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### Triangular Numbers

```
Date: 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/
```
Associated Topics:
College Number Theory

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