Recursive Summation of the nth Powers Consecutive Congruent Numbers

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1 Int. Journal of Math. Analysis, Vol. 7, 013, no. 5, 19-7 Recursive Summation of the nth Powers Consecutive Congruent Numbers P. Juntharee and P. Prommi Department of Mathematics Faculty of Applied Science King Mongut s University of Technology North Bango 1518 Phacharat 1 Road, Bangsue, Bango 10800, Thailand pjt@mutnb.ac.th, ptpm@mutnb.ac.th Abstract The main purpose of this paper is to find a recursive form for the summation of nth powers of consecutive congruent numbers. Explicit summation formulas for the first, second and third powers of consecutive congruent numbers are derived. These formulas are applied to some problems involving the nth powers of consecutive congruent numbers. Keywords: Recursive Summation, Consecutive Numbers, Congruent Numbers, Modulo 1 Introduction Calculating sums of nth powers of consecutive congruent numbers can tae considerable time. However, the sums can be easily obtained if a recursive form is available for the summation of nth powers of consecutive congruent numbers. With that motivation, we first revisit previous studies by Shiflett and Shultz [7] and Juntharee and Prommi [6]. Shiflett and Shultz showed that a positive number n can be expressed as a sum of consecutive congruent numbers j + 1, j + 3,..., s 1 for some integers j and s with j 0 and s > j + 1 if and only if n s js + j. On the other hand, Juntharee and Prommi [6] described a theorem giving conditions under which a positive number n can be expressed as a sum of consecutive positive congruent numbers. It was shown that the required conditions are as follows: For a given positive number m and for r 0, 1,, 3,..., m 1, a positive number n can be expressed as a sum of consecutive positive congruent numbers which are congruent to r modulo m if and only if

2 0 P. Juntharee and P. Prommi case 1 if m is an even positive number and m m 0 for some positive number m 0 then n can be factorized as n ab for some positive numbers a, b, where a b and b m 0 i + a 1 + r for some i 0, 1,, 3,... case if m is an odd positive number then n can be factorized as n ab for some positive numbers a, b, where a b and b mi + a 1 + r for some i 0, 1,, 3,... These results stimulated us to undertae this study. The outline of the present paper is as follows. In section, basic definitions and results required in this research are summarized. These include definitions of consecutive congruent numbers, and the binomial theorem, Pascal s triangle and sums of a finite series. In section 3, a recursive form is derived for calculating the sums of nth powers of consecutive congruent numbers and explicit summation formulas are obtained for sums of first, second and third powers of consecutive congruent numbers. In section 4, the applications of the results to two practical problems are described and conclusions are made. Preliminaries and Notation In this section, we summarize some general mathematical bacground required in this wor. For definition.1 see [1],[],[3],[5]. A proof of Theorem. The binomial theorem is given in [4]. Throughout this paper, Z, Z + and Z + 0 denote the sets of integers, positive integers and positive integers with zero respectively. Definition.1. Let m Z +, r Z + 0 and a, b Z such that r m 1. 1 We say that a is congruent to b modulo m if m divides a b. We use the notation a bmod m to indicate that a is congruent to b modulo m. Define the set N m, r {q Z : q rmod m}. Note that: since N m, r is an infinitely countable set, we can label elements in N m, r as follows:..., q i 1, q i, q i+1,... where q i im + r. One says that q i and q i+1 are consecutive congruent numbers in N m, r. Theorem.. The binomial theorem Let x and y be variables and let n be a positive integer. Then x+y n C n x n y, where C n n!!n!. Proof. See the proof in [4]. 0

3 Recursive summation of the nth powers consecutive congruent numbers 1 Lemma.3. Let m Z + and r Z + 0 such that r m 1. If q i and q i+1 are consecutive congruent numbers in N m, r then q i < q i+1 and q q i + mj, for all j Z. Proof. The results follow immediately from the fact that q i and q i+1 are consecutive congruent numbers in N m, r. 3 Main Results In this section, we prove the main Theorem 3.4 for recursive summation of the nth powers of consecutive congruent numbers. In Corollary 3.5 we use Theorem 3.4 to obtain explicit summation formulas for the first, second and third powers of consecutive congruent numbers. We begin by stating and proving some lemmas. Lemma 3.1. Let m Z + and r Z + 0 such that r m 1. If q i N m, r then q n q 1 n 1 +1 C n q n m, whenever n Z + and j Z. Proof. According to the binomial theorem and q i N m, r, we have q n q 1 n q n q m n q n C n q n m C n q n m, whenever n Z + and j Z. Lemma 3.. Let m Z + and r Z + 0 such that r m 1. If q i N m, r then q n q 1 n qi+s n qi 1, n Proof. By expanding s j0 j0 q n q n 1, we obtain where s Z + 0 and n Z +. q n q 1 n qi n qi 1 n + qi+1 n qi n qi+s n qi+s 1 n j0 q n i+s q n i 1, whenever s Z + 0 and n Z +.

