intsu011h1s

Intsu011h1s - Introductory Number Theory Homework#1 Some solutions and grading notes 1 Textbook#1.1(p 11 One can make a list of triangular numbers

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Introductory Number Theory. Homework #1 Some solutions and grading notes. 1. Textbook, #1.1 (p. 11) One can make a list of triangular numbers and squares, and see what numbers appear in both, but that might take a lot of time and actually tell us little. The question is really: For what natural numbers n,m do we have n ( n + 1) 2 = m 2 . (1) Another way of writing this is as n ( n + 1) = 2 m 2 . The factor of 2 suggests dividing the quest into two cases, n odd, n even. Since the textbook left it as a somewhat open ended question, I’ll just give you my solution and leave it up to you, and your discretion, to verify that I covered all possibilities, and that all works. I am not assuming that any of you got as complete a solution ; though it will be nice if some of you did. It took me a while to figure all this out; things like these require work. They don’t drop like manna from heaven. Case n odd. In my solution (your’s could be simpler!) I generate inductively two sequences of numbers, { ‘ ν } and { r ν } as follows. ‘ 1 = 1. Assuming ‘ ν defined for some ν ≥ 1 I define r ν = ‘ ν + p 2 ‘ 2 ν- 1 , and then ‘ ν +1 = r ν + p 2 r 2 ν + 1 . It is not at all obvious at first sight that all these numbers are integers, but an inductive proof shows they are. All solutions of the equation (1) for n odd are given by n = 2 ‘ 2 ν- 1 ,ν = 1 , 2 , 3 ,... ; an infinity of solutions. The corresponding m is of course found by taking the square root of n ( n +1) / 2. Here are the few first ones: • with ν = 1, have ‘ 1 = 1, so n = 1. We get the triangular number 1 , which is also a square number. • I’ll omit the subscript from ‘,r . With ‘ = 1 we get r = 1 + √ 2 · 1 2- 1 = 2; thus the next ‘ is ‘ = 2 + √ 2 · 2 2 + 1 = 5, thus n = 2 · 5 2- 1 = 49, we get the triangular/square number 49 · 50 2 = 1225 = 35 2 . • The next value of r is r = 5 + √ 49 = 12; then ‘ = 12 + √ 289 = 29; this gives n = 1681, and we get the triangular/square number 1681 · 1682 2 = 1 , 413 , 721 = 1189 2 . • The next r is given by r = 29+ √ 1681 = 70, then ‘ = 70+ √ 9801 = 169. This gives a value of n = 57 , 121 and the triangular/square number 57121 · 57121 2 = 1 , 631 , 432 , 881 = 40 , 391 2 . And so it goes. There is an infinity of triangular/square numbers with odd n . The case n even. Here we simply need to swap the ‘ ’s and r ’s from the odd case. With the same definition, all the triangular/square numbers with n even are obtained by taking n = 2 r 2 ν , ν = 1 , 2 ,... . Here are the first few ones, recalling that the first few values of r were 2 , 12 , 70: n = 2 · 2 2 = 8 gives the triang/sq. number 8 · 9 2 = 36 = 6 2 , n = 2 · 12 2 = 288 gives the triang/sq. number 288 · 289 2 = 41 , 616 = 204 2 , n = 2 · 70 2 = 9800 gives the triang/sq. number 9800 · 9801 2 = 48 , 024 , 900 = 6930 2 ....
View Full Document

This note was uploaded on 07/13/2011 for the course MAS 3202 taught by Professor Staff during the Summer '11 term at FAU.

Page1 / 7

Intsu011h1s - Introductory Number Theory Homework#1 Some solutions and grading notes 1 Textbook#1.1(p 11 One can make a list of triangular numbers

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online