If $a^{n}-1$ is prime then $a=2$ and $n$ is prime?

Question

I was doing some basic Number Theory problems and came across this problem :

Show that if $a$ and $n$ are positive integers with $n\gt 1$ and $a^{n} - 1$ is prime, then $a = 2$ and $n$ is prime

My Solution : (Sloppy)

$a^{n}-1$ = $(a-1)$ . $(a^{n-1} + a^{n-2} + ... + a + 1)$

This means that $(a-1)$ | $(a^{n}-1)$

But $(a^{n}-1)$ is prime

So , $(a-1)$ = 1 $\Rightarrow$ $a = 2$

Now , let $n$ be composite

$n = kl$ , where $1 \lt k \lt n$ and $1 \lt l \lt n$

Now , $a^{kl}-1$ = $(a^{k}-1)$ . $(a^{k.{(l-1)}} + a^{k.{(l-2)}} + ... + a^k + 1)$

This means that $2^{n} -1$ is composite

Hence , we have achieved a contradiction

My Question : Am I correct ?

seems ok to me. See Mersenne numbers. http://en.wikipedia.org/wiki/Mersenne_prime — Ofir Schnabel, Feb 18 '15 at 14:47
I think you need to say that $a-1<a^n-1$ becouse $(a-1)|(a^n-1)$ can mean that $a-1=a^n-1$ — Antony, Feb 18 '15 at 14:52
Hi , @Antony , if I am wrong (even slightly) can you please explain in detail as an answer -- would be grateful :) — pranav, Feb 18 '15 at 15:11
@pranav , you say that if $p$-prime and $k|p$ than $k=1$ . But there is also case $k=p$ and you lost it . You only need to say that $a-1<a^n-1$ for $n>1$ . — Antony, Feb 18 '15 at 15:15
See also: http://math.stackexchange.com/questions/1559994/prime-numbers-of-form-an-1 — Martin Sleziak, Dec 11 '15 at 16:09
There is also the similar, if $2^k + 1$ is prime then $k = 2^n$ as a matter of fact :) — Mr Pie, Jan 16 '18 at 06:56

score 8 · Accepted Answer · answered Feb 18 '15 at 15:20

The proof is alright there are two or three details though (the same issue twice actually), oe was already pointed out in comments:

Likely you should exclude the case $a=1$ right away. Just by saying $1^n -1 = 0$ is not prime so assume $a>1$.
You cannot derive from $a^n -1$ being a prime and $(a-1) \mid (a^n-1)$ directly that $(a-1)= 1$. What you can do is say $a-1=1$ or $a-1 = a^n-1$. The latter is impossible as $n > 1$ (and $a >1$); note here you use $n>1$.
When you assert that $a^{kl}-1$ is composite, you should state explicitly that both factors you exhibit are not $1$.

These are not major problems, but if one is picky one could insist on them.

2

Thanks @quid , I got it :) – pranav Feb 18 '15 at 15:22

score 3 · Answer 2 · answered Jun 03 '21 at 18:55

I'm a little late to the party, but here is a way to justify this using a contrapositive statement.

Let $a,n\in\mathbb{Z}^+$, where $n>1$, and assume $a^n-1$ is prime. Show $n$ is prime by the contrapostive, i.e., show if $n$ is composite, then $a^n-1$ is composite. Since $n$ is composite, $n=ks,$ where $k,s,\in\mathbb{Z}$. Now, $a^n-1=a^{ks}-1=(a^k)^s-1 =(a^k-1)[(a^k)^{s-1}+(a^k)^{s-2}+\cdots+a+1)$. Clearly $a^n-1$ is composite.

We can directly show $a=2$. Since $a^n-1$ is prime, $a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)$ is also prime. Since the R.H.S. expression is prime, $a-1$ must equal $1$, showing $a=2$.

$\therefore$ if $a^n-1$ is prime, $n>1$ is prime and $a=2$.

score -4 · Answer 3 · edited Jan 16 '18 at 07:31

-4

Since $a$ and $n$ are both positive integers so $a>0$ and $n>0$ then taking $a=1$ putting in $a^n-1$ we get $1^n-1=0$ which is not prime hence take $a=2$ where $2^n-1$ is a prime if and only if $n$ is a prime

edited Jan 16 '18 at 07:31

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27,005

answered Jan 16 '18 at 06:52

Agni Chakraborty

1

If $a^{n}-1$ is prime then $a=2$ and $n$ is prime?

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