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I'm currently tackling a few problems about divisibility and gcd, and I'm stumped by a few things. The arguments aren't that long for most of them so I'll write them in a list-type structure.

1: Let $a$ and $n$ be positive integers with $n \neq 1$. Prove that, if $a^n -1$ is a prime number, then $a=2$ and $n$ is a prime number.

Using the factorization $x^n - y^n = (x-y) \sum_{i=1}^n x^{n-i}y^{i-1}$, we see that, choosing $x=a$ and $y=1$ that, in order for the number not to be composite, $x$ must equal $2$. But why must $n$ be a prime number for $a^n-1$ to be prime?

2: Let $a$ and $n$ be positive integers with $a > 1$. Prove that, if $a^n+1$ is a prime number, then $a$ is even and $n$ is a power of $2$.

Suppose $a$ is odd, then $a^n$ is odd and $a^n+1$ is even and hence not a prime. But why must $n$ be a power of $2$?

Thanks! I hope you can help.

Numbersandsoon
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  • These are a lot of questions, and the answers are going to come from different directions. It's likely better if you were to split off 3 and 4 into a different question. –  Sep 15 '13 at 00:55
  • Most definitely! Thank you. – Numbersandsoon Sep 15 '13 at 00:59

2 Answers2

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Hint for the 1st one:

You're on the right track to use the identity $x^n-1 = (x-1)(x^{n-1}+\cdots+1)$. Now, suppose that $n$ is not a prime number, i.e. it can be decomposed into $n=ab$ for $a,b>1$. You can write $x^{n}=x^{ab}=(x^{a})^b$.

Hint for the 2nd one:

Notice that if $n$ is odd, we can use the identity: $x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1})$. You can prove the identity by distributing $(x+y)$ over $(x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1})$ and then canceling the terms. In particular, if $x=2$ and $y=1$ then $2^n+1$ is divisible by 3. Now what if $n=2k$ where $k$ is odd? Can you apply the same identity in some way?

Remark: The numbers of the form $2^n-1$ and $2^n+1$ are important type of numbers from historical point of view. The former type is called a Mersenne prime while the later one is called Fermat prime and there are still many questions open about them.

user66733
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  • Aha, so for 1st (using $pq$ instead of $ab$): $2^n=2^{pq}=(2^{p})^{q}$, which changes the value of $a$ away from $2$, which contradicts what we just found by using the identity $x^n-1 = (x-1)(x^{n-1}+\cdots+1)$. Thus $n$ must be prime. Correct?

    For 2nd: Don't we need to choose $n=2^m k$? This effectively describes all even integers not a power of $2$ for a suitable choice for $m$ and $k$, $k$ odd. Then we have $a^n+1=a^{2^m k}+1=(a^{2^m})^k+1=(a^{2^m}+1)(a^{2^m-1}-a^{2^m-2}+\cdots-a+1)$, which is again a composite number. Thus, $n$ must be a power of $2$.

     – Numbersandsoon Sep 16 '13 at 01:01
  • @BoSchmidt: Yes, that sounds good. Also, for the 2nd one, yes, you should choose $n=2^mk$ to describe all even integers that are not a power of 2. But I was just trying to show you what happens if $m=1$ to make it easier to understand and then I was sure you could find out why $n$ should be of the form $2^m$ by an inductive argument on your own. Well done. – user66733 Sep 16 '13 at 03:20
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Hint for 1: Think about the factorization of composite exponents $n$ in $x^n-y^n$

Hint for number $2$: Using $1$, what happens if $n$ has an odd factor?

chubakueno
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