Triangular NumbersDate: 04/03/97 at 20:47:42 From: Barry M. Baker Subject: Infinite Square Triangles How do I prove that there is an infinite number of triangular numbers that are equal to a square number? I.e., 1+2+3+4+5+6+7+8 = 36 and 6*6 = 36. Such numbers are 1, 36, 1225, 41616, 1413721, 48024900... Date: 04/04/97 at 10:14:23 From: Doctor Mitteldorf Subject: Re: Infinite Square Triangles Dear Barry, Funny you should ask - about 3 years ago, I had the same query, worked on it a bit by asking my computer to search for square triangles, then posed the question to my friend Uri Wilensky. Uri is a mathematician in the Tufts U Dept of Education. Here's what Uri wrote back to me: -------------------------------------------------------------- I got distracted from my work and started computing square triangles. There are infinitely many. Here is the formula: a(n) = sqrt of the nth square triangle then a(n) = ((3 + sqrt (8))^(n+1) - (3 - sqrt (8))^(n+1)) / 2 * sqrt (8) Try it. It works! ------------------------------------------------------ Flabbergasted, I wrote back and asked Uri how he came up with it? Here's what he said: ----------------------------------------------------- Here's a quickie insight: The numbers a(i) I computed are likely to be recursively representable. Since the numbers themselves are second order, they are likely to form a second order recurrence relation. Generating 4 numbers is sufficient to see that they satisfy a(n) = 6a(n-1) - a(n-2). Solving this is like solving a linear differential equation. Consider the homogeneous part and note that linear combinations preserve solutions. ------------------------------------------------------ The next step in this saga for me was to try this technique myself. I used the same kind of reasoning to search for a formula that would generate the nth Fibonacci number. You might have fun doing the same! -Doctor Mitteldorf, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 02/05/2000 at 17:30:39 From: Gary Sivek Subject: Triangular numbers I was looking through your number theory archive and I saw one question asking of a proof that there are infinitely many triangular numbers that are also perfect squares. The answer given contained a formula, but no proof. I believe I have a proof and a simpler method to obtain these numbers. Let n and m denote integer values. We are solving for: (n^2 + n)/2 = m^2 n^2 + n = 2m^2 4n^2 + 4n + 1 = 8m^2 + 1 (2n+1)^2 = 8m^2+1 Let N = 2n + 1. Then: N^2 - 8m^2 = 1 a Pell equation with trivial solution (3,1), so there are infinitely many solutions. All we are concerned about are the ones where N is odd, because otherwise n is not an integer. The first several solutions of (n, m) are (1, 1), (8, 6), (49, 25), (288, 204), and (1681, 1189). Finally, this brings me to my question. I know how to solve for solutions to Pell equations recursively, but how do you derive explicit formulae for solutions to Pell equations? Thanks, Gary Sivek Date: 02/07/2000 at 01:08:43 From: Doctor Schwa Subject: Re: Triangular numbers Nicely done! The recursive formulae you mention are linear recurrences, so they can be solved explicitly using the same method as in the Fibonacci sequence: that is, suppose that the nth term is given by r^n, plug into the recursive formula to solve for r (there will be more than one solution), and then the general solution is a linear combination of the various r^n that you get. To find the particular solution, you find the coefficients of the linear combination using the first few terms (one term for each coefficient you need to find). If that brief summary doesn't get you a solution, please write back so I can fill in some more details. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/