Stronger prime-divisor congruence for Fermat numbers ($p\mid F_n,n\ge 2⟹p\equiv 1(\mathrm{mod}{2}^{n+2})$)

For $n\ge 2$, the first order bound already gives $p\equiv 1(\mathrm{mod}8)$. The second supplementary law for quadratic reciprocity says that two is then a quadratic residue modulo $p$. Euler's criterion gives $2^{(p-1)/2}\equiv 1$, so its multiplicative order $2^{n+1}$ divides $(p-1)/2$. Thus $2^{n+2}\mid p-1$, further reducing the candidate factors. At $n=5$, $641=5\cdot 128+1$ meets this bound. The restriction matters: $F_1=5$ has factor five, which is not one modulo eight.