c is the difference of two squares.

These two squares are (d+n)(d+n) and (x+n)(x+n), the latter being the smaller square. You can visualize c as an L-shape surrounding the smaller x+n square, like in the first picture.

If we found (d+n)(d+n) or (x+n)(x+n), we'd be able to factor c.

Odd squares are equal to eight times a triangle number plus one (8T(u) + 1). Even squares are equal to four times a triangle number plus four times the previous triangle number (4(T(u)+T(u-1)). This is shown in the second picture. If we represent either (d+n)(d+n) or (x+n)(x+n) as squares made of triangles like this, and if we can find the base of these triangles, we can factor c. As stated >>7656 (You) here, we can find the parity of x+n based on e%4. We then therefore know the parity of (d+n)(d+n), since we know the parity of c, and odd-odd=even, even-odd=odd, odd-even=odd, even-even=even. This means we always know whether (x+n)(x+n) and (d+n)(d+n) are 8T(u)+1 or 4(T(u)+T(u-1)). That means we know what configuration of triangles they are.