Pythagoras' Theorem Continued

This post builds on the last one which
was a response to a comment on one of
my older posts. In the last post, I showed
that given an odd integer, you could
generate a Pythagorean triplet with the
odd integer as the smallest side. This
post is to do the same with even integers.

In the last post, I used the fact that
consecutive squares have an odd difference.
To make use of even numbers, I would have
to use squares, whose roots have a difference
of 2. For example: 6² - 4² = 20. Now, we
need to generalize the rule.


(Where c = b + 2)
a² + b² = c²
a² = c² - b²
a² = (c + b)(c - b)
a² = 2(c + b)
a² = 2(b + 2 + b)
a² = 2(2b + 2) a² = 4b + 4.

Isolating b, we get:

b = (a/2)² - 1
c = (a/2)² + 1

These values should satisfy Pythagoras' Theorem.
Trying a few out will yield the truth.

(a = 8)
b = 15
c = 17

64 + 225 = 289

(a = 12)
b = 35
b = 37

144 + 1225 = 1369

While I suspect this is the end of this series
on Pythagoras' Theorem, I hope to add another
post, with a wider range of possibilities, as currently,
b and c are separated by either 1 or 2 only.

Edit (On 2 March 2016):

I have begun preparation of a new post,
which I hope will contain the generalization of
this post, or in other words, an algorithm to
generate all the Pythagorean triplets of a number,
given the number.