If prime $p$ such $p|F_{n}(n\ge 2)$, show that $p\equiv \pm 2 \pmod 5$

Question

Let $p$ be a prime number and $$p|F_{n}, n\ge 2$$ where $F_{n}=2^{2^n}+1.$

Show that $$p\equiv \pm 2\pmod 5.$$

I have proved $$F_{n}\equiv 2\pmod 5,$$ because $$F_{n}=2^{2^n}+1=(2^2)^{2^{n-1}}+1\equiv (-1)^{2^{n-1}}+1=2\pmod 5,$$

but I can't show $p\equiv \pm2\pmod 5$. Thanks.

Answer 1

No. $641$ is a prime divisor of $F_5$

other examples at https://en.wikipedia.org/wiki/Fermat_number#Factorization_of_Fermat_numbers