Let $D_5$ be the dihedral group of order ten, i.e. $D_5=\langle x,y\mid x^5=y^2=1=(xy)^2\rangle$. Let $\mathbb{F}_{2^k}$ be the field with $2^k$ elements. Consider the following subsets of the group ring $\mathbb{F}_{2^k}[D_5]$ $$A = \bigg\{\sum_{i=0}^4a_i x^{i}+b_i x^{i}y \ \big| \sum_{i=0}^{4} a_i+b_i = 0 : a_i, b_i \in \mathbb{F}_{2^k}\bigg\}$$ and $$B = \bigg\{\sum_{i=0}^4a_i x^{i}+b_i x^{i}y \ \big|\bigg( \sum_{i=0}^{4} a_iw^i\bigg) \bigg( \sum_{i=0}^{4} a_iw^{-i}\bigg) = 0 = \bigg( \sum_{i=0}^{4} b_iw^i\bigg) \bigg( \sum_{i=0}^{4} b_iw^{-i}\bigg) : a_i, b_i \in \mathbb{F}_{2^k}\bigg\}.$$ Here $w$ is primitive $5^{th}$ root of unity in $\mathbb{F}_{2^k}$, so we assume that $5\mid2^k-1$. Equivalently, we assume that $4\mid k$.
Now if $|A| = 2^{9k}$ and $|B| = (2^{4k+1}-2^{3k})^2$, then what is the $|A\cap B|$?
$|B|$ is calculated as in How to find no. of polynomials over a finite field satisfying given condition.$. Any hint.
