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Arithmetic Progressionsin the Polygonal Numbers

Scott M. Dunn

(Joint work with Kenny Brown and Josh Harrington)

Department of MathematicsUniversity of South CarolinaColumbia, SC 29208, USA

June 22, 2012Canadian Number Theory Association XII Meeting

Motivating Question

Definition: Triangular Number

The n-th Triangular Number T (n) is the number of points that areneeded to fill an equilateral triangle with sides of length n − 1.This number is given by

T (n) =n(n + 1)

Consider the sequence {T (n)}∞n=1.

Do infinitely-long arithmetic progressions exist in thetriangular numbers? No.

Do arbitrarily-long arithmetic progressions exist in thetriangular numbers? ???

What can we say about arithmetic progressions in thetriangular numbers? ???

Scott M. Dunn (University of South Carolina) Arithmetic Progressions in Polygonal Numbers 2/14

Motivating Question

T (n) =n(n + 1)

Motivating Question

T (n) =n(n + 1)

Do infinitely-long arithmetic progressions exist in thetriangular numbers?

Motivating Question

T (n) =n(n + 1)

Motivating Question

T (n) =n(n + 1)

Do arbitrarily-long arithmetic progressions exist in thetriangular numbers?

Motivating Question

T (n) =n(n + 1)

Motivating Question

T (n) =n(n + 1)

What can we say about arithmetic progressions in thetriangular numbers?

Motivating Question

T (n) =n(n + 1)

A Quick Aside

Definition: Arithmetic Progression

An arithmetic progression (AP) with a common difference d is asequence of numbers, finite or infinite, such that the difference ofany two consecutive terms is a constant d .

For the purposes of this talk, we will assume that d is a positiveinteger.

We will also assume that our sequence has at least three terms.

A Quick Aside

Some Examples

Question: Do arithmetic progressions even exist in the triangularnumbers?

T (1) = 1,T (7) = 28, and T (10) = 55 form an arithmeticprogression with common difference d = 27

T (2) = 3,T (72) = 2628, and T (102) = 153 form anarithmetic progression with common difference d = 2625

Some Examples

Question: Do arithmetic progressions even exist in the triangularnumbers? Yes.

Some Examples

Four-Term AP’s in the Triangular Numbers

Theorem (Mordell 1969; Sierpinski 1964)

There cannot be four squares in arithmetic progression withcommon difference d 6= 0.

Theorem (Brown, D., Harrington)

There cannot be four triangular numbers in arithmetic progressionwith common difference d 6= 0.

Three-Term AP’s in the Triangular Numbers

Let n be an arbitrary positive integer. Then there exist infinitelymany integers d > 0 such that there is a three-term arithmeticprogression with common difference d in the triangular numbersbeginning with T (n).

Proof Idea:

Suppose that we do have a three-term arithmetic progression inthe triangular numbers. T (n)− T (a) = T (b)− T (a)

Then (2b + 1)2 − 2(2a + 1)2 = −(2n + 1)2.

Letting B = 2b + 1, A = 2a + 1, and N = 2n + 1, we have

B2 − 2A2 = −N2.

Supposing that B = NX and A = NY , we can reduce this to

X 2 − 2Y 2 = −1.

This Pell equation has infinitely many solutions (X ,Y )(and it is easy to show that both X and Y are odd).

Then b = (2n+1)X−12 and a = (2n+1)Y−1

2 for any positive integer n.

Proof Idea:

Then (2b + 1)2 − 2(2a + 1)2 = −(2n + 1)2.

B2 − 2A2 = −N2.

X 2 − 2Y 2 = −1.

Then b = (2n+1)X−12 and a = (2n+1)Y−1

Proof Idea:

Then (2b + 1)2 − 2(2a + 1)2 = −(2n + 1)2.

B2 − 2A2 = −N2.

X 2 − 2Y 2 = −1.

