International Journal of Pure and Applied Mathematical Sciences. ISSN 0972-9828 Volume 9, Number 2 (2016), pp. 123-131 © Research India Publications https://dx.doi.org/10.37622/IJPAMS/9.2.2016.123-131

CONGRUUM PROBLEM

Manju Somanath1 and J. Kannan2

1Assistant Professor, Department of Mathematics, National College, Trichy - 01, India

2Research Scholar, Department of Mathematics, National College, Trichy - 01, India

Abstract The ternary quadratic Diophantine equation 푋2 + 푍2 = 2푌2 is considered and a few interesting properties among the solutions are presented. Keywords – Arithmetic progression, Diophantine equation, Integral solution, Congruum.

NOTATIONS USED: (i) 푃(푛) = 푛(푛 + 1) Pronic number of rank 푛. 2 (ii) 푆푛 = 6푛 − 6푛 + 1 of rank 푛. (iii) 퐍∗ = {0}∪ 퐍 Non negative .

INTRODUCTION Number theory is the branch of Mathematics concerned with studying the properties and relations of integers. In Number Theory, a 푐표푛𝑔푟푢푢푚 (plural congrua) is the difference between successive square numbers in an arithmetic progression of three squares equally spaced apart from each other. Then the space between them, 푍2 − 푌2 = 푌2 − 푋2 is called a congruum.

For instance, the number 96 is a 푐표푛𝑔푟푢푢푚 since it is the difference between each pair of three squares 4, 100 & 196. The first few 푐표푛𝑔푟푢푎 are 24, 96, 120, 216, 240, 124 Manju Somanath and J. Kannan

336, 384, 480,… If 푋, 푌, 푍 represents the sides of the Pythagorean triangle then the congruum is itself four times the area of the Pythagorean triangle. The two right triangles with leg and hypotenuse (7, 13) and (13, 17) have equal third sides of length √120. The square of this side, 120, is a congruum: it is the difference between consecutive values in the Arithmetic progression of squares 72, 132, 172 .

Figure 1: A geometrical example of a Congruum

Many mathematicians of the ancient years Fibonacci, Fermat, Oystein Ore etc [1, 4, 6] studied the 푐표푛𝑔푟푢푢푚 problem. This problem can be formalized as a Diophantine equation 푋2 + 푍2 = 2푌2.

Diophantine problems are important because of their rich variety. In this communication, the ternary quadratic Diophantine equation 푋2 + 푍2 = 2푌2 is analyzed for various patterns of solutions.

METHOD OF ANALYSIS In 푐표푛𝑔푟푢푢푚 problem, we want to find a number ℎ such that 푌2 + ℎ = 푋2 and 푌2 − ℎ = 푍2 which gives 2ℎ = 푋2 − 푍2 = (푋 + 푍)(푋 − 푍). since left hand side is even, 푋 − 푍 must be both odd or both even. Therefore, 푋 − 푍 is even, ℎ ℎ ℎ 푋 − 푍 = 2푘. Hence 푋 + 푍 = . Therefore 푋 = + 푘 and 푍 = − 푘.which can 푘 2푘 2푘 ℎ 2 be written as 푌2 = ( ) + 푘2. 2푘

ℎ i.e.,( 푌, , 푘) forms a Pythagorean triangle with sides 2푘 푡(푚2 + 푛2), 2푚푛푡, 푡(푚2 − 푛2). Congruum Problem 125

In this communication, the ternary quadratic Diophantine equation 푋2 + 푍2 = 2푌2 is analyzed for various patterns of integer solutions.

Figure 2: Pictorial representation of the equation

Considering the 푐표푛𝑔푟푢푢푚 problem as a ternary quadratic Diophantine equation with 3 unknown

푋2 + 푍2 = 2푌2 (1)

We present below different patterns of integral solution of (1): A. Pattern The linear transformation

푋 = 푢 + 푣, 푍 = 푢 − 푣 in (1) leads to the Pythagorean equation

푢2 + 푣2 = 푌2 (2) with solutions

푢 = 푝2 − 푞2 126 Manju Somanath and J. Kannan

푣 = 2푝푞

푌 = 푝2 + 푞2 Thus the solutions of (1) are

푋 = 푝2 + 2푝푞 − 푞2

푌 = 푝2 + 푞2

푍 = 푝2 − 2푝푞 − 푞2 Some interesting properties of the above solutions are

(i) If 푝 = 푞, then 푋 − 푍 is a perfect square.

