 Prime divisor of a Fermat number

If a prime number $p$ divides the Fermat number $F_n$, it is odd and $2^{2^n}\equiv -1(\mathrm{mod}p)$. The multiplicative order of $2$ divides $2^{n+1}$ but not $2^n$, so it is exactly $2^{n+1}$. Fermat's little theorem then gives $2^{n+1}\mid p-1$. For $n\ge 1$, every prime divisor is $1$ modulo $4$. A prime need not divide any Fermat number merely because it is $1$ modulo $4$: order $12$ excludes the prime $13$ from the whole family.