Then b = (2n+1)X−12 and a = (2n+1)Y−1

Proof Idea:

Then (2b + 1)2 − 2(2a + 1)2 = −(2n + 1)2.

B2 − 2A2 = −N2.

X 2 − 2Y 2 = −1.

Then b = (2n+1)X−12 and a = (2n+1)Y−1

Proof Idea:

Then (2b + 1)2 − 2(2a + 1)2 = −(2n + 1)2.

B2 − 2A2 = −N2.

X 2 − 2Y 2 = −1.

This Pell equation has infinitely many solutions (X ,Y )

(and it is easy to show that both X and Y are odd).

Then b = (2n+1)X−12 and a = (2n+1)Y−1

Proof Idea:

Then (2b + 1)2 − 2(2a + 1)2 = −(2n + 1)2.

B2 − 2A2 = −N2.

X 2 − 2Y 2 = −1.

Then b = (2n+1)X−12 and a = (2n+1)Y−1

Proof Idea:

Then (2b + 1)2 − 2(2a + 1)2 = −(2n + 1)2.

B2 − 2A2 = −N2.

X 2 − 2Y 2 = −1.

Then b = (2n+1)X−12 and a = (2n+1)Y−1

Some Notes on Pell Equations:

We reduced our problem to the equation B2 − 2A2 = −N2.

For any divisor qi of N, we can let B = qiXi , A = qiYi ,Qi = N/qi , and consider

X 2i − 2Y 2

i = −Q2i .

If this equation has a relatively prime solution, we get infinitelymany solutions.

This allows us to find all three-term arithmetic progressionsbeginning with T (n).

X 2i − 2Y 2

i = −Q2i .

X 2i − 2Y 2

i = −Q2i .

X 2i − 2Y 2

i = −Q2i .

X 2i − 2Y 2

i = −Q2i .

Polygonal Numbers

Definition: Polygonal Number

Let s be a fixed integer with s ≥ 3. For a natural number n, then-th Polygonal Number Ps(n) is the number of points that areneeded to create a regular s-gon with each side being of lengthn − 1. This number is given by

Ps(n) =( s

2− 1)

n2 −( s

2− 2)

Examples: P3(4) = 10, P4(4) = 16, P5(4) = 22

Examples of Polygonal Numbers for s = 3, 4, 5 and n = 1, 2, 3, 4.

Polygonal Numbers

Ps(n) =( s

2− 1)

n2 −( s

2− 2)

Examples: P3(4) = 10, P4(4) = 16, P5(4) = 22

Polygonal Numbers

Ps(n) =( s

2− 1)

n2 −( s

2− 2)

Examples: P3(4) = 10, P4(4) = 16, P5(4) = 22

Polygonal Numbers

Ps(n) =( s

2− 1)

n2 −( s

2− 2)

Examples: P3(4) = 10, P4(4) = 16, P5(4) = 22

Arithmetic Progressions in the Polygonal Numbers

Let s be a fixed integer with s ≥ 3. Then there cannot be fours-gonal numbers in arithmetic progression with common differenced 6= 0.

Let s be a fixed integer with s ≥ 3. Let n be an arbitrary positiveinteger. Then there exist infinitely many integers d > 0 such thatthere is a three-term arithmetic progression with a commondifference d in the s-gonal numbers beginning with Ps(n).

Some Remarks about AP’s in the Polygonal Numbers

The proof of the last theorem is actually very explicit. It providesan algorithm for finding all three-term arithmetic progressions inthe polygonal numbers.

Not all solutions to the associated Pell equation generatearithmetic progressions.

For example, three-term arithmetic progression starting with P5(n).We need a solutions to (6b − 1)2 − 2(6a− 1)2 = −(6n − 1)2.With B = 6b − 1, A = 6a− 1, and N = 6n − 1, we haveB2 − 2A2 = −N2.Supposing B = NX and A = NY , we have X 2 − 2Y 2 = −1.(X = 7 and Y = 5 is a solution.)