Let $m=4k+1$ where $k$ is a non-negative integer. Show that $$a=\binom{n}{1}+m\binom{n}{3}+m^2\binom{n}{5}+\cdots+m^{\frac{n-1}{2}}\binom{n}{n}$$
is divisible by $2^{n-1}$, where $n$ is an odd number.
First, we claim $a=\frac{1}{2\sqrt{m}}\left((1+\sqrt{m})^n - (1-\sqrt{m})^n\right)$. This is because $$\begin{array}{rl} (1+\sqrt{m})^n &= C_n^0 + C_n^1\cdot m^\frac{1}{2} + C_n^2\cdot m + C_n^3\cdot m^\frac{3}{2} + C_n^4\cdot m^2+\cdots+C_n^n\cdot m^\frac{n}{2}\\ (1-\sqrt{m})^n &= C_n^0 - C_n^1\cdot m^\frac{1}{2} + C_n^2\cdot m - C_n^3\cdot m^\frac{3}{2} + C_n^4\cdot m^2+\cdots+(-1)^nC_n^n\cdot m^\frac{n}{2} \end{array}$$
Subtracting the $2^{nd}$ relation from the $1^{st}$ and dividing $2\sqrt{m}$ immediately leads to the claim. Rewriting $a$ as $$\begin{array}{rl} a &=\frac{1}{2\sqrt{m}}((1+\sqrt{m})^n - (1-\sqrt{m})^n\\ &=\frac{1}{2}\cdot\frac{1}{\sqrt{m}}\cdot\Big(\Big({2}\cdot\frac{1+\sqrt{m}}{2}\Big)^n -\Big({2}\cdot\frac{1-\sqrt{m}}{2}\Big)^n \Big)\\ &=2^{n-1}\cdot\frac{1}{\sqrt{m}}\cdot\Big(\Big(\frac{1+\sqrt{m}}{2}\Big)^n -\Big(\frac{1-\sqrt{m}}{2}\Big)^n\Big) \end{array}$$
Therefore, to prove the original claim that $2^{n-1}|a$ is equivalent to show that $\frac{1}{\sqrt{m}}\cdot\Big(\Big(\frac{1+\sqrt{m}}{2}\Big)^n -\Big(\frac{1-\sqrt{m}}{2}\Big)^n\Big)$ is an integer where $m=4k+1$. This indeed holds. Please see # 3862.