Existence of solution for a system of quadratic diophantine equations / symmetric quadratic froms

Question

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.

Answer 1

This sketch of a half-answer is based on and is developing the ideas of Max’ answer. He works with $\mathbb Q[I,x]$ with $I^2=-1$ modulo the cyclotomic polynomial $\Phi_p(x)$. Writing $x$ as $z^4$ and $I$ is $z^p$, this is identified with the $4p$-cyclotomic field $K$, i.e., $\mathbb Q[z]$ modulo $\Phi_{4p}(z)$. In particular, Max’ $F$ is identified with an element $f\in K$.

Since his $F^\star(x) = x^{p-2}F(1/x)$, his $x^{2-p}F^\star$ is identified with $\sigma\cdot f$ for a suitable involution $\sigma$ of $K$. One can see that $σz=z^{2p-1}=-z^{-1}$ (so $σI=I$, $σx=1/x$). Denote by $N$ the norm $N(g)≔g·σg$ of $K$ over the fixed points $K_2$ of $σ$. Conclusion: Max’ equation is equivalent to $N(f) = -2(z^2+1/z^2)^2$ (with integer $f$).

Since I do not recollect what Class Field/Iwasawa Theories say exactly about this equation, I use ad hoc method: I consider solutions in cyclotomic units instead.

Since $N(I)=N(z)=-1$, $N(1+I)=2I$, and $N(1+x)=(z^2+1/z^2)^2$, it is enough to solve $N(g)=\pm I$. Recall that since $4p$ is not a power of a prime, cyclotomic units are generated (multiplicatively) by the units $1-ζ$ (for primitive roots $ζ$ of $1$ of degree $4p$), and roots of $1$ in $K$. Modulo roots of $1$, they form a lattice spanned by such $1-ζ$ with $\Im ζ>0$ and the only relation¹⁾ $\prod (1-ζ)=I^{(p-1)/2}$. (Their importance is in the fact that they have finite index in units of $K$.)

¹⁾ Indeed, if $\Pi$ is this product, then $|\Pi|^2 = \Phi_{4p}(1) = 1$ (since $\Phi_{4p}(x)(x^2+1)(x^{2p}-1)=x^{4p}-1$), and combining $1-ζ$ and $1+1/ζ$ together gives $N(1-ζ)=1/ζ-ζ$ with argument $-π/2$. Hence argument of $\Pi$ is $(p-1)π/4$.

Lemma: If $p=2r+1$, then $U:≔\prod_{k=1}^r (1-z^{2k-1})$ solves $N(U)=±I$ provided $r$ is odd. Update: moreover, if $Λ$ is the (multiplicative) lattice generated by $(1-ζ)/(1-(-ζ^{-1}))$ with primitive roots ζ in the first quadrant. then $N(Λ)=1$, and any solution modulo this lattice is $U^s$ with an odd $s$.

Indeed, the argument in the footnote above shows that $U$ is the product over primitive roots $ζ$ of $1$ of degree $4p$ in the first quadrant, hence $N(U)$ is $\Pi$. (Update: this also describes all the solutions to $N(f)=\text{unit}$ for any $p$. In particular, if $4|p-1$, then $s∉ℤ$, hence there is no solution in cyclotomic units.)

Update: here was a completely wrong “theorem” about the case $4|p-1$. The arguments above show only this:

If the index of cyclotomic units inside units of $K$ is odd, there is no solution. (While vol.1 of Lang’s Cyclotomic fields contains some info about the 2-part of this index, I’m not fluent enough to use this info. I do not even know how to find it in PARI/gp.)

Answer 2

UPDATE. Using factorization $-2(x+1)^2x^{p-3}$ over the corresponding number field, I established that there are no solutions for $p=17$. Furthermore, I computationally verified that for primes $p<30$ we have solutions for all $p\equiv 3\pmod{4}$ and do not have any for $p\equiv 1\pmod{4}$.

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This is just an extended comment, giving reformulation of the problem and reducing it to just $p-1$ unknowns and $p-1$ quadratic equations over the Gaussian integers.

Consider the generating polynomials: \begin{split} A(x) &:= \sum_{i=0}^{p-1} \alpha_i x^i, \\ B(x) &:= \sum_{i=0}^{p-1} \beta_i x^i. \end{split}

The linear equations $\sum_j \alpha_j = \sum_j \beta_j = 0$ are equivalent to $A(1)=B(1)=0$, i.e., both $A(x)=(x-1)\bar A(x)$ and $B(x)=(x-1)\bar B(x)$ are multiples of $x-1$.

Viewing indices modulo $p$ is equivalent to viewing the polynomials modulo $x^p - 1 = (x-1)\Phi_p(x)$, where $\Phi_p(x) := 1 + x + \dots + x^{p-1}$ is $p$-th cyclotomic polynomial.

For reciprocal polynomials (of fixed degree $p-1$) we have $A^\star(x):=x^{p-1}A(x^{-1})\equiv x^{p-1}A(x^{p-1})\pmod{x^p-1}$ and $B^\star(x):=x^{p-1}B(x^{-1})\equiv x^{p-1}B(x^{p-1})\pmod{x^p-1}$. Then the quadratic equations (under the condition $A(1)=B(1)=0$) translate into $$\begin{cases} A(x)B^\star(x) + A^\star(x)B(x) \equiv 0 \pmod{x^p-1},\\ -A(x)A^\star(x) + B(x)B^\star(x) \equiv -4x^{p-1} + 2x + 2x^{p-3} \equiv 2(x^2-1)^2x^{p-3} \pmod{x^p-1} \end{cases} $$ Dividing both congruences by $(x-1)x(\frac1x-1)=-(x-1)^2$, we get $$\begin{cases} \bar A(x)\bar B^\star(x) + \bar A^\star(x)\bar B(x) \equiv 0 \pmod{\Phi_p(x)},\\ -\bar A(x)\bar A^\star(x) + \bar B(x)\bar B^\star(x) \equiv -2(x+1)^2x^{p-3} \pmod{\Phi_p(x)}. \end{cases} $$

In terms of polynomials over Gaussian integers, we have $$F(x)F^\star(x) \equiv -2(x+1)^2x^{p-3}\pmod{\Phi_p(x)},$$ where $$F(x) := \bar B(x) + I\cdot \bar A(x)$$ is a polynomial of degree $p-2$ over the Gaussian integers.

The last congruence can be viewed as a system of $p-1$ quadratic equations on the coefficients of $F(x)$ as unknowns.

Alternatively, it can also be viewed as the identity of palindromic polynomials: $$F(x)F^\star(x) + 2(x+1)^2x^{p-3} = G(x)\cdot \Phi_p(x),$$ where the left-hand side, $G(x)$, and $\Phi_p(x)$ are palindromic polynomials of degree $2p-4$, $p-3$, and $p-1$, respectively.