Small zeros of quadratic congruences to a prime power modulus

Abstract

Let m be a positive integer, p be an odd prime, and Zpm=Z/(pm) be the ring of integers modulo $p^m $. Let Q(x)=Q(x1,x2,...,xn)=∑1⩽i⩽j⩽naijxixj,be a quadratic form with integer coefficients. Suppose that n is even and detAQ≢0(modp). Set Δ=((−1)n/2detAQ/p), where (⋅/p) is the Legendre symbol and ‖x‖=max|xi|. Let V be the set of solutions the congruence $ Q(x)≡0(modpm)(1)$, contained in $\mathbb{Z}^n $ and let B be any box of points in $\mathbb{Z}^n $of the type B={x∈Zn|ai⩽xi<ai+mi,1⩽i⩽n},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.