Proving that $a=2$ and $n$ is a prime given $p$ is a prime in $p= a^n - 1$

Question

Let $a$ and $n$ be integers greater than $1$.

Suppose that $a^n - 1$ is prime. Show that $a = 2$ and that $n$ is prime.

First of all let $$p = a^n - 1$$

Since both $a$ and $n$ are integers greater than one then $$p \ne 2$$

Thus $p$ is an odd prime

Rearanging: $$a^n = p + 1$$

Thus $a^n$ is an even integer, and then $a$ is and even integer.

Taking $\log_2$ of both sides: $$ n\log_2(a) = \log_2(p+1) $$ and $$n = \frac{\log_2(p+1)}{\log_2(a)}$$

Since n is an integer then $$\log_2(a) = 1$$ and $$a=2$$

I managed to get thus far, but when it comes to proving n is a prime I don't seem to know how to approach it.

How do you get $n = \frac{\log_{2}(p+1)}{\log_{2}(a)}\implies\log_2(a)=1$? — Eclipse Sun, Oct 17 '17 at 23:54
But it sure implies that $$a=2^k$$ and $$p+1=2^r$$ where $$r>k$$ — Sul, Oct 17 '17 at 23:57
$a^n-1$ is divisible by $a-1.$ Thus if $a\neq 2,$ $a^n-1$ can't be prime. — RideTheWavelet, Oct 17 '17 at 23:58
Similarly, $a^{pq}-1$ is divisible by $a^{p}-1$ and $a^{q}-1,$ so $n$ can't be composite. — RideTheWavelet, Oct 17 '17 at 23:59

Eclipse Sun · Answer 1 · 2017-10-18T00:01:17.587

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Hint: $$x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+\cdots+y^{k-1}).$$ And you may want to use the formula for $x=a^t$ and $y=b^t$ for some integer $t$ to prove that $n$ is prime.

edited Oct 18 '17 at 00:01

answered Oct 18 '17 at 00:00

Eclipse Sun

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Proving that $a=2$ and $n$ is a prime given $p$ is a prime in $p= a^n - 1$

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