23
$\begingroup$

A little bird told me that if $2^n+1$ is prime, then $n$ is a power of $2$. I tend not to trust talking birds, so I'm trying to verify that statement independently.

Suppose $n$ is not a power of $2$. Then $n = a \cdot 2^m$ for some $a$ not a power of $2$ and some integer $m$. This gives $2^n+1 = 2^{a \cdot 2^m}+1$. Now I suspect there's a way to factor that, but I don't see how. Can someone give me a hint?

  • 7
    A proof of this fact is also given in Wikipedia article on [Fermat primes](http://en.wikipedia.org/wiki/Fermat_prime#Other_theorems_about_Fermat_numbers).2012-05-04
  • 0
    Neat. I didn't know it had a name.2012-05-04
  • 2
    Disproof: $2^0+1=2$$n=2^k=0$?2016-02-18
  • 0
    Related: https://math.stackexchange.com/questions/12138002018-11-26

1 Answers 1

25

Hint. For any odd natural number $a$, the polynomial $x+1$ divides $x^a+1$ evenly.

In particular, we have $$ \frac{x^a+1}{x+1}=\frac{(-x)^a-1}{(-x)-1}=1-x+x^2-\cdots+ (-x)^{a-1}$$ by the geometric sum formula. In this case, specialize to $x=2^{2^{\large m}}$ and we have a nontrivial divisor. (Also, $x^a+1\equiv(-1)^a+1\equiv-1+1\equiv0\mod x+1$ inside $\Bbb Z[ x]$ is pretty slick.)

  • 0
    Hi, sorry for a late post, but could you explain your answer a bit more for me. Did you mean to say, set $x = 2^m$ instead of $x = 2^{2^m}$2013-01-28
  • 4
    @TylerHilton No, I meant $x=2^{2^m}$ as I wrote. If $n=a2^m$ with $a$ odd, then this shows that $2^{2^m}+1$ divides $2^{a2^m}+1=2^n+1$, and hence $2^n+1$ is composite when $n$ is not a power of $2$.2013-01-28