= Stronger prime-divisor congruence for Fermat numbers
{title2=$p\mid F_n,\ n\geq2\quad\Longrightarrow\quad p\equiv1\pmod{2^{n+2}}$}
For $n\geq2$, the first order bound already gives $p\equiv1\pmod8$. The <second supplementary law for quadratic reciprocity> says that two is then a <quadratic residue> modulo $p$. <Euler's criterion> gives $2^{(p-1)/2}\equiv1$, so its <multiplicative order> $2^{n+1}$ divides $(p-1)/2$. Thus $2^{n+2}\mid p-1$, further reducing the candidate factors. At $n=5$, $641=5\cdot128+1$ meets this bound. The restriction matters: $F_1=5$ has factor five, which is not one modulo eight.
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