Consider the following problem: The difference of the squares of two numbers is the same as their sum. What numbers satisfy this condition?

We might begin by expressing the problem as an equation:

$$a^2 - b^2 = a + b$$
Then we can look for possible solutions. Let's try to simplify the expression:

$$(a + b)(a - b) = a + b \\ a - b = 1, ~~~ a + b \ne 0$$

Solving for $$a$$, we get $$a = b + 1$$. So, any adjacent natural numbers, $$n$$ and $$n + 1$$, will satisfy the conditions of the problem. Thus,

$$(n + 1)^2 - n^2 = (n + 1) + n \\ n^2 + 2n + 1 - n^2 = 2n + 1$$

For example,

$$4^2 - 3^2 = 4 + 3 \\ 16 - 9 = 7$$

or

$$33^2 - 32^2 = 33 + 32 \\ 1089 - 1024 = 65$$

Since $$a + b$$ is always odd, if we pick odd squares, then we can generate an unlimited number of Pythagorean triples, for example:

$$2n + 1 = 15^2 = 225 \\ n = 112,~~ n + 1 = 113 \\ 113^2 - 112^2 = 113 + 112 \\ 15^2 + 112^2 = 113^2$$

or

$$2n + 1 = 23^2 = 529 \\ n = 264,~~ n + 1 = 265 \\ 265^2 - 264^2 = 265 + 264 \\ 23^2 + 264^2 = 265^2$$