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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 $B(x,x)=c$ 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 $c$. 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 $B=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1\\ 1 & 0 & \cdots & 0 & 0\\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 1 &\cdots &1 &0 \end{pmatrix}$.

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 $B=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1\\ 1 & 0 & \cdots & 0 & 0\\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 1 &\cdots &1 &0 \end{pmatrix}$.

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 $B(x,x)=c$ 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?

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 $B=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1\\ 1 & 0 & \cdots & 0 & 0\\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 1 &\cdots &1 &0 \end{pmatrix}$.

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 $B(x,x)=c$ 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 $c$. 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 $B=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1\\ 1 & 0 & \cdots & 0 & 0\\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 1 &\cdots &1 &0 \end{pmatrix}$.

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 $B(x,x)=c$ does not depend on $B$. Also, what is intuition that this is happening for odd rank $V$, but not even ones?

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 $B=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1\\ 1 & 0 & \cdots & 0 & 0\\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 1 &\cdots &1 &0 \end{pmatrix}$.

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 $B(x,x)=c$ 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?

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 $B=\begin{pmatrix} 0 & 0 & \cdots & 0 & 1\\ 1 & 0 & \cdots & 0 & 0\\ 1 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots &\vdots \\ 0 & 1 &\cdots &1 &0 \end{pmatrix}$.