Given $n\equiv 1\pmod 8$, show that the number of subsets of a $n$-element set, whose size is $0\pmod 4$ is $2^{n-2}+2^ \frac{n-3}{2}$ Given $n\equiv 1\pmod 8$, show that the number of subsets of a $n$-element set, whose size is $0\pmod 4$ is $2^{n-2}+2^ \frac{n-3}{2}$
I don't get this question. If I have an n-element set, then it's size has to be $1\pmod 8$ and therefore cannot be $0\pmod 4$. Isn't this correct? How does this make sense? Any help would be appreciated!

What you want is 
$$\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$$
for $n\equiv 1\pmod 8$.
Note that we have the followings :
$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}+\cdots+\binom{n}{n}=2^n$$
$$\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\binom{n}{4}-\cdots+(-1)^n\binom{n}{n}=0$$
$$\binom{n}{0}+\binom{n}{1}i-\binom{n}{2}-\binom{n}{3}i+\binom{n}{4}+\binom{n}{5}i-\cdots=(1+i)^n$$
$$\binom{n}{0}-\binom{n}{1}i-\binom{n}{2}+\binom{n}{3}i+\binom{n}{4}-\binom{n}{5}i-\cdots=(1-i)^n$$
Adding these gives us
$$4\left(\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots\right)=2^n+(1+i)^n+(1-i)^n$$$$=2^n+2^{n/2}\left(\cos (n\pi/4)+i\sin (n\pi/4)\right)+2^{n/2}\left(\cos (n\pi/4)-i\sin (n\pi/4)\right)$$$$=2^n+2\cdot 2^{n/2}\cos(n\pi/4).$$
Hence, we have
$$\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots=2^{n-2}+2^{(n-2)/2}\cos (n\pi/4)=2^{n-2}+2^{(n-3)/2}.$$
Here, note that $\cos (n\pi/4)=2^{-1/2}$ for $n\equiv 1\pmod 8$.
