Divisibility question

Let $r$ be an integer greater than $2$. Is there a simple way of showing that $2^r$ divides $\left(\begin{array}{c} {2}^{r-2} \\ k \end{array}\right) 2^{2k}$ but it does not divide $\left(\begin{array}{c} 2^{r-l} \\ k \end{array}\right) 2^{2k}$ for $l>2$ and for all $k$ greater than $0$ and less or equal than $2^{r-2}$?

Equivalently, is there a simple way of showing that the element $5$ has order $2^{r-2}$ in the group of units of the ring $\mathbb{Z}_{2^r}$?

The two questions are connected by writing $5 = 4 + 1$ and using the binomial expansion formula to study the order of $5$.

Solution 1:

I don't know off-hand about the binomial coefficients, but here's a proof of the order of $5$.

For $r=3$, the order of $5$ is $2$, which agrees with what we want. So assume that $r\gt 3$.

Assume that the order of $5$ has order $2^{r-2}$ modulo $2^{r}$. The order of $5$ modulo $2^{r+1}$ is a multiple of $2^{r-2}$. It cannot be of order $2^r$ (the order of the group), because both $2^{r}-1$ and $2^r+1$ have order $2$ and they are not congruent modulo $2^{r+1}$ so the group of units is not cyclic. So the order of $5$ is either $2^{r-2}$ or $2^{r-1}$.

Since the order of $5$ modulo $2^{r}$ is $2^{r-2}$, and the order modulo $2^{r-1}$ is $2^{r-3}$, then $5^{2^{r-3}}\equiv 1 \pmod{2^{r-1}}$, so $5^{2^{r-3}}\equiv 1 + k2^{r-1}\pmod{2^{r}}$, with $k=0,1$. Can't be $k=0$, because then the order of $5$ modulo $2^r$ would divide $2^{r-3}$, so $5^{2^{r-3}}\equiv 1 + 2^{r-1}\pmod{2^r}$. Hence $5^{2^{r-3}}\equiv 1 + 2^{r-1} + k2^r\pmod{2^{r+1}}$, with $k=0$ or $1$. Squaring both sides, we have: $$\begin{align*} 5^{2^{r-2}} &\equiv (1 + 2^{r-1} + k2^r)^2\pmod{2^{r+1}}\\ &\equiv 1 + 2^{2r-2} + k2^{2r} + 2^r + k2^{r+1} + k2^{2r}\pmod{2^{r+1}}\\ &\equiv 1+2^r\pmod{2^{r+1}}. \end{align*}$$ So the order of $5$ is $2^{r-1}$, as desired.