Pythagorean Triples: Consecutive Legs

Almost done now. Honestly. Maybe one more entry after this one. Maybe two.

On page 100 of Mathematical Recreations, by Maurice Kraitchik, Second Revised edition, Dover Publishing, 1953, the author poses this question:

Find a right triangle whose legs are consecutive integers

The answer is not the 3 - 4 - 5 triangle (again!), but an infinite set of such triples (of which 3-4-5 is a member). I have to admit that I didn't follow the logic the first time through and had to re-read the entire section, but I could sum it up with my own method once I discovered that was involved, along with the fraction:
, which converges on .

It made sense to me that would be involved with Pythagorean triples because of the numbers association with right isosceles triangles. As the legs of the triangle increase in size, the ratio of two consecutive numbers comes closer to 1:1.

Followed by: 99/70 and 239/169

If you look at the fractions with the odd denominators, you have a Pythagorean Triple in the form of (a+b)/c, where a and b are consecutive numbers (i.e., b = a + 1) 7/5 gives you 3, 4, 5. 41/29 gives you 20, 21, 29 239/169 gives you 119, 120, 169

There has to be an easier way to find these triples!

There is ... and that's on the next page of my journal.