TITLE: Generators of the multiplicative group modulo $2^k$
QUESTION [9 upvotes]: In most books and lecture notes that explicitly give generators of the multiplicative group of the odd integers modulo $2^k$, the set $\{-1, 5\}$ is offered.
However, the number 5 can be replaced by 3 which seems more logical for a standard choice. The proof I know do not suffer from these change.
Is there a particular reason for this traditional choice of generator?

REPLY [3 votes]: I've seen this proved by proving that $5^{2^{k-3}} \equiv 1 + 2^{k-1} \not\equiv 1\bmod 2^k$, for $k\ge 3$, using induction. (See for instance Elementary Number Theory by Bolker.)
On the other hand, $3^{2^{k-3}} \equiv 1 + 2^{k-1} \bmod 2^k$, for $k\ge 4$ but not for $k=3$, even though it is still true that $3^{2^{k-3}} \not\equiv 1\bmod 2^k$ for $k=3$.
So, using $5$ allows a slightly shorter proof.