Small zeros of quadratic congruences to a prime power modulus

Abstract

Let $m$ be a positive integer, $p$ be an odd prime, and $\mathbb{Z}{p^m } = \mathbb{Z}/(p^m )$ be the ring of integers modulo $p^m $. Let $$Q({\mathbf{x}}) = Q(x_1 ,x_2 ,...,x_n ) = \sum\limits{1 \leqslant i \leqslant j \leqslant n} {a_{ij} x_i x_j } ,$$ be a quadratic form with integer coefficients. Suppose that $n$ is even and $\det A_Q \not \equiv 0;(\bmod p)$. Set $\Delta = (( - 1)^{n/2} \det A_Q /p)$, where $( \cdot /p)$ is the Legendre symbol and $\left| {\mathbf{x}} \right| = \max \left| {x_i } \right|$. Let $V$ be the set of solutions the congruence $ $Q({\mathbf{x}}), \equiv ;0\quad (\bmod p^m ) \quad(1)$$, contained in $\mathbb{Z}^n $ and let $B$ be any box of points in $\mathbb{Z}^n $of the type $$B = \left{ {{\mathbf{x}} \in \mathbb{Z}^n \left| {,a_i \leqslant x_i < a_i + m_i ,;,1 \leqslant i \leqslant n} \right.} \right},$$ where $a_i ,m_i \in \mathbb{Z},;1 \leqslant m_i \leqslant p^m $. In this dissertation we use the method of exponential sums to investigate how large the cardinality of the box $B$ must be in order to guarantee that there exists a solution ${\mathbf{x}}$of (1) in $ B$. In particular we will focus on cubes (all $m_i $equal) centered at the origin in order to obtain primitive solutions with $\left\| {\mathbf{x}} \right\|$ small. For $m = 2$ and $n \geqslant 4$ we obtain a primitive solution with $\left\| {\mathbf{x}} \right\| \leqslant \max \left\{ {2^5 p,2^{18} } \right\}$. For $m = 3$, $n \geqslant 6$, and $\Delta = + 1$, we get $\left\| {\mathbf{x}} \right\| \leqslant \max \left\{ {2^{2/n} p^{(3/2) + (3/n)} ,2^{(2n + 4)/(n - 2)} } \right\}$. Finally for any $m \geqslant 2$, $n \geqslant m,$ and any nonsingular quadratic form we obtain $\left\| {\mathbf{x}} \right\| \leqslant \max \{ 6^{1/n} p^{m[(1/2) + (1/n)]} ,2^{2(n + 1)/(n - 2)} 3^{2/(n - 2)} \} $. Others results are obtained for boxes $B$ with sides of arbitrary lengths.