When I wrote this a few years ago, I didn't know that anyone else had ever tried to find all the triangular squares. Turns out lots of folks have been interested in it, and the problem was solved by Euler in 1778. I guess that's what happens when an amateur without much background in number theory strikes off on his own. For more information about triangular squares see the Wikipedia article and its references.

I recently revisited the problem because I discovered it was related to the problem of finding
a general solution to Pythagorean right triangles where the legs differ by 1. (Examples are the familiar 3,4,5 triangle and
the 20,21,29 triangle). It turns out that a Pythagorean triangle whose legs differ by 1 exists for every triangular square. It gives
a geometric interpretation to the fact that the ratio between the t_{k} and s_{k} approaches the square root
of 2 as k increases. When I get a chance, I'll write an entry about Pythagorean triangles whose legs differ by 1 and their relation
to triangular squares.

Simply put, triangular squares are triangle numbers that are
also
squares. For example, T_{8} = ½·
8 · 9 = 36 = 6^{2}. In general a triangular square
is
any number *n*, such that *n* = ½ *a*(*a*
+ 1) and *n *= *b*^{2},
where *a* and *b *are
whole numbers. So,

In
order for *b *to be a whole number either is
a perfect square *and *
is
a perfect square (for *a *even), or *a *is
a perfect square
*and *
is
a perfect square (for *a *odd). For example, in the
case of T_{8}
above, which
is a perfect square, and *a *+ 1 = 9, which is also
a perfect
square. The next triangular square is T_{49} = ½ ·
49 · 50 = 1225 = 35^{2}. In this case, *a *=
49
is a perfect square and is
a perfect square. Is there a way to generate the number *a*
so
that we can find any triangular square without having to search to
list of triangle numbers for squares (or the list of squares for
triangle numbers)? The fourth triangular square—the first one
is T_{1} = 1 = 1^{2}—is T_{288}
= ½
· 288 · 289 = 41,616 = 204^{2}. So we have
three series of numbers to work with: the triangle numbers (1, 8, 49,
288, ...), the squares (1, 6, 35, 204, ...), and the triangular
squares themselves (1, 36, 1225, 41616, ...). Of course, it's easy to
find more triangular squares using a computer. (The next one is
1,413,721. See for yourself.) But I would like to find a general
method for generating triangular squares from the natural numbers (1,
2, 3, 4, ...).

As
a first step toward our goal, we take note of the fact that the ratio
between successive squares is always between 5 and 6. In fact, it
appears that 35 = 6 · 6 – 1 and 204 = 6 · 35 –
6. Let's see if 6 · 204 – 35 = 1189 also leads to a
solution. To do so, we need to solve the equation given above for *a*.

For
*b *= 1189, *a *= 1681. A quick
check of squares between
204^{2} and 1189^{2} (and
triangle numbers between
T_{288} and T_{1681}) shows
that we didn't miss any
in between. This makes a good working method for generating
triangular squares, but we are still missing a rigorous proof that
the method works for all triangular squares and captures all of them.
We also don't have a method for finding the *n*th
triangular
square without generating all the 1 through *n –* 1
triangular squares.