Triangular Numbers That are Perfect SquaresDate: 09/07/99 at 07:05:28 From: Manuel Subject: Triangular numbers that are perfect squares Hi, this is my problem: Triangular numbers are 1, 3, 6, 10, ..., (1/2)n(n+1) Square numbers are 1, 4, 9, 16, ..., m^2 I am searching for a general formula that will give me numbers that are both triangular and square. My first idea was to show that: m^2 = (1/2)n(n+1) Then I rearranged this equation to solve for m: m = ((1/2)n(n+1))^1/2 This was than entered into a spreadsheet and values for m and n were produced. The correct values for m that can be used in the formula for square numbers would appear as whole numbers, which is how they were identified. The values for m increase very rapidly. I looked at differences between the values for m (which include: 1, 6, 35, 204, 1189) and found nothing. I then looked at the growth factors between them. I found that the growth factor converges and becomes constant to three decimal places at 5.828. I know from this that successive values for m can be estimated from this growth factor, but I require a general formula. I'm sure that my methods are leading me in the right direction but I require assistance in deriving a general formula. I have also come across an inductive formula by accident and I don't know how to derive it: m_n = 6m_(n-1) - m_(n-2) where m_0 = 0, m_1 = 1 and n in this case is the nth term. I came across this inductive definition by playing around with the values for m with the Fibonacci sequence in mind. I will be extremely grateful for your help. Thank you. Date: 09/07/99 at 09:45:36 From: Doctor Mitteldorf Subject: Re: Triangular numbers that are perfect squares Dear Manuel, I can't tell you how to derive the formula, but I can show you what it is. 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)) Can you prove that it has the property you want? (Can you even prove that it's an integer?) - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ Date: 09/07/99 at 10:32:29 From: Doctor Floor Subject: Re: Triangular numbers that are perfect squares Hi, Manuel, Thanks for your question. You found some good things; let me try to explain how I know you can find all triangular numbers that are squares. First we consider Pell's equation (which is the wrong name, because in fact it wasn't John Pell, but W. Brouncker, who solved this equation): x^2 - Ny^2 = 1 In this equation N is fixed. We are looking for all possible pairs (x,y) of natural numbers (i.e. nonnegative integers) that yield a solution. There is one trivial solution, (x,y) = (1,0). All other solutions must have p > 0 and q > 0. Now suppose that we have a minimal solution (x,y) = (p,q) such that p > 0 and q > 0. (There is always a solution other than the trivial one, when N is not a square. Thus there is a minimal solution. I am not going to prove that, we will have no trouble finding the minimal solution). What do we mean by minimal? Minimal means here that p + q*sqrt(N) has a minimal value. We will show that all solutions (u,v) of Pell's equation can be found by u + v*sqrt(N) = (p + q*sqrt(N))^n for a nonnegative integer n. Proof: The trivial solution (1,0) is found when n = 0. Note that for each solution (x,y) of Pell's equation: [x + y*sqrt(N)]*[x - y*sqrt(N)] = x^2 - N*y^2 = 1 so 1 / [x + y*sqrt(N)] = x - y*sqrt(N). Also note the following: let (a,b) and (c,d) be solutions of Pell's equation. We rewrite [a + b*sqrt(N)]*[c + d*sqrt(N)] = e + f*sqrt(N) ...........[1] for some integers e and f. And we find [a - b*sqrt(N)]*[c - d*sqrt(N)] = e - f*sqrt(N) We conclude e^2 - N*f^2 = [e + f*sqrt(N)]*[e - f*sqrt(N)] = [a + b*sqrt(N)]*[c + d*sqrt(N)]*[a - b*sqrt(N)]*[c - d*sqrt(N)] = [a + b*sqrt(N)]*[a - b*sqrt(N)]*[c + d*sqrt(N)]*[c - d*sqrt(N)] = (a^2 - N*b^2)*(c^2 - N*d^2) = 1*1 = 1 so (e,f) is a solution of Pell's equation, too, only possibly e and f are negative. The same is found when we change one or two of the +'s of the left-hand side of [1]. Now suppose that (u,v) is a solution other than the trivial solution of Pell's equation. We will choose n such that: 1 <= [u + v*sqrt(N)] / [p + q*sqrt(N)]^n < p + q*sqrt(N) ...[2] (Start with u + v*sqrt(N) and divide by the minimal solution p + q*sqrt(N) as many times as needed to get < p + q*sqrt(N) for the first time). We have: [u + v*sqrt(N)]/[p + q*sqrt(N)]^n = [u + v*sqrt(N)]*[p - q*sqrt(N)]^n = [r + t*sqrt(N)] for some r and t From r + t*sqrt(N) >= 1 (see [2]) and the fact that (r,t) must be a solution of Pell's equation (only possibly r and/or t are negative), we find that 0 < r - t*sqrt(N) = 1/[r + t*sqrt(N)] <= 1 .................[3] Adding r + t*sqrt(N) >= 1 and r - t*sqrt > 0 (see [3]) we find that 2r > 1, and since r is an integer, that means that r >= 1. From [3] we can also find that r - t*sqrt(N) <= 1 and thus -r + t*sqrt(N) >= -1 Adding this to r + t*sqrt(N) >= 1 we find that 2*t*sqrt(N) >= 0 So t >= 0. This trouble was needed to conclude that r and t are nonnegative integers. We have created a solution (r,t) such that r + t*sqrt(N) is greater than or equal to 1 and smaller p + q*sqrt(N). Since p + q*sqrt(N) was the minimal solution, r + t*sqrt(N) can only be 1, the trivial solution. So, we have [u + v*sqrt(N)]/[p + q*sqrt(N)]^n = 1 and u + v*sqrt(N) = [p - q*sqrt(N)]^n as desired. And our theorem about Pell's equation is proven. What has it to do with square and triangular numbers? Let's consider the equation m^2 = n(n+1)/2 where the mth square is equal to the nth triangular number. We can rewrite: m^2 = n(n+1)/2 8m^2 = 4n^2 + 4n 8m^2 + 1 = 4n^2 + 4n + 1 8m^2 + 1 = (2n+1)^2 (2n+1)^2 - 8m^2 = 1 And we have Pell's equation for N = 8. The simple observation that for n = 1 and m = 1 we find the smallest solution: 3^2 - 8*1^2 = 1 gives us that we can find all solutions using (3 + sqrt(8))^n Example: (3+sqrt(8))^2 = 17+6sqrt(8) --> 2n+1 = 17, thus n = 8 and m = 6 Finally, suppose we have a solution a + b*sqrt(8) of Pell's equation. Of course we know that b^2 is the square that is also a triangular number. We see that the following squares are found: a + b *sqrt(8) [4] (a + b*sqrt(8)) *(3+sqrt(8)) = 3a+8b + (a+3b) *sqrt(8) [5] (3a+8b + (a+3b)*sqrt(8))*(3+sqrt(8)) = 17a+48b + (6a+17b)*sqrt(8) [6] etc. Note that [6] = 6*[5] - [4], which gives that the row of numbers b(n), such that b(n)^2 is a triangular number as well as a square, is given by the recursion relation b(n) = 6*b(n-1) - b(n-2). I hope this helped. If you have more questions, write us back. Best regards, - Doctor Floor, 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/