0
$\begingroup$

I understand that this is true.

Primes are always odd so a would naturally have to be even, if $1$ is being added to it.

For example, $4^2+1 = 17$.

But I'm not quite sure how to prove this.

  • 0
    Do you mean the $a$ is even part? You have just proved that. If a>1 is odd, then $a^n+1$ is even and greater than $2$, so cannot be prime. (One could add, because it is divisible by $2$ and greater than $2$, but I think that hardly needs saying.) The somewhat harder part is showing that $n$ is a power of $2$? Is that the part you are asking about?2012-12-10

2 Answers 2

3

Let $P(x) = x^n + 1$ then when $n$ is odd we have $P(-1) = 0$ therefore $x+1|P(x)$.

As a consequence of this we have $a^{(2^r)}+1|P(a^{(2^r)})$ i.e. $P(a^{(2^r)}) = a^{2^r \cdot n} + 1$ has a non-trivial factor unless $n=1$. This shows that for it to be prome the exponent must be a power of two.

2

If $a$ is an odd that is greater than $1$ then $a^n+1>2$ and $a^n+1$ is even (contradicting the fact tat it is prime).

If $n=2^jm$ (where $m$ is an odd integer greater than $1$), then $a^n+1=(a^{2^j})^m-(-1)^m$ which is divisible by $a^{2^j}--1$ (contradicting the fact that $a^n+1$ is prime).