# Cardinality of $A_n=\{ x \in \mathbb{F}_{5}^{n} : x\cdot x=0\}$ [duplicate]

Let $$A_n=\{ x \in \mathbb{F}_{5}^{n} : x\cdot x=0\}$$. Show that $$|A_n|=(1/5+f(n))5^{n}$$ where $$\lim_{n \to \infty} f(n) = 0$$.

Notation: The ring $$(\mathbb{F}_5^{n},+,*)$$ have elements of the form $$(x_1,\ldots,x_n)$$,$$x_i \in \mathbb{F}_5$$, $$\mathbb{F}_{5}$$ is the field $$\mathbb{Z}/5\mathbb{Z}$$, $$|A|$$ denotes the cardinality of the set $$A$$ and $$x\cdot x = x_1^2+\ldots+x_n^2$$ . Life would have been easy if the map $$\phi: \mathbb{F}_{5}^{n} \to \mathbb{F}_5$$ defined by $$\phi(x) = x\cdot x$$ was a ring homomorphism, but sadly that is not the case. I am not even sure how to start. Any hints?

• Studied here for a general finite field $\Bbb{F}_q$ of which $q=5$ is a special case. As $5\equiv1\pmod4$ the simpler answer works. Commented Sep 6, 2021 at 12:42
• Thanks a lot @JyrkiLahtonen.
– Sam
Commented Sep 6, 2021 at 12:48

Denote by $$M_n(b) = |\{x \in \mathbb{F}_5^n:x\cdot x = b\}|$$ the set of vectors with inner product equal to $$b \in \mathbb{F}_5$$. We will establish a recursion on $$M_n(b)$$ as follows. We can write, for example, $$M_{n+1}(0) = M_n(0) + 2M_n(1) + 2M_n(4),$$ since $$x_{n+1}^2$$ can only take values in $$\{0,1,4\}$$, where the values $$1$$ and $$4$$ are assumed with multiplicity 2 for $$x_{n+1}=1$$, $$x_{n+1}=4$$ and for $$x_{n+1}=2$$, $$x_{n+1}=3$$. In general, we have $$M_{n+1}(b) = M_n(b) + 2M_n(b-1) + 2M_n(b+1)$$ and thus we can deduce the following recursion, $$\begin{pmatrix}M_{n+1}(0)\\M_{n+1}(1)\\M_{n+1}(2)\\M_{n+1}(3)\\M_{n+1}(4) \end{pmatrix} = \begin{pmatrix} 1&2&0&0&2 \\2&1&2&0&0\\0&2&1&2&0 \\ 0&0&2&1&2 \\ 2&0&0&2&1 \\ \end{pmatrix} \begin{pmatrix}M_{n}(0)\\M_{n}(1)\\M_{n}(2)\\M_{n}(3)\\M_{n}(4) \end{pmatrix}.$$ The claim then follows from the standard analytical combinatorics arguments using the fact that the largest Eigenvector of the recursion matrix is 5.