My guess was that there would be an infinite number of them and it seemed to me that the obvious way to attack this problem was to use Heron's formula for the area of a triangle.
where are the lengths of the three sides of the triangle and
. I let the sides of the triangle be represented by the three consecutive integers
. I then found, after simplification, that
Upon examination of this result it was clear that had to be an even integer in order for to be an integer. For if were odd, would be odd and even if were a perfect square, would not divide
. Also, by observation, it is clear that must be equal to a number of the form . I did a quick search and found that the integers all worked. Since my assumption was that there were going to be an infinite number of these triangles, I looked for a recurrence relation for which were the first three terms that generated and infinite sequence of values for satisfying (1). I would then test this sequence of to see if the values always determined a Super-Heronian triangle.
Looking at the three terms, , my guess was that these values satisfied the sequence given by the recurrence relation
I used this sequence to generate more terms, all of which seemed to work (see the table on the next page).
The sequence generated by the recurrence relation (2) can also be represented by
Here are the first terms terms which do, in fact, generate Super-Heronian triangles. The first row generates a line which represents a degenerate triangle.
As one can see, the sides get large very fast. It is left to show that every term of the sequence yields a solution; that is, an area that is an integer. Let
Let us examine another interesting feature of these triangles. Suppose we use the area formula
. We then have the area of the triangle given by
One might observe that the s can be generated on their own from the recurrence relation
Bill Richardson 2010-12-10