Triangular Numbers That are Perfect Squares

Date: 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/