Let $p>2$ be a prime number, $V=\left(\mathbb{Z}/p^n\mathbb{Z}\right)^{2k+1}$. The bilinear form $$B:V\times V \rightarrow \mathbb{Z}/p^n\mathbb{Z}$$ is a perfect pairing. That is, mapping $x\in V$ to $B(x,-)\in V^*$ is an isomorphism between $V$ and $V^*$.

Is it true that the number of solutions to $B(x,x)=0$ does not depend on $B$. Also, what is intuition that this is happening for odd rank $V$, but not even ones? How about over general rings?

Edit: It has been pointed out that this won't hold for general $B(x,x)=c$ (which was the original version of this question). Looks like it is true for $c=0$, though. Still want to ask the intuition behind.

I believe this invariance can lead to some interesting facts. Like this MO post, tries to count the number of solutions to the quadratic equation $$𝑥^2_1+⋯+𝑥^2_m=0.$$ If $m=2k+1$ is odd, This is indeed our case when $B$ is the identity matrix. Using the invariance, one can compute it by counting the number of solutions to $$x_1(x_2+\ldots+x_{2k+1})+x_2(x_3+\ldots+x_{2k+1})+\ldots + x_{2k}x_{2k+1}=0,$$ which comes from the case $B=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1\\ 1 & 0 & \cdots & 0 & 0\\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 1 &\cdots &1 &0 \end{pmatrix}$.


2 Answers 2


This is already false for $k=0$ (and $n$ arbitrary)

Le $B_0(x,y)=xy$ and $B_1(x,y)=-xy$, and $c=-1$. We then have two equations $x^2=-1$ and $-x^2=-1$. The second one always has at least two solutions (maybe more), while the first have no solutions if $p\equiv 3 \mod 4$.

  • $\begingroup$ Thank you. It looks like it is only true for c=0. Any intuition why it is happening? $\endgroup$
    – Ted Mao
    Jul 9, 2021 at 18:31

Your claim is never true for $n=1$, assuming nondegeneracy. This gives also many counterexamples for $n>1$ using Hensel's lemma.

In Lidl and Niederreiter's book "Finite Fields", 2nd edition, Chapter 6, section 2 (quadratic forms) you'll find plenty of information on the $n=1$ case. In particular, from their Theorem 6.27 it follows that for fixed $k$ and nondegenerate $B\colon (\mathbb{Z}/p\mathbb{Z})^{2k+1}\times (\mathbb{Z}/p\mathbb{Z})^{2k+1} \to \mathbb{Z}/p\mathbb{Z} $, the number of solutions to $B(x,x)=c$ is a non-constant (explicit) function of the Legendre symbol of $c$ mod $p$.

The reason for the parity difference is essentially their Lemma 6.24, showing that quadratic forms in two variables are well-behaved, namely the function $b \mapsto \#\{(x_1,x_2): a_1 x_1^2 + a_2 x_2^2 =b\}$ is essentially constant (depends only on whether $b=0$ or not), and then some linear algebra allows you to reduce the study of $B(x,x)=c$ when $x \in (\mathbb{Z}/p\mathbb{Z})^{2k}$ to $k=1$ (which behaves almost like a constant) and the study of $B(x,x)=c$ when $x \in (\mathbb{Z}/p\mathbb{Z})^{2k-1}$ to $k=1$, that is, counting $b$ with $ax^2 = b$, which clearly depends on the Legendre symbol of $b$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .