There exist triangular numbers that are also square
This fact was no longer news to me when in September 2000 I received a message from Armando Guarnaschelli, Argentina. A Microsoft Word file has been attached to the message with a penetrating observation into the problem. Unfortunately, I could not convert the Word file into HTML - the language of the Web, whereas retyping it into HTML in its entirety seemed to be too arduous a job. With my apologies to Armando, I here present his results in my own words.
Closely studying the table of the numbers that are both triangular and square, Armando made an interesting observation. Let x be the side of the triangular T(x), y the side of the square Q(y). The numbers in the same row of the table have the property that
|x1 + y1 = x2 - y2.|
In fact, the second pair of the numbers lies in the row immediately following that of the first pair.
Let's prove that this is always the case. In other words, let's assume that
|x1 + y1 = x2 - y2, and T(x2) = Q(y2).|
Introduce u = x1 + y1 and assume that x2 and y2 do exists. Let's see where this assumption may lead us. (Note that the deductive steps will be reversible.) Then y2 = x2 - u, and
|x22 - x2(4u + 1) + 2u2 = 0,|
from which x2 is expressed as
Having in mind that T(x1) = Q(y1), one can show that
|x2 = ((4x1 + 4y1 + 1) ± (2x1 + 4y1 + 1))/2.|
The minus sign leads to x2 = x1, from which y2 would be equal to
|x2 = ((4x1 + 4y1 + 1) + (2x1 + 4y1 + 1))/2 = 3x1 + 4y1 + 1|
From y2 = x2 + u we easily find y2. I'll write the two expressions together:
x2 = 3x1 + 4y1 + 1|
y2 = 2x1 + 3y1 + 1
This is how simple it finally becomes. Starting with one pair
This is indeed so. To see why, solve (*) for x1 and y1:
x1 = 3x2 - 4y2 + 1|
y1 = -2x2 + 3y2 - 1
(*) yields increasing sequences of x and y. (**) yields decreasing sequences. However, using
Assume on the contrary that x and y with
I remove my hat to Armando Guarnaschelli.