4 P. Juntharee and P. Prommi Lemma 3.3. Let m Z + and r Z + 0 such that r m 1. If q i N m, r then 1 +1 C n q n m nm q n C n q n, m j0 whenever s Z + 0 and n Z + with n. Proof. The result is obtained by expanding 1 +1 C n q n. m We find 1 +1 C n q n m j0 j0 j0 j0 C1 n q n 1 m C n q n j0 nm j0 q n 1 + j0 m 1 +1 C n q n m, whenever s Z + 0 and n Z + with n. Theorem 3.4. Let m Z + and r Z + 0 such that r m 1. If q i N m, r then n+1 mn + 1 q n q n+1 i+s qn+1 i C n+1 m j0 j0 n+1 j0 j0 q n+1 whenever s Z + 0 and n Z +. Proof. For s Z + 0, n Z + and q i N m, r, we can apply the result in Lemma 3.3 and obtain n+1 mn + 1 q n C n+1 q n+1 m j C n+1 q n+1 m. 1 If we use Lemma 3.1 the left hand side term of 1 can be simplified as follows: mn + 1 q n + j0 j0 n C n+1 q n+1 m j0 q n+1, qn+1 1.

5 Recursive summation of the nth powers consecutive congruent numbers 3 Again, by applying Lemma 3. to the right hand side term of, we obtain, n+1 mn + 1 q n C n+1 q n+1 m q n+1 i+s qn+1 i 1, so that j0 mn + 1 j0 j0 q n q n+1 i+s n+1 qn+1 i C n+1 m j0 q n+1 Corollary 3.5. Let m Z + and r Z + 0 such that r m 1. If q i N m, r then for each s Z + 0, the following formulas hold: 1 m 3m 3 4m q qi+s qi 1 + m s + 1 and more precisely, j0 q j0 j0 s + 1 [mi + s + r]. q qi+s 3 qi m s + 1 [m6i + 3s + 6r]. q 3 qi+s q 4 i 1+mq 4 i+s q 3 i 1+m 3 3 s+1[mi+s 1+r]. j0 Proof. Proof of 1. Using the results of Theorem 3.4, we deduce, m q qi+s qi 1 + m j0 Then, we can write m 1 qi+s qi 1 + m s + 1. j0 q qi+s qi 1 + m s + 1 j0 q i+s q i 1 q i+s q i 1 + m s + 1 [ms + 1][mi + s 1 + r] + m s + 1 ms + 1[mi + s + r]. Thus, finally q j0 s + 1 [mi + s + r]. 3

6 4 P. Juntharee and P. Prommi Proof of. We use the results of Theorem 3.4 and Equation 3. Using Equation 3, we obtain, 3 1 Cm 3 3m q m 3 1 j0 q 3 j0 j0 s + 1 3m [mi + s + r] m 3 s + 1 m s + 1 [m6i + 3s + 6r]. 4 Finally, using the results of Theorem 3.4 and Equation 4, we deduce 3m j0 q qi+s 3 qi m s + 1 [m6i + 3s + 6r]. 5 Proof of 3. The proof uses the results of Theorem 3.4 and Equations 3 and 5. From the results of 3 and 5, we obtain 4 1 C 4 m j0 q 4 6m q 4m 3 j0 q + m 4 j0 1 j0 m qi+s 3 qi m s + 1 [m6i + 3s + 6r] m 3 s + 1[mi + s + r] + m 4 s + 1 m q 3 i+s q 3 i 1 + m 3 s + 1 [mi + s 1 + r]. 6 Then using Theorem 3.4 and Equation 6, we obtain 4m q 3 qi+s 4 qi mqi+s 3 qi m 3 s + 1[mi + s 1 + r]. j0 4 Conclusions and Applications For given positive integers m and r such that 0 r < m, we define the set N m, r {q Z : q rmod m}. The elements in N m, r can be labelled consecutively as q i im + r, i Z. The important conclusion of this study is

