Prime factors of Fermat numbers

Idea. The order of $2$ modulo $p$ is exactly $2^{k+1}$, and orders divide $p-1$.

Proof. Let $p\mid F_k$. Then $p$ is odd and $2^{2^k}\equiv-1\pmod p$. Squaring, $2^{2^{k+1}}\equiv1\pmod p$, so the order $d$ of $2$ divides $2^{k+1}$, hence $d=2^j$ with $j\le k+1$. If $j\le k$, then $2^{2^k}\equiv1$ (as $d\mid2^k$), but we know $2^{2^k}\equiv-1\not\equiv1\pmod p$ since $p>2$. Thus $d=2^{k+1}$. By Fermat's little theorem $d\mid p-1$, so $p\equiv1\pmod{2^{k+1}}$.

Case $k=5$. The primes dividing $F_5$ are among $64m+1$: $193,257,449,577,641,\dots$ Only $641=10\cdot64+1$ is needed.
We use two expressions for $641$: $641=5\cdot2^7+1$ and $641=2^4+5^4$. The first gives $5\cdot2^7\equiv-1\pmod{641}$, so $5^4\cdot2^{28}\equiv1$. The second gives $5^4\equiv-2^4$. Substituting, $-2^4\cdot2^{28}\equiv1$, i.e. $2^{32}\equiv-1\pmod{641}$, so $641\mid F_5$ ✓.

Remark. $F_5=641\cdot6\,700\,417$. Euler found $641$ in 1732 using exactly this restriction on divisors, refuting Fermat's belief that all $F_k$ are prime.

Lesson. Compute the exact order of an element from a congruence like $a^m\equiv-1$; it restricts the possible primes to an arithmetic progression.