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

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