M)=1$.

Let $R$ to be an arbitrary natural number.

Then we have:

$$ M \mid a^R-1 \ \ \ \Longleftrightarrow \ \ \ \text{ord}_M(a) \mid R \ \ \ . $$

By the $v_p(t)$, we mean the highest power of $p$ which divides $t$.

Now see the **Lemma1** from here:

$p$-adic valuation of $x^n+1$[or how many times does a prime number divides $x^n+1$]

From the mentioned **lemma1** and **Lemma(II)** we have the following lemma:

**Lemma(III)**: Let $p$ to be an odd prime and let $a$ be a natural number such that $\gcd(a,p)=1$.

Let $r:=\text{ord}_p(a)$; i.e.

the minimum power $r$ for which $a^r-1$ is divisible by $p$.

Also let $t:=v_p(a^r-1)$; i.e. the highest power of $p$ which divides $a^r-1$.

If $k \leq t$ let $s:=0$, else let $s:=k-t$.

Then we have:

$$\text{ord}_{p^k}(a)=rp^s.$$

**Lemma(IV)**: Let $p=2$

and assume that $a\overset{4}{\equiv} 1$.

Then we have $1=\text{ord}_4(a)$.

Also let $t:=v_2(a-1)$; i.e. the highest power of $2$ which divides $a-1$.

If $k \leq t$ let $s:=0$, else let $s:=k-t$.

Then we have:

$$\text{ord}_{2^k}(a)=2^s.$$

**Lemma(V)**: Let $p=2$

and assume that $a\overset{4}{\equiv} 3$.

Then we have $2=\text{ord}_4(a)$.

Also let $t:=v_2(a^2-1)$; i.e. the highest power of $2$ which divides $a^2-1$.

If $k=1$ let $s:=-1$, if $1 < k \leq t$ let $s:=0$, else let $s:=k-t$.

Then we have:

$$\text{ord}_{2^k}(a)=2.2^s \ .$$

**First case**: Assume that $p$ be an odd prime.

then let $r:=\text{ord}_p(a)$; i.e.

the minimum power $r$ for which $a^r-1$ is divisible by $p$.

Also let $t:=v_p(a^r-1)$; i.e. the highest power of $p$ which divides $a^r-1$.

If $k \leq t$ let $s:=0$, else let $s:=k-t$.

Then**Lemma(III)**implies the following proposition :

**Proposition3**: $n$ satisfies the **above divisibility diophantine equation**

if and only if it is divisible by $rp^s$.

**Second case**: Assume that $p=2$

and assume that $a\overset{4}{\equiv} 1$,

then we have $1=\text{ord}_4(a)$.

Also let $t:=v_2(a-1)=v_2(a^1-1)$;

i.e. the highest power of $2$ which divides $a-1$.

If $k \leq t$ let $s:=0$, else let $s:=k-t$.

Then**Lemma(IV)**implies the following proposition :

**Proposition4**: $n$ satisfies the **above divisibility diophantine equation**

if and only if it is divisible by $2^s$.

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