Composite Fermat number implies Poulet number

Statement

Suppose $n$ is a nonnegative integer. Let $F_n$ be the $n^{th}$ Fermat number:

$F_n=2^{2^n}+1$.

Then:

$2^{F_n-1}\equiv 1(\mathrm{mod}{F}_n)$.

In particular, if $F_n$ is composite, then $F_n$ is a Poulet number.

Related facts

• Prime divisor of Fermat number is congruent to one modulo large power of two
• Mersenne number for prime or Poulet is prime or Poulet

Proof

Given: A nonnegative integer $n$, $F_n=2^{2^n}+1$.

To prove: $2^{F_n-1}\equiv 1(\mathrm{mod}{F}_n)$.

Proof: For any nonnegative integer $n$, $n+1\le 2^n$. Thus, $2^{n+1}|2^{2^n}$.

Now, for a Fermat number $F_n$, the order of $2$ modulo $F_n$ is precisely $2^{n+1}$. From the above, we conclude that the order of $2$ modulo $F_n$ divides $2^{2^n}=F_n-1$, so $2^{F_n-1}\equiv 1(\mathrm{mod}{F}_n)$.