A Family of Solution and Number of Palindromic

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Keywords: Almost Pythagorean Equation, almost Pythagorean triples, Pythagorean equation, primitive. Pythagorean triples, component, numeric palindromes. 1.
A Family of Solution and Number of Palindromic Solutions to Almost Pythagorean Equation John Rafael M. Antalan* Department of Mathematics and Physics, College of Arts and Sciences, Central Luzon State University Science City of Munoz, Nueva Ecija, Philippines *Corresponding author: [email protected]

Received Month X, XXXX; revised Month X, XXXX; accepted Month X, XXXX

Abstract In this article, we give a particular family of solution to the Diophantine equation π‘₯ 2 + 𝑦 2 = 𝑧 2 + 1 which we call almost Pythagorean equation. We then consider the number of palindromic solutions to the given Diophantine equation. In particular, we show that there are infinitely many triples with one and three numeric palindrome components that solves the Diophantine equation. We also give some interesting open problems and present some preliminary results about one of these open problems.

Keywords: Almost Pythagorean Equation, almost Pythagorean triples, Pythagorean equation, primitive Pythagorean triples, component, numeric palindromes.

For example, given the PPT (3,4,5), it can easily be verified that the triples:

1. Introduction A Pythagorean triple (PT) is an ordered triple (π‘₯, 𝑦, 𝑧) that satisfies the Pythagorean equation: π‘₯ 2 + 𝑦 2 = 𝑧 2 (1).

π‘₯ = 3𝑑 + 1 {𝑦 = 4𝑑 + 3 𝑧 = 5𝑑 + 3

(5)

π‘₯ β€² = 3𝑑 + 2 {𝑦 β€² = 4𝑑 + 1 𝑧 β€² = 5𝑑 + 2

(6)

and If π‘₯, 𝑦 and 𝑧 are pairwise relatively prime then the triple is a primitive Pythagorean triple (PPT). Modifying (1) slightly we have: π‘₯2 + 𝑦2 = 𝑧2 + 1

(2).

Let us agree to call (2) as almost Pythagorean equation. An ordered triple that satisfies (2) was named almost Pythagorean triple (APT) by O. Frink in [1].

solves (2) where 𝑝, 𝑝′ , π‘ž, π‘ž β€², π‘Ÿ and π‘Ÿβ€² are respectively the integers 1, 2, 3, 1, 3 and 2. However, for PPT (π‘Ž, 𝑏, 𝑐) whose components were large, finding integers 𝑝, 𝑝′, π‘ž, π‘ž β€² , π‘Ÿ and π‘Ÿβ€² such that (3) and (4) becomes solution to (2) is not an easy task.

He also showed that if (π‘Ž, 𝑏, 𝑐) is a PPT, then the triple: π‘₯ = π‘Žπ‘‘ + 𝑝 {𝑦 = 𝑏𝑑 + π‘ž 𝑧 = 𝑐𝑑 + π‘Ÿ

(3)

and π‘₯ β€² = π‘Žπ‘‘ + 𝑝′ {𝑦 β€² = 𝑏𝑑 + π‘žβ€² (4) 𝑧′ = 𝑐𝑑 + π‘Ÿβ€²

In this article we show that for the family of PPT in which (3,4,5) belongs, the integers 𝑝, 𝑝′ , π‘ž, π‘ž β€² , π‘Ÿ and π‘Ÿβ€² such that (3) and (4) becomes solution to (2) is uniquely given by 1,2π‘˜ βˆ’ 2,2π‘˜ βˆ’ 1, 2π‘˜ 2 βˆ’ 4π‘˜ + 1,2π‘˜ βˆ’ 1 and 2π‘˜ 2 βˆ’ 4π‘˜ + 2 respectively for an integer π‘˜ β‰₯ 2. Thus, we generate a family of solution to the almost Pythagorean equation.

are APT for all positive integer 𝑑 and for unique integers 𝑝, 𝑝′ , π‘ž, π‘ž β€² , π‘Ÿ and π‘Ÿβ€² that depends on the PPT (π‘Ž, 𝑏, 𝑐) satisfying 𝑝 + 𝑝′ = π‘Ž, π‘ž + π‘ž β€² = 𝑏 and π‘Ÿ + π‘Ÿ β€² = 𝑐. Likewise, all APT’s takes the form (3) and (4).

Also in this paper, we determine the number of palindromic solutions to (2) having a single and triple numeric palindrome components. We also present some preliminary results about the finitude or infinitude of triples that solves (2) containing two numeric palindrome components. This endeavour was highly motivated by the work of Antalan and Tagle in [2].

Thus, given a PPT, finding solutions to (2) highly depends on finding the integers 𝑝, 𝑝′ , π‘ž, π‘ž β€² , π‘Ÿ and π‘Ÿβ€².

We end this article by stating a conjecture and giving out some open problems that one might be interested with.

2. Preliminaries Before presenting the results, we first state a preliminary lemma from [3] and a preliminary notation that will be used in proving our claims.