(ii) 3[푌(1, 푛) − 푋(1, 푛)] + 1 is Star number of rank 푛.

푋(푛,1)+푌(푛,1) (iii) is a Pronic number of rank 푛. 2

(iv) If 푝 = 푞, then 6(푋 − 푍) is a Congruum.

(v) 3(푋 + 푌 + 푍) is a Nasty number, when 푞 = 푝.

B. Pattern (1) can be rewritten as

푋2 − 푌2 = 푌2 − 푍2 (3) 푋 − 푌 푌 + 푍 푝 = = (푠푎푦) 푌 − 푍 푋 + 푌 푞 Expressing this as a system of simultaneous equations:

푞푋 − (푝 + 푞)푌 + 푝푍 = 0 (4)

푝푋 + (푝 − 푞)푌 − 푞푍 = 0 (5) (4) and (5) solving we get

푋 = −푝2 + 푞2 + 2푝푞

푌 = 푝2 + 푞2

푍 = 푝2 − 푞2 + 2푝푞 Congruum Problem 127

Choice 1: (3) can be also written as 푋 − 푌 푌 − 푍 푝 = = (푠푎푦) 푌 + 푍 푋 + 푌 푞 we obtain the solution of (1) as

푋 = 푝2 − 푞2 − 2푝푞

푌 = −푝2 − 푞2

푍 = 푝2 − 푞2 + 2푝푞 Choice 2: (3) can be also written as 푋 + 푌 푌 + 푍 푝 = = (푠푎푦) 푌 − 푍 푋 − 푌 푞 we obtain the solution of (1) as

푋 = 푝2 − 푞2 + 2푝푞

푌 = 푝2 + 푞2

푍 = 푝2 − 푞2 − 2푝푞

C. Pattern The substitution

푋 = 푢 + 푣, 푌 = 푢 − 푣 in (1) gives 푢 = 3푣 ± √(8푣2 + 푍2)

Assume 훼2 = 8푣2 + 푍2 (6) Case (I) Solving (6), we get

푣 = 2푝푞

푍 = 8푝2 − 푞2

훼 = 8푝2 + 푞2 128 Manju Somanath and J. Kannan

For the corresponding two values of 푢, we get the solutions of (1) as

푋 = 8푝푞 + 8푝2 + 푞2

푌 = 4푝푞 + 8푝2 + 푞2

푍 = 8푝2 − 푞2 and

푋 = 8푝푞 − 8푝2 − 푞2

푌 = 4푝푞 − 8푝2 − 푞2

푍 = 8푝2 − 푞2 Case (II) Proceeding as in the above case for the choice 푣 = 2푝푞

푍 = 푝2 − 8푞2

훼 = 푝2 + 8푞2 The corresponding solutions of (1) are

푋 = 8푝푞 + 푝2 + 8푞2

푌 = 4푝푞 + 푝2 + 8푞2

푍 = 푝2 − 8푞2 and

푋 = 8푝푞 − 푝2 − 8푞2

푌 = 4푝푞 − 푝2 − 8푞2

푍 = 푝2 − 8푞2

Case (III)

Letting 푍 = 1, in (6) we get 훼2 = 8푣2 + 1 (7)

Congruum Problem 129

Solving (7) as a Pell’s equation, we get

푢푛 = 3푣푛 ± 훼푛 where

1 푛+1 푛+1 훼 = [(3 + √8) + (3 − √8) ] 푛 2

1 푛+1 푛+1 푣푛 = [(3 + √8) − (3 − √8) ] 2√8 Sub case I For the choice of

3 푛+1 푛+1 1 푛+1 푛+1 푢푛 = [(3 + √8) − (3 − √8) ] + [(3 + √8) + (3 − √8) ] 2√8 2 We have the sequence of solutions of (1) as

1 푛+1 푛+1 푋푛 = [(4 + √8)(3 + √8) − (4 − √8)(3 − √8) ] 2√8

1 푛+1 푛+1 푌푛 = [(2 + √8)(3 + √8) − (2 − √8)(3 − √8) ] 2√8 A few numerical examples are presented in the following table:

푿 풀 풉 = 푿ퟐ − 풀ퟐ

= 풀ퟐ − 풁ퟐ

X0 = 7 Y0 = 5 24

X1 = 41 Y1 = 29 840

X2 = 239 Y2 = 169 28560

X3 = 1393 Y3 = 985 970224

X4 = 8119 Y4 = 5741 32959080

X5 = 47321 Y5 = 33461 1119638520

10 X6 = 275807 Y6 = 195025 3.803475062 × 10 130 Manju Somanath and J. Kannan

From the above solutions we observe some interesting properties:

(i) 푋푛, 푌푛, 푍푛 all are odd . ∗ (ii) (푋푛, 푌푛) = 1, for all 푛 ∈ 푵 . 2 2 (iii) 푋푛푌푛(푋푛 − 푌푛 ) is a area of a Pythagorean triangle. 3 3 (iv) 4(푋푛푌푛 − 푋푛푌푛 ) is a congruum.

(v) (푋푛, 푌푛, 푍푛) is any one of the solution then,

(−푋푛, −푌푛, −푍푛), (푋푛, 푌푛, −푍푛)(푋푛, −푌푛, 푍푛), (−푋푛, 푌푛, 푍푛), (푋푛, −푌푛, −푍푛),

(−푋푛, −푌푛, 푍푛), (−푋푛, 푌푛, −푍푛), is also solution of above equation. ∗ (vi) 푋6(푛+1) − 푋6푛 ≡ 0 (푚표푑 10), for all 푛 ∈ 푵 .

(vii) 푋2(푛+1) − 푋2푛 ≡ 0 (푚표푑 10), for all 푛 such that 푛 푎푛푑 푛 + 1 ≠ 3푚, for any integer 푚. ∗ (viii) 푌3푛 ≡ 0 (푚표푑 5), for all 푛 ∈ 푵 . ∗ (ix) 푌3(푛+1) − 푌3푛 ≡ 0 (푚표푑 10), for all 푛 ∈ 푵

(x) 푌푛+1 − 푌푛 ≡ 0(푚표푑 10), for all 푛 such that 푛 ≡ 1(푚표푑 3).

Sub Case (II)

Substituting these values in

푢푛 = 3푣푛 − 훼푛 Giving the solutions of (1) as,

1 푛+1 푛+1 푋푛 = [(4 − √8)(3 + √8) − (4 + √8)(3 − √8) ] 2√8

1 푛+1 푛+1 푌푛 = [(2 − √8)(3 + √8) − (2 + √8)(3 − √8) ] 2√8

The solutions in the above two sub cases satisfy the following recurrence relations.

(i) 푋푛+2 − 6푋푛+1 + 푋푛 = 0 (ii) 푌푛+2 − 6푌푛+1 + 푌푛 = 0 (iii) 푍푛 = 1.

Congruum Problem 131

OBSERVATION 퐶표푛𝑔푟푢푎 is closely related to congruent numbers which is defined as the area of a Pythagorean triangle with rational sides. Every is a 푐표푛𝑔푟푢푢푚 multiplied by a square of a rational number. The problem of three squares in A.P is also closely related to congruent numbers.

CONCLUSION To Conclude, One may search for other patterns of solutions.

REFERENCES [1] L.E. Dickson, History of Theory of Numbers, Vol.2, Chelsea Publishing Company, New York, 1952. [2] Dr. Manju somanath, J. Kannan and Mr. K. Raja, “Gaussian Pythagorean Triples 푋2 + 푌2 = 푍2 ”, International Journal of Engineering Research and Management, Vol. 03 Issue 04 (April 2016). [3] Carmichael. R. D, Theory of Numbers and Diophantine Analysis, Dover Publication Inc., New York, 1952. [4] Niven. J, Zuckerman and Montgomery, An introduction to the theory of numbers, fifth edition, John We lay, New York 1991. [5] M. A. Gopalan, Manju somanath and N.Vanitha, “On Space Pythagorean equation 푋2 + 푌2 + 푍2 = 푊2”, International Journal of Mathematics, Computer Science and Information Technology Vol. 1, pp129-133, January- June 2007. [6] Oystein Ore, Number Theory and its History, Dover publication, New York. [7] Conway J H and Guy R K, "The book of Numbers", Springer Science and Business Media, 2006.

132 Manju Somanath and J. Kannan