7 Recursive summation of the nth powers consecutive congruent numbers 5 that the summation of the nth powers of consecutive congruent numbers can be carried out using the recursive form mn + 1 j0 q n q n+1 i+s n+1 qn+1 i C n+1 m j0 where s Z + 0, n Z + and C n+1 q n+1 n + 1!!n + 1!. Using this recursive form, we obtained the following explicit summation formulas for the first, second and third powers of consecutive congruent numbers: 1 q j0 3m 3 4m j0 s + 1 [mi + s + r]. q qi+s 3 qi m s + 1 [m6i + 3s + 6r]. q 3 qi+s 4 qi 1 4 +mqi+s 3 qi 1+m 3 3 s+1[mi+s 1+r]. j0 The above results can be used to solve some practical problems. For example, to find the total surface or volume or mass of a set s objects in which all of them are cubes but of different sizes, and the difference in lengths of sides of the consecutive objects is equal to m units. Those objects have been shown in the following figure., In general, the calculation of sums of an nth power of consecutive congruent numbers can be a long and tedious tas. However, the calculation can be carried out easily by using the recursive summation form of the nth powers of consecutive congruent numbers and the summation formulas for the first, second and third powers. We now give two examples.

8 6 P. Juntharee and P. Prommi Example 4.1. Suppose a plastic factory wants to produce a pacage containing 50 different plastic boxes in which all boxes are cubes but of different sizes. When arranged in the pacage, the differences in length of sides of boxes that are adjacent or consecutive will be equal to 10 centimeters and the smallest box will be assumed to have a volume 7 cubic centimeters. The question is what will be the total volume of these 50 plastic boxes. Solution. We can analyze and solve this problem as follows: Obviously, we see that n 3, s 50, m 10, r 3 and i 0, then we obtain q i r, q 4 i+s q 4 i 1 64, 013, 551, 680, m q 3 i+s q 3 i 1, 545, 77, 400, m 3 s , 000 and mi + s 1 + r 496. Therefore, we find 4m q 3 qi+s 4 qi mqi+s 3 qi m 3 s + 1[mi + s 1 + r] j0 Thus, q 3 j0 66, 558, 880, , 663, 97, 014 cubic centimeters and we conclude that the total volume of all plastic boxes is 1, 663, 97, 014 cubic centimeters. Example 4.. Suppose one wants to design a shelf whose shape loos lie a pyramid. It requires some conditions as follows: total surface area of all stages is 45, 75 square centimeters, the area of a stage on the upper level is less than that on the lower level, numbers of levels are 10, each stage on any level is square, side differences of stages that are adjacent or consecutive are 10 centimeters. In addition, the area of the highest stage is 5 square centimeters. The question is whether it is possible to design this shelf in a pyramid form with the above conditions. Solution. We can solve this problem as follows: Clearly, we see that n, s 10, m 10, r 5 and i 0. Then we have q i r, q 3 i+s q 3 i 1 1, 157, 750, m s + 1 Hence, we infer the results, 3m j0 5, 500 and [m6i + 3s + 6r] 310. q qi+s 3 qi m s + 1 [m6i + 3s + 6r], 86, 750.

9 Recursive summation of the nth powers consecutive congruent numbers 7 Therefore, it implies that q 95, 45 square centimeters. Thus, the j0 total area of all stages of the shelf would be 95, 45 square centimeters and not 45, 75 square centimeters as required. Hence the given conditions cannot be satisfied. ACKNOWLEDGEMENTS The authors would lie to than King Mongut s University of Technology North Bango for financial support during preparation of this paper. Also, the authors than Dr. Elvin James Moore, Department of Mathematics, King Mongut s University of Technology North Bango for comments on the writing of this manuscript. References [1] C. V. Eynden, Elementary Number Theory, nd ed, The McGraw-Hill Companies, Inc., New Yor, 001. [] D. M. Burton, Elementary Number Theory, 5 th ed, The McGraw-Hill Companies, Inc., New Yor, 00. [3] I. Niven, H.S. Zucerman and H. L. Montgomery, An Introduction to the Theory of Numbers, 5 th ed, John Wiley & Sons, Inc., Hoboen, [4] Kenneth H. Rosen, Discrete Mathematics and Its Applications, 3 rd ed., The McGraw-Hill, Inc., New Yor, [5] P. D. Schumer, Introduction to Number Theory, PWS Publishing Company, Inc., Boston, [6] P. Juntharee and P. Prommi, Sums of consecutive positive congruent numbers, The Journal of KMUTNB., Vol., no., p , May- August 01. [7] R. C. Shiflett and H. S. Shultz, An odd Sum, Mathematics Teacher Journal, Vol. 95, no.3, p , March 00. Received: September, 01