π‘₯β€² = (2π‘˜ βˆ’ 1)𝑒 + (2π‘˜ βˆ’ 2) { 𝑦′ = (2π‘˜ 2 βˆ’ 2π‘˜)𝑒 + (2π‘˜ 2 βˆ’ 4π‘˜ + 1) 𝑧′ = (2π‘˜ 2 βˆ’ 2π‘˜ + 1)𝑒 + (2π‘˜ 2 βˆ’ 4π‘˜ + 2)

(9)

where π‘˜, 𝑒 ∈ β„€+ satisfying π‘˜ β‰₯ 2 and 𝑒 > 0. Lemma 2.1. All of the solutions of the Pythagorean equation π‘Ž2 + 𝑏2 = 𝑐 2 satisfying the condition: i. ii. iii.

gcd(π‘Ž, 𝑏, 𝑐 ) = 1 2|𝑏 π‘Ž, 𝑏, 𝑐 > 0

π‘₯ 2 + 𝑦 2 = 2 βˆ’ 4π‘˜ + 4π‘˜ 2 βˆ’ 2𝑑 + 8π‘˜π‘‘ βˆ’ 12π‘˜ 2 𝑑 + 8π‘˜ 3 𝑑 + 𝑑 2 βˆ’ 4π‘˜π‘‘ 2 + 8π‘˜ 2 𝑑 2 βˆ’ 8π‘˜ 3 𝑑 2 + 4π‘˜ 4 𝑑 2

is given by:

while working on the right side and simplifying we have: π‘Ž = 𝑠2 βˆ’ 𝑑 2 { 𝑏 = 2𝑠𝑑 𝑐 = 𝑠2 + 𝑑 2

𝑧 2 + 1 = 2 βˆ’ 4π‘˜ + 4π‘˜ 2 βˆ’ 2𝑑 + 8π‘˜π‘‘ βˆ’ 12π‘˜ 2 𝑑 + 8π‘˜ 3 𝑑 + 𝑑 2 βˆ’ 4π‘˜π‘‘ 2 + 8π‘˜ 2 𝑑 2 βˆ’ 8π‘˜ 3 𝑑 2 + 4π‘˜ 4 𝑑 2 .

(7)

for relatively prime integers 𝑠 > 𝑑 > 0 and 𝑠 β‰’ 𝑑(π‘šπ‘œπ‘‘ 2). If 𝑑 ∈ β„€10 and 𝑛 is a positive integer, the expression 𝑑𝑛 means 𝑛 copies of 𝑑. For instance, if we have the expression 22 this refers to the integer 22. If we have 103 1 we are referring to the integer 10001. We are now ready to show and prove our results.

3. Results 3.1. A Family of Solution Pythagorean Equation

to

Almost

Definition 3.1.1. Two primitive Pythagorean triples (π‘Ž, 𝑏, 𝑐) and (π‘Žβ€² , 𝑏′ , 𝑐′) are said to be in the same family if they were generated by the same arithmetic sequence π‘ π‘˜ and π‘‘π‘˜ in Lemma 2.1. Definition 3.1.2. Two almost Pythagorean triples (π‘₯, 𝑦, 𝑧) and (π‘₯ β€² , 𝑦 β€² , 𝑧′) is in the same family whenever they were generated by primitive Pythagorean triples that belong to the same family. Lemma 3.1.1. The primitive Pythagorean triple (3,4,5) belongs to the family of primitive Pythagorean triple (2π‘˜ βˆ’ 1,2π‘˜ 2 βˆ’ 2π‘˜, 2π‘˜ 2 βˆ’ 2π‘˜ + 1) for integers π‘˜ β‰₯ 2. Proof. Let π‘ π‘˜ = π‘˜ and π‘‘π‘˜ = π‘˜ βˆ’ 1 where π‘˜ = 2,3,4, … . Note that for any π‘˜, gcd(π‘ π‘˜ , π‘‘π‘˜ ) = 1 and π‘ π‘˜ β‰’ π‘‘π‘˜ (π‘šπ‘œπ‘‘ 2). By using Lemma 2.1 it follows that we generated the family of primitive Pythagorean triple of the form (2π‘˜ βˆ’ 1,2π‘˜ 2 βˆ’ 2π‘˜, 2π‘˜ 2 βˆ’ 2π‘˜ + 1). If π‘˜ = 2, we then have the PPT (3,4,5). This completes the proof of the Lemma.∎ Theorem 3.1.1. For the family of primitive Pythagorean triple (2π‘˜ βˆ’ 1,2π‘˜ 2 βˆ’ 2π‘˜, 2π‘˜ 2 βˆ’ 2π‘˜ + 1), π‘˜ β‰₯ 2, the corresponding family of solution to the almost Pythagorean equation (2) are given by: π‘₯ = (2π‘˜ βˆ’ 1)𝑒 + 1 { 𝑦 = (2π‘˜ 2 βˆ’ 2π‘˜)𝑒 + (2π‘˜ βˆ’ 1) 𝑧 = (2π‘˜ 2 βˆ’ 2π‘˜ + 1)𝑒 + (2π‘˜ βˆ’ 1) and

Proof. Substituting (8) in (2) and working on the left side we have:

(8)

Since the right hand side and the left hand side are equal, we conclude that (8) is a solution to (2). Now, since (8) is a solution to (2), it follows from (4) that (9) is also a solution to (2), and we are done.∎ Illustration: The PPT (7,24,25) were generated with 𝑠 = π‘˜ = 4 and 𝑑 = 3 in Lemma 2.1. Using Theorem 3.1.1 with 𝑑 = 1, it follows that the ordered triple (8,31,32) and (13,41,43) are solutions to the almost Pythagorean equation and thus they are almost Pythagorean triples which can easily be verified.

