Math 341: Number Theory Section 2.1: Polygonal Numbers

Mckenzie West

Last Updated: February 21, 2021

We acknowledge that UW-Eau Claire occupies the sacred and ancestral lands of Indigenous Peoples. We honor the land of the Ojibwe and Dakota Nations. 2

Last Time. – The Binomial Theorem – Pascal’s Triangle – Properties of Binomial Coeﬃcients

Today. – Triangular Numbers – Pentagonal Numbers 3

Deﬁnition. Recall the deﬁnition of the nth triangular number,

n(n + 1) T = 1 + 2 + 3 + ··· + n = . n 2

Similarly, we call m ∈ N triangular if there is some n for which m = Tn.

Exercise 1. n(n + 1) n + 1 Verify that = . 2 2 4

Exercise 2. What are the ﬁrst 10 triangular numbers?

n 1 2 3 4 5 6 7 8 9 10

Tn 111 111 111 111 111 111 111 111 111 111 5

Exercise 3. Go to https://oeis.org/A000217 to see a longer list of triangular numbers. (a) Can you ﬁnd two triangular numbers whose sum and diﬀerence are also triangular?

(b) How about three consecutive triangular numbers whose product is a perfect square?

Exercise 4.

Show that for all n ≥ 1, 8Tn + 1 is a perfect square. 7

Exercise 5. Follow the steps below to show that if 8m + 1 = n2 for some nonzero n ∈ Z, then m is triangular. (Do write your work on your own paper, this will be hard to write entirely out on the slide.) (a) Explain why n must be odd.

n−1 (b) Let k = 2 . What is Tk ?

2 (c) Verify that 8Tk + 1 = n .

(d) Conclude that m = Tk (why), so m is triangular. 8

Exercise 6. Let n be a triangular number. Show that 9n + 1 is also triangular. k(k+1) (a) Let n = Tk = 2 . j(j+1) (b) Find the integer j ∈ N such that 2 = 9Tk + 1.

(c) Conclude that 9n + 1 is triangular.(WHY??) Example. 9 Here’s how we would write the formal proof for the fact that if n is triangular then so is 9n + 1.

Proof. Let n ∈ N be a triangular number. We want to show 9n + 1 is also triangular. Since n is triangular, there is a k ∈ N such that k + 1 k(k + 1) n = T = = . k 2 2

Via algebraic manipulation, we see

k(k + 1) 9k2 + 9k + 2 (3k + 1)(3k + 2) 9n + 1 = 9 + 1 = = . 2 2 2

Let j = 3k + 1 ∈ N. Substituting this value, (3k + 1)(3k + 2) j(j + 1) 9n + 1 = = = T . 2 2 j

Therefore, we conclude that 9n + 1 = Tj is also triangular. 10

Exercise 7. Show that the diﬀerence between the squares of two consecutive triangular numbers is always a cube. 11

Exercise 8. Prove that for all n, 1 1 1 1 1 + + + + ··· + < 2. 1 3 6 10 tn

2 1 1 (Hint: Observe that n(n+1) = 2( n − n+1 ).) (a) Use the hint to ﬁnd an algebraic formula for

1 1 1 1 1 + + + + ··· + . 1 3 6 10 tn

(b) Verify that the formula you found will be less than 2 no matter what the value of n ∈ N. Deﬁnition. 12 Pentagonal numbers are integers that represent the number of dots that can be arranged evenly in a pentagon:

In general, p1 = 1 and pn = pn−1 + (3n − 2) for n ≥ 2.

Exercise 9. What are the ﬁrst 10 pentagonal numbers?

n 1 2 3 4 5 6 7 8 9 10 111 111 111 111 111 pn 1 5 12 22 35 111 111 111 111 111 13

Exercise 10.

Use induction, and the inductive deﬁnitions: p1 = 1 and pn = pn−1 + (3n − 2), to show that n(3n − 1) p = , n ≥ 1. n 2 14

Exercise 11. For n ≥ 2, verify 2 pn = tn−1 + n . 15

Exercise 12.

Show that for all n ≥ 2, 2tn−1 + tn is a pentagonal number. 16

Deﬁnition.

The generating function for a sequence a1, a2, a3,... is the formal power series

∞ 2 3 X n a1x + a2x + a3x + ··· = anx . n=1

Note. Generating functions don’t have to start at n = 1, but this one does. We call this a “formal power series” because we have no intention of plugging in a value for x. Instead, we use x as a place marker. 17

Exercise 13.

Write out the ﬁrst few terms of the generating function for Tn, the triangular numbers, and for pn, the pentagonal numbers. ∞ X n (a) Tnx = n=1

∞ X n (b) pnx = n=1 18

Recall. 1 2 3 In Calculus II, you learned 1−x = 1 + x + x + x + ··· , for all x ∈ R with |x| < 1.

Exercise 14.

I claim that the generating function for Tn satisﬁes

∞ X x T xn = . n (1 − x)3 n=1

Verify that the coeﬃcients of x1, x2, x3, x4 do match. Here’s the right-hand side, re-written using the Calc II identity. x = x(1+x+x2+x3+··· )(1+x+x2+x3+··· )(1+x+x2+x3+··· ) (1 − x)3