Lower Bounds on Solutions of Quadratic Polynomials Defined over Finite Rings

Let be a positive integer and let denote the ring , and let denote the Cartesian product of copies of . Let be a quadratic polynomial in . In this paper, we are interested in giving lower bounds on the number of solutions of the quadratic polynomial over the ring .


Introduction
Let   = Z/() be a finite ring with  positive integer.Let (x) be a quadratic polynomial in Z[ 1 , . . .,   ].Write  (x) =  (x) + a ⋅ x + , where a ∈ Z  ,  ∈ Z, and (x) is a quadratic form given by where  is a symmetric  ×  matrix with integer entries.Throughout this paper (with the exception of Lemma 4), we will assume that gcd(det , ) = 1.Let  be the algebraic subset of    defined by  (x) ≡ 0 (mod ) .
Let  denote the Euler phi-function, let () denote the number of distinct positive divisors of , and for positive integers  and  set Considerable attention has been given to the problem of finding zeros of any polynomial over finite fields; see for example [1][2][3][4][5][6][7][8].A special case of particular interest when the polynomial is quadratic over finite rings is studied in [9] by the author who obtained the following.Theorem 1.For any subsets  and  of    , with || ≤ ||, we have In this paper, we will make the above result more precise when  +  is a box of points in    , that is, the image B of a box B in Z  under the canonical mapping of Z  onto    , where for some   ,   ∈ Z with 0 <   ≤ , 1 ≤  ≤ .In this case we obtain the following.
The second part of Theorem 2 follows immediately from Lemma 3 of the next section.The first part of the theorem will be proven in Section 3.

Auxiliary Lemmas
where the product is over all primes  dividing .In particular, if  ≥ 6, then for all  > 1, The last inequality follows since  ≥ 2. The first part of the lemma now follows from the multiplicative property of ().Now suppose that  ≥ 6.Again, letting  = /2 we can say that since () < . Thus, To prove Theorem 2, we make use of exponential sums.Let   () =  (2/) .We will abbreviate complete sums ∑ ∈   () by simply ∑  ().Also we will need to use the following fundamental identity: for any y ∈    , Let (x) and (x) be as defined by ( 1) and (2).By viewing    as a Z-module, the Gauss sums are well defined whether we take y ∈ Z  or y ∈    .For any  ×  matrix  with integer entries, we define ker  () by We need the following lemmas.

Proof of Theorem 2
Let  and  be subsets of    , and let  be the set of points in    satisfying (x) ≡ 0 (mod ).Let  be the number of triples (s, t, k) ∈  ×  ×  such that s + t ≡ k (mod ).Lemma 6.Let  be the number of triples (s, t, k) ∈  ×  ×  such that s + t ≡ k (mod ).Then, for any subsets  and  of    , with || ⩽ || we have Proof.By the fundamental identity (14), where Peeling off the  = 0 term yields where the sum ∑ * on y is over all y ≡ 0 (mod ).

Theorem 2 .
(7)pose that  ≥ 4. Let B be a box in    whose sides are all of the same length  < ; that is, let B be the 2 International Journal of Mathematics and Mathematical Sciences image of a box B as given in(7), where   = , 1 ≤  ≤ .