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 T(x) = Q(y). Let x1 and y1 be two such numbers. Then there exists two other numbers x2 and y2 in the table such 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 T(x1) = Q(y1). Then there exist x2 and y2 such 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 T(x2) = Q(y2) yield a quadratic equation

  x22 - x2(4u + 1) + 2u2 = 0,

from which x2 is expressed as

 

Having in mind that T(x1) = Q(y1), one can show that 8u2 + 8u + 1 = (2x1 + 4y1 + 1)2, which leads to a more straightforward expression for x2:

  x2 = ((4x1 + 4y1 + 1) ± (2x1 + 4y1 + 1))/2.

The minus sign leads to x2 = x1, from which y2 would be equal to -y1 < 0. Thus we are left with the only meaningful

  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 (x, y) such that T(x) = Q(y), we can use the recurrence (*) to obtain infinitely many such pairs. Let's for example start with the first row: (x, y) = (1, 1). (*) gives successively (8, 6), (49, 35), (288, 204), etc. - row after row. It appears that the recurrence (*) generates the whole table of all possible pairs (x, y).

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 T(x) = Q(y) it can be shown

that if x > 1 and y > 1, then 3x - 4y + 1 > 0 and also -2x + 3y - 1 > 0.

Assume on the contrary that x and y with T(x) = Q(y) have not been obtained from (*) starting with (1, 1). From (**) we get decreasing sequences of x and y none of which belongs to the table but all are positive. Since neither of the sequences may be infinite the pair (1, 1) is bound to appear sooner or later. Contradiction.

I remove my hat to Armando Guarnaschelli.

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