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  • Level: GCSE
  • Subject: Maths
  • Word count: 2104

About Triangular Square Numbers

Extracts from this document...

Introduction

About Triangular Square Numbers

By August Pieres

January 18th, 2003

 I believe I have discovered an algorithm which generates an infinity of triangular squares.  "Triangular squares" are triangular numbers which are also perfect squares.  These are triangular numbers: 1,3,6,10,15,21,28,36,45,55,66,78,91,105,120,136,153,171,190,210,…

Notice that 120=5! (and 6=3!) and that 1,3,21, and 55 are also Fibonacci numbers; one might call them "Fibonacci triangles."  Are there any more Fibonacci triangles?  These are the perfect squares: 1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400,…

        The only triangular squares listed so far are 1 and 36.  Earlier today I thought that these were the only existing triangular squares, but I found out that there are more and quite possibly an infinity of them.  I made a program on my programmable Texas Instruments TI-86 calculator.  Here it is:

PROGRAM:TRISQUAR

1N

Lbl A

N*(N+1)/2M

If M==iPartM

Then

Disp M

End

1+NN

Goto A

I ran the simple program above and it found the following additional triangular squares: 1225, 41616, 1413721.  Then I "played" with these new numbers -- with the help of the calculator -- trying to find patterns.

        To each triangular square corresponds a pair of parameters: s and t, such that a triangular square N is the sthperfect square and the tthtriangular number, i.e.

N=s2=Tt.

So N=N(s,t), or s=s(N) and t=t(N), where

image00.png

image01.png.

Here I list the s and t parameters for the first five triangular squares:

1.                s=1,                t=1.                t/s=1.

36.                s=6,                t=8.                t/s=1.333…

1225.                s=35,                t=49.                t/s=1.4.

41616.                s=204,                t=288.                t/s=1.411764705…

1413721.        s=1189,        t=1681.        t/s=1.413793103…

I also listed the ratio t(N)

...read more.

Middle

,

and

image13.png,

so N7 = 1631432881 is the 40391th perfect square, and it is the seventh triangular square.

        Repeating the paradigm shown above, it can be found that the eighth triangular square is:

N8 = 55420693056,

that it is the square of 235416, and that it is the 332928th triangular number.  Also,

image14.png

which is closer to image08.png = 1.41421356237…

        My conjecture is that this algorithm generates a countable infinity of triangular squares, and that any and all triangular squares which exist will be generated by this algorithm.

        But I have not proven that the algorithm works, nor do I know why it works, nor have I proven that there are no triangular squares lurking somewhere in the numerical space between, say, N7 and N8, or generally, between Ni and Ni+1.  I still have faith in this algorithm, though.

        NOVAM RATIŌNEM NUMERŌRUM HODIĒ DISCĒBAM.

January 20th

I applied the G-transformation to the finite difference formula

image15.png,

and letting x0 = 1, x1 = 1, x2 = 3, solved for xn, obtaining

image16.png

But since p(N) has the property that

image17.png,

with p(N1) = 1, p(N2) = 3, p(N3) = 7, it follows that

image18.png.

Also, I found a general formula for t in terms of p; viz.

image19.png.

It follows that

image20.png

Note that: i is not -- here -- the square root of -1, but just an index variable (or a pointer) as used in computer science.

        Then one can derive the general formula for the ithtriangular square:

image22.png.

This simplifies to

image23.png.

        I just induced another relation: notice that p(Ni) always divides into s(Ni).  E.g.

image24.png,

image25.png,

image26.png

...read more.

Conclusion

        The terms in image58.png which did not cancel out are the result of "constructive interference", i.e. they show up twice, once in image49.pngand once again in image60.png; so all the terms of image58.png will have even coefficients.  When this difference is divided by image08.png, the terms of image61.pngwill all be even, and so

image37.png

will produce only natural numbers.  Since one now knows that p(Ni) and image62.png are both natural (for natural i), it follows that

image63.png

is also natural for any i.  Therefore the formula for Ni yields only triangular squares.

        This means that there is an algorithm which I have now proven to produce an infinity of triangular squares.  The only thing that would remain to be proved is that this algorithm produces all the possible triangular squares.

        As far as I know, this is an original problem in number theory: I invented it (ILLUD INVENIĒBAM), I induced a solution, and then I deduced it.  I never did that before.

        I imagine that somebody, somewhere, some time ago, already contemplated the infinitude of triangular squares, and wrote about it; I have read many times about triangular numbers and about perfect squares, but never about triangular squares.  This is the best I have ever done in number theory.

        If I were to show this to Pythagoras, he would admit me into his secret society: tetraktys all the way!

January 28th, 2003

I looked up "triangular square numbers" with Google and found at least one website which not only mentioned triangular squares, but also analyzed them, in a fashion similar to mine, so my discovery is sublunary.

NIHIL SUB SŌLE NOVUM  (Ecclesiastes.)

...read more.

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