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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 $n$ 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 $b$.

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 $b$.

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 $n$ 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 $b$.

By Hensel's Lemma, this produces countless example for $n>1$ as well.

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 $n$ 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 $b$.

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 $n$ 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 an (explicit) function of the Legendre symbol of $c$ mod $p$. So only for degenerate quadratic forms this may be independent of $p$.

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 and induction 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 constantly) 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 $b$.

By Hensel's Lemma, this produces countless example for $n>1$ as well.

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 $n$ 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 $b$.

By Hensel's Lemma, this produces countless example for $n>1$ as well.