International Journal of Mathematics and Mathematical Sciences
Volume 2015 (2015), Article ID 305730, 5 pages
http://dx.doi.org/10.1155/2015/305730

Research Article

Lower Bounds on Solutions of Quadratic Polynomials Defined over Finite Rings

Ali H. Hakami

Department of Mathematics, Faculty of Science, Jazan University, P.O. Box 277, Jazan 45142, Saudi Arabia

Received 17 January 2015; Revised 16 April 2015; Accepted 20 April 2015

Academic Editor: Harvinder S. Sidhu

Copyright © 2015 Ali H. Hakami. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

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 .

1. Introduction

Let be a finite ring with positive integer. Let be a quadratic polynomial in . Write

where , , and 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 . Let be the algebraic subset of defined by

If and , we will say that if is congruent to modulo the ideal . For any subset of and divisor of , let

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 [1–8]. 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 of a box in under the canonical mapping of onto , where

for some with , . In this case we obtain the following.

Theorem 2. Suppose that . Let be a box in whose sides are all of the same length ; that is, let be the image of a box as given in (7), where , . Put . Then,

In particular, if and , then is nonempty.

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.

2. Auxiliary Lemmas

Lemma 3. If , then for any integer ,

where the product is over all primes dividing . In particular, if , then for all , .

Proof. Let and set , so that . Since and are both multiplicative, is multiplicative. If is a prime and , then

The last inequality follows since . The first part of the lemma now follows from the multiplicative property of .
Now suppose that . Again, letting we can say that

since . But

Thus,

To prove Theorem 2, we make use of exponential sums. Let . We will abbreviate complete sums by simply . Also we will need to use the following fundamental identity: for any ,

Let and be as defined by (1) and (2). By viewing as a -module, the Gauss sums

are well defined whether we take or .

For any matrix with integer entries, we define by

We need the following lemmas.

Lemma 4 (see [10, Theorem4]). Let be a quadratic form given by (2), where now we allow to be any symmetric integral matrix with even diagonal entries. Then, given , the Gauss sum is zero unless for all , in which case .

Lemma 5. Let be a quadratic polynomial as given by (1) with . Let , , , and set . Then,

Proof. By (1), we have

Now , so that by Lemma 4, unless satisfies the following condition:

Now

but as we conclude that

Thus, setting , we see that (19) is equivalent to saying that, for all ,

But and similarly so that (22) simplifies to the congruence ; that is, . Hence, satisfies (19) if and only if, for all , , that is . If then by Lemma 4, and (21), we have

3. Proof of Theorem 2

Let and be subsets of , and let be the set of points in satisfying . Let be the number of triples such that .

Lemma 6. Let be the number of triples such that . Then, for any subsets and of , with we have

Proof. By the fundamental identity (14),

where

Peeling off the term yields

Thus (from (27)),

so that by Lemma 5,

where the sum on is over all . On replacing by and by , the sum over all , we obtain

Now,

Setting and letting run through a complete set of representatives for , we can say

Thus, by (31),

and therefore, by (30),

Proof of Theorem 2. Let

for some , , where . Let , , and be the images of , , and in under the canonical mapping of onto . Then, and . We claim that for any divisor of ,

and that the same inequality holds for . Let be a fixed point in . If is a point in such that , then , for some , . Since , there are at most choices for each , and thus at most choices for . Since , we obtain (36). It is clear that so that (36) holds also for .
We now apply Theorem 1 with and replaced by the sets and just defined. We will abbreviate the sum by simple . From Lemma 6 and (36) we have

Now, since , we have

The first sum on the right-hand side of (38) is just . We make crude estimates for the remaining sums. For we have

Thus, by (38), we see that

Theorem 2 now follows from the observation that

4. Remarks

(1)It is clear from (39) and Lemma 3 that if then in the statement of Theorem 2 we can replace by .(2)Let be given by (1), let be the set of zeros of in , and again suppose that . Lemma 5 provides us with an easy means of estimating . For , we obtain

where the product is over all primes dividing , and for each such , is a real number of absolute value ≤1. Equation (42) follows from the observation that

By Lemma 5, we then have

for some , with . Since

we obtain

To obtain (42), we apply (46) in turn to each prime power dividing and use the Chinese remainder theorem to compute . That is, for each divisor of , we let be the number of points in satisfying the congruence . Consequently, if then . Thus, by (46), we have

for some with , , and by Lemma 3 we have

for any prime power , when . Equation (42) is now immediate.

Equation (42) indicates that we obtain roughly the expected quota of zeros for , namely, , when . When , this is no longer the case. For example, suppose that , where and are distinct primes. Let be quadratic nonresidue , let be a quadratic nonresidue , and let be defined by

If is an integral solution of the congruence then and , so that and . Thus, if is the set of points in satisfying , then , rather than expected quota of . This example indicates that Theorem 2 does not hold when .

We have not been able to obtain an analogue of Theorem 2 when . The main difficulty is that no longer leads to such a simple description as in the case when ; see (21). To overcome this difficulty, one may be able to use the description of given in section 4 of [11] which involved the invariant factors of .

Conflict of Interests

The author has no competing interests.

Acknowledgments

The author is very grateful to the referees and the editors for their valuable comments and some helpful suggestions that improved the clarity and readability of the paper. The author is thankful to Jazan University (JU), Jazan, Saudi Arabia, for providing excellent research facilities to carry out this research.

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