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I have a few questions about the proof that $n \nmid 2^n - 1$ for any natural number $n > 1$

I found the solution, but I have some questions and wanted to make sure I understood everything correctly. Thank you in advance.

Proof By Contradiction :

Suppose, to the contrary, that an integer $n > 1$ exists such that $n \mid 2^n - 1$. Since $2^n - 1$ is odd, $n$ must be odd too.

Let's take $p$ to be the smallest prime dividing $n$ (Q1 : Why do we take a smallest prime number?).

Then, $p \mid 2^n - 1$ (rule: if $c \mid b$ and $b \mid a$, then $c \mid a$) and $p \mid 2^{(p-1)} - 1$ (Q2 : Why the exponent here is $p-1$?) .

Hence, $p \mid 2^d - 1$, where $d := \gcd(n,p-1)$ (Because $\gcd(a^k - 1, a^l - 1) = a^{\gcd(k,l)} - 1$. Since $p \mid 2^n - 1$ and $p \mid 2^{(p-1)} - 1$, then $p \mid \gcd(2^n - 1, 2^{(p-1)} - 1) = 2^{\gcd(n, p-1)} - 1$).

However, as $p$ is the smallest prime divisor of $n$, we have $gcd(n,p-1) = 1$. (Q3 : If I understood correctly, this is always true?)

Hence, $p \mid 2^d - 1 = 1$. For $p$ to divide $1$, it has to be smaller or equal to $1$, which contradicts that $p$ is a prime number.

We get a contradiction and therefore $n$ cannot divide $ 2^n - 1$ for any natural number $n > 1$.

Prem
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Kamile
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  • Exponent is p-1 from Fermat's little theorem – stranger Feb 11 '24 at 19:54
  • If $p\not| x$, then $p|x^{p-1}-1$ always. This is because it is sufficient to work in the multiplicative group $\mathbb{Z}_p\setminus {0}$. The order of the group is $p-1$, and in any finite group $G$, for any $x\in G$, $x^{|G|}=e$. –  Feb 11 '24 at 19:54
  • We take the smallest prime number because that guarantees that $\text{gcd}(n,p-1)=1$. This isn't necessarily true if $p$ is not the smallest prime divisor of $n$. For example, if $n=6$ and $p=3$, $\text{gcd}(6,p-1)=2$. –  Feb 11 '24 at 19:57
  • If any questions remain please post them in comments on ther answers in the linked dupe. Only if you get no answer should you then post a new question. This helps to improve prior answers. – Bill Dubuque Feb 12 '24 at 04:19

1 Answers1

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This solution is overkill but I can answer your queries.

(The proof is based on contradiction.)

First, by Fermat's Little Theorem

$$a^{p-1} \equiv 1 \pmod{p},(a,p) = 1$$

Therefore, as $n$ is odd, the smallest prime can not be $2$.

$a = 2$ will give $2^{p-1} \equiv 1 \pmod{p}$.

(The next part can be done by the method in the solution. I think you don't have any query with that.)

Now, if you know what $orders$ are then it is clear that if $ord_p(2) = d$ then $2^d \equiv 1 \pmod{p}$.

This implies, $d|n$ and $d|p-1$.

As GCD is the largest common divisor of both, $d|(n,p-1)$

But as $p$ is the smallest prime,

$$(n,p-1) = 1$$

As all other divisors of $n$ are equal or greater than $p$.

This means $order$ has to be $1$.

$$2^1 \equiv 1 \pmod{p}$$ $$1 \equiv 0 \pmod{p}$$

Contradiction.

Krave37
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