3.2. Number of Palindromic Solutions to Almost Pythagorean Equation. Theorem 3.2.1. There are infinitely many almost Pythagorean triples with three numeric palindrome components. Proof. Note that for the almost Pythagorean equation (2), the two trivial solution is given by the triples (1, 𝑦, 𝑦) and (π‘₯, 1, π‘₯). By choosing 𝑦 to be an arbitrary palindrome in the first triple, while π‘₯ to be an arbitrary palindrome in the second triple we always generate a solution of the form (1, 𝑝, 𝑝) and (𝑝, 1, 𝑝)… triples with three numeric palindrome component. Since the set of palindromic numbers is infinite there are also infinite number of such triples.∎ Theorem 3.2.2. There are infinitely many almost Pythagorean triples with a single numeric palindrome component. Proof. In (6) let 𝑑 = {4𝑛 |𝑛 ∈ 𝑍 + }. As 𝑛 runs over the set of positive integers, we generate the triple (13π‘›βˆ’1 4,17𝑛 , 2𝑛+1 ) which is a solution to (2) with a single numeric palindrome component. Since the set of positive integers is infinite, we conclude that there are infinitely many such triples.∎ For the case of triples with two numeric palindrome components that solves the almost Pythagorean equation, it is currently not known whether there are infinitely or finitely many of them. We state it in the next section as one of our open problem.

4. Open Problems

Generate other family of PPT that yields a family of APT. Prove Theorem 3.2.1 by using non – trivial solutions. Prove or disprove that there are infinitely many almost Pythagorean triples with two numeric palindrome component.

There are two possible ways to start from here to prove our conjecture. First, fix π‘˜ and show that corresponding to π‘˜ there are infinitely many pairs of the form (π‘˜, 𝑒) that gives almost Pythagorean triples with two numeric palindrome components. Second is that, show that for almost all (infinitely many) π‘˜, there corresponds an integer 𝑒 such that (π‘˜, 𝑒) gives almost Pythagorean triples with two numeric palindrome components. However, this is a laborious task to do that requires a good computing software or program to arrive at a proof that is similar to the proof of Theorem 3.2.2.

5. Concluding Remarks and Some Preliminary Results on an Open Problem

In fact if we fix π‘˜, Mathematics Stack Exchange User mjqxxxx estimates a very rough probability that β€˜it is likely that some value of 𝑒 will produced a double palindrome just by chance”. However for a larger values of π‘˜, one β€œneed to look at greater and greater values for 𝑒” [4].

The results and observations in this article suggest the following extensions and future works: ο‚· ο‚· ο‚·

In this article, we have successfully given a family of solution to the almost Pythagorean equation. Also we established the infinitude of triples that contain a single and triple numeric palindrome components that solves the almost Pythagorean equation. Lastly, we gave some open problems related to the topic. For the open problem involving the infinitude or finitude of triples that contain a double numeric palindrome components that solves the almost Pythagorean equation, we conjecture that the case is that there are infinitely many of them. In proving our conjecture, the original plan was to use the family of solutions generated from Theorem 3.1.1 particularly equation (8) to yield an infinitely many almost Pythagorean triples with two numeric palindrome components. With the help of Mathematics Stack Exchange User 2012ssohn we have for now the combinations for (π‘˜, 𝑒) with π‘˜ given and a minimum output 𝑒. [4] Table 5.1 Pair (π’Œ, 𝒖) given π’Œ with a minimum output 𝒖 that solves the almost Pythagorean equation that leads to almost Pythagorean triple with double numeric palindrome component. k u 2

1

3

13

4

14

7

17

8

118

14

31

15

2197715

20

1312826

21

12924364

26

117094

27

683854

To end this article we present another attack on proving our conjecture which emphasizes the importance on finding a particular or a family of solutions to (2). Show that all (or infinitely many) PPT generates APT with two numeric palindrome using the result of mjqxxxx.

Acknowledgement The author highly appreciates the help of the head of their department Dr. Amelia Z. Daquioag in the creation of this short article. Also the author is highly indebted to the MSE users 2012ssohn and mjqxxxx for giving him insights about the third open problem. Lastly he wants to say many thanks to his love ones for the inspiration and encouragement.

References [1] Frink, O., Almost Pythagorean Triples, Mathematics Magazine, 60 (1987), No. 4, pp.234-236. [2] Burton, D., Elemenrary Number Theory Revised printing, Allyn and Bacon Inc., (1980), pp. 243-249. [3] Antalan, J. and Tagle, R., JP Journal of Algebra, Number Theory and Applications. (to appear) [4]https://math.stackexchange.com/questions/1298586/expressionsevaluating-to-palindromes.