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

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