Prove that the order of $5$ mod $2^k$ is equal to $2^{k-2}$ where k is any integer greater than or equal to 3.

Question

Here is a proof: Prove that $\text{ord}_{2^k}5=2^{k-2}$ where $k$ is any integer $\geq3$

I do understand that $n_k$ is odd but how does that relate to the order of 5 mod $2^k$? Why does that show that $2^{k-2}$ is the least positive integer stastifying the required property? Thanks.

Answer 1

By Lifting the Exponent lemma $v_2 (5^{2^n} -1 ) = v_2(5-1)+v_2 (2^n) = n+2$

Hence the highest exponent of $2$ that divides $5^{2^{k-2}}-1$ is $k$.

Conversely, the order of $5$ mod $2^k$ will therefore be $2^{k-2}$