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Triangular Triples: Means that Are Not So Average

Date: 04/28/2012 at 17:01:15
From: Reema
Subject: Averages of Triangular Numbers

How many triangular numbers can be written as the average of two other
triangular numbers?

I have found a few:

    190 is the average of   55 and 325
    465 is the average of  300 and 630
   1275 is the average of  780 and 1770
   1891 is the average of 1081 and 2701
   4851 is the average of 3486 and 6216

But I do not see any pattern in them to make a generalization.

So, are there a certain number? or are there infinitely many? If the
latter, what is the generating equation?



Date: 04/29/2012 at 06:05:01
From: Doctor Jacques
Subject: Re: Averages of Triangular Numbers

Hi Reema,

This is a very interesting question. As you will see, there are infinitely
many solutions, and there is a formula that produces all the solutions.
The algebra is a bit complicated, but it only involves simple concepts.

We want to find integer solutions of

   k(k + 1) + m(m + 1) = 2n(n + 1)

This is a second degree equation. In such problems, it is always a good
idea to try to simplify the equation by completing the squares. To do
that, we multiply everything by 4 and do some manipulations.

             4k^2 + 4k + 4m^2 + 4m = 2(4n^2 + 4n)
   (2k + 1)^2 - 1 + (2m + 1)^2 - 1 = 2((2n + 1)^2 - 1)
           (2k + 1)^2 + (2m + 1)^2 = 2(2n + 1)^2

Make this change of variables:

   u = 2k + 1
   v = 2m + 1
   w = 2n + 1                                                    [1]

With these substitutions, the result of completing the square becomes:

   u^2 + v^2 = 2w^2                                              [2]

We must find solutions of this equation where u, v, and w are odd
integers.

It may help to look at rational solutions. If we divide [2] by w^2, and
let x = u/w, y = v/w, the equation becomes:

   x^2 + y^2 = 2                                                 [3]

This is the equation of a circle of radius sqrt(2), and we must find
rational solutions of that equation, such that the numerator and
denominator are both odd.

The good news is that there are obvious solutions, like ((+/-)1, (+/-)1).
We will use the solution x = y = -1. This means that the point 
P = (-1, -1) is on the circle. We will draw a straight line through P; 
that line will intersect the circle in another point Q.

The general equation of a straight line through P is:

   (y + 1) = t(x + 1)                                            [4]

Here, t is the slope of the line. If we extract y from [4] and substitute
into [3], we get:

   (t^2 + 1)x^2 + (2t^2 - 2t))*x + (t^2 - 2t - 1) = 0            [5]

The sum of the roots of that equation in x is -(2t^2 - 2t)/(t^2 + 1). We
already know one root, namely x = -1 (you can substitute that value into
the equation to check). This means that the other root (the x-coordinate
of the point Q) is:

   x = -(2t^2 - 2t)/(t^2 + 1) + 1
     = (-2t^2 + 2t + 1)(t^2 + 1)                                 [6]

We can compute y from [4]:

   y = (t^2 + 2t - 1)(t^2 + 1)                                   [7]

You can check that x and y are solutions of [3].

If t is rational, x and y will be rational, too. Conversely, if Q is a
point on the circle with rational coordinates, the slope of the line PQ
will be rational, and we will find Q using [6] and [7], except for the
point (-1, 1) where t would be infinite.

To summarize, to find a solution of [3], we pick a rational value for t,
and compute x and y from [6] and [7].

For example, with t = 2/3, we get x = 17/13 and y = 7/13, corresponding to
the solution of [3]:

   (17/13)^2 + (7/13)^2 = 2

To get a solution of [2], we multiply by 13^2:

   17^2 + 7^2 = 2*13^2

That is, 

   u = 17      v = 7      w = 13
   
Recovering our original variables from the substitution [1], we find 

   k = 8       m = 3      n = 6
   
This gives the triangular numbers 36, 6, and 21.

There could be a small issue here. We must have u, v, and w odd. However,
as the whole calculation is based on the fractions u/w and v/w, we may
assume that these fractions are in lowest terms; i.e., that 
gcd(u, v, w) = 1. It can be shown that, in such a case, the three integers
must be odd. (You can try to prove this by looking at [2] as a congruence 
modulo 4.)

On the other hand, there is no requirement that the fractions u/w and v/w
be in lowest terms, as long as the terms are odd. This means that we may
multiply all three numbers u, v, and w by any odd factor, and this will
give us another solution.

For example, multiplying our solution (17, 7, 13) by 3, we get

   u = 51      v = 21      w = 39
   k = 25      m = 10      n = 19

This gives the three triangular numbers 325, 55, and 190, which you 
discovered and listed first among your results.

It may happen that, for some values of t, the values of x and/or y are
negative. However, as the equation of the circle only involves x^2 and
y^2, you may ignore the signs: if the point (x, y) is on the circle, the
points (-x, y), (x, -y) and (-x, -y) are also on the circle; they
correspond to other values of t.

There are many more interesting aspects in this problem. Here are a
couple of ideas you may want to investigate, if you find this interesting.

The formulas [6] and [7] use a rational parameter t. By writing t = p/q,
where p and q are integers, and simplifying the formulas, you can get
formulas for u, v, and w that use only the two integer parameters p and q.

As I said before, there are may "equivalent" points on the circle: given a
point (x, y), there are also points ((+/-)x, (+/-)y). You can also
interchange x and y; this will simply swap two of the triangular numbers.
You could try to find a range of values for t to eliminate those
redundancies. To do that, you could require the point Q to be in the first
45 degrees of the circle, and use a little geometric reasoning to find the
corresponding range of values for t.

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 04/29/2012 at 08:36:04
From: Reema
Subject: Thank you (Averages of Triangular Numbers)

Wow!

Thank you so much for all your help. It is really appreciated.

I wasn't even expecting to get a response, but I got one and it was very
detailed and explicit.

Again, thanks so much!
Associated Topics:
High School Basic Algebra
High School Number Theory
High School Statistics

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