Exercise 10.2.7

Proof. $F_0=2^{2^0}+1=3,F_1=2^{2^1}+1=5,F_2=2^{2^2}+1=17,F_3=2^{2^3}+1=257,F_4=2^{2^4}+1=65537$.

$3,5,17$ are well-known primes.

Since $3^{(F_3-1)∕2}=3^{128}\equiv -1(\mathit{mod}257)$, and $3^{(F_4-1)∕2}=3^{32768}\equiv -1(\mathit{mod}65537)$, the order of $[3]$ is $2^{2^n}=F_n-1$ in ${(\mathbb{Z}∕F_n\mathbb{Z})}^{\text{∗}}$ for $n=3,4$, so the Lucas test proves that $F_3,F_4$ are prime numbers.

(Thanks to SAGE for the numerical results.) □