I am interested in solving, or even just deciding the existence of a solution, for a system of quadratic diophantine equations.

Let $p$ be a prime congruent to 1 modulo 8, so $ p =17$ is the first case. We want to solve the following equations inside the integers.

Let $\alpha_1,\alpha_2,...,\alpha_p$ and $\beta_1,\beta_2,...,\beta_p$ be unknowns. We understand the indices of the $\alpha$'s and $\beta$'s modulo $p$, so that e.g. $\alpha_{-1} = \alpha_{p-1}$. We have the linear dependence between the unknowns:

\begin{align} & \sum_{j = 1}^p \alpha_j = 0, \\ & \sum_{j = 1}^p \beta_j = 0 \end{align}

Moreover for all $i$ between $1$ and $p-1$ we have two quadratic equations given as: \begin{align} & \sum_{j=1}^p \alpha_j\beta_{j-i} + \alpha_{j-i}\beta_j = 0, \\ & \sum_{j=1}^p - \alpha_j\alpha_{j-i} + \beta_j\beta_{j-i} = \left\{ \begin{array}{ll} 2, & i = \pm 2 \\ 0, & i \neq \pm 2 \end{array}\right. \end{align} It is easy to see that the equations for $\pm i$ are actually the same. Additionally we have conditions on the parities: \begin{align} & \alpha_1 \equiv \alpha_{p-1} \equiv \beta_1 \equiv \beta_{p-1} \equiv 1 \mod 2, \\ & \alpha_i \equiv \beta_i \equiv 0 \mod 2, \ \ \text{if} \ \ i \neq \pm 1 \end{align}

Alternative formulation: The problem can also be formulated using symmetric quadratic forms. For this let $A$ be the $p \times p$-permutation matrix $$\begin{pmatrix} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{pmatrix} $$

Note that $A$ has order $p$. Let $B_i = A^i + (A^i)^T$ for each $1 \leq i \leq \frac{p-1}{2}$. Writing in block form, in particular $0$ for a $0$-matrix, define for $1 \leq i \leq \frac{p-1}{2}$ the quadratic forms $$Q_i: \mathbb{Z}^{2p} \rightarrow \mathbb{Z}, \ \ x \mapsto x^T \begin{pmatrix} 0 & B_i \\ B_i & 0 \end{pmatrix}x $$ and $$R_i: \mathbb{Z}^{2p} \rightarrow \mathbb{Z}, \ \ x \mapsto x^T \begin{pmatrix} -B_i & 0 \\ 0 & B_i \end{pmatrix}x $$ Then we can formulate the equations as $$Q_i(x) = 0 \ \ \text{and} \ \ R_i(x) = \delta_{i,2} $$ for all $1 \leq i \leq \frac{p-1}{2}$ where $\delta_{i,j}$ is the Kronecker delta.

If this is of any help, we can completely solve the equations modulo $4$. This gives that \begin{align} & \alpha_1\beta_{p-1} + \alpha_{p-1}\beta_1 \equiv 1 \mod 4, \\ & \alpha_i + \beta_i + \alpha_{-i} + \beta_{-i} \equiv 0 \mod 4 \ \ \text{if} \ \ i \neq 0, \pm 1 \\ & \alpha_0 + \beta_0 \equiv 2 \mod 4 \end{align}

By the origin of the problem from a question on group rings, we also know that solutions exist when p is congruent to 3 modulo 4, but we have no clue when p is congruent to 1 modulo 8. We have tried some computer experiments, but found no solution and the system seems too big for a complete solution by the programs we tried.

I am interested in solving, or even just deciding the existence of a solution, for a system of quadratic diophantine equations.

Let $p$ be a prime congruent to 1 modulo 8, so $ p =17$ is the first case. We want to solve the following equations inside the integers.

Let $\alpha_1,\alpha_2,...,\alpha_p$ and $\beta_1,\beta_2,...,\beta_p$ be unknowns. We understand the indices of the $\alpha$'s and $\beta$'s modulo $p$, so that e.g. $\alpha_{-1} = \alpha_{p-1}$. We have the linear dependence between the unknowns:

\begin{align} & \sum_{j = 1}^p \alpha_j = 0, \\ & \sum_{j = 1}^p \beta_j = 0 \end{align}

Moreover for all $i$ between $1$ and $p-1$ we have two quadratic equations given as: \begin{align} & \sum_{j=1}^p \alpha_j\beta_{j-i} + \alpha_{j-i}\beta_j = 0, \\ & \sum_{j=1}^p - \alpha_j\alpha_{j-i} + \beta_j\beta_{j-i} = \left\{ \begin{array}{ll} 2, & i = \pm 2 \\ 0, & i \neq \pm 2 \end{array}\right. \end{align} It is easy to see that the equations for $\pm i$ are actually the same. Additionally we have conditions on the parities: \begin{align} & \alpha_1 \equiv \alpha_{p-1} \equiv \beta_1 \equiv \beta_{p-1} \equiv 1 \mod 2, \\ & \alpha_i \equiv \beta_i \equiv 0 \mod 2, \ \ \text{if} \ \ i \neq \pm 1 \end{align}

Alternative formulation: The problem can also be formulated using symmetric quadratic forms. For this let $A$ be the $p \times p$-permutation matrix $$\begin{pmatrix} 0 & 0 & \cdots & 0 & 1 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{pmatrix} $$

Note that $A$ has order $p$. Let $B_i = A^i + (A^i)^T$ for each $1 \leq i \leq \frac{p-1}{2}$. Writing in block form, in particular $0$ for a $0$-matrix, define for $1 \leq i \leq \frac{p-1}{2}$ the quadratic forms $$Q_i: \mathbb{Z}^{2p} \rightarrow \mathbb{Z}, \ \ x \mapsto x^T \begin{pmatrix} 0 & B_i \\ B_i & 0 \end{pmatrix}x $$ and $$R_i: \mathbb{Z}^{2p} \rightarrow \mathbb{Z}, \ \ x \mapsto x^T \begin{pmatrix} -B_i & 0 \\ 0 & B_i \end{pmatrix}x $$ Then we can formulate the equations as $$Q_i(x) = 0 \ \ \text{and} \ \ R_i(x) = 4\delta_{i,2} $$ for all $1 \leq i \leq \frac{p-1}{2}$ where $\delta_{i,j}$ is the Kronecker delta.

If this is of any help, we can completely solve the equations modulo $4$. This gives that \begin{align} & \alpha_1\beta_{p-1} + \alpha_{p-1}\beta_1 \equiv 0 \mod 4, \\ & \alpha_i + \beta_i + \alpha_{-i} + \beta_{-i} \equiv 0 \mod 4 \ \ \text{if} \ \ i \neq 0, \pm 1 \\ & \alpha_0 + \beta_0 \equiv 2 \mod 4 \end{align}

By the origin of the problem from a question on group rings, we also know that solutions exist when p is congruent to 3 modulo 4, but we have no clue when p is congruent to 1 modulo 8. We have tried some computer experiments, but found no solution and the system seems too big for a complete solution by the programs we tried.