The method of descent: understanding Pierre de Fermat's proof argument

Imagine a square room, of dimensions n metres by n metres, where n is a whole number. We have two identical square carpets, both having dimensions p metres by p metres, where p is also a whole number. The area of the room is twice the area of one of the carpets. We ask the question: what are the smallest possible values for n and p, for which this is true?

For example, let us try n is seven metres and p is five metres. Then the area of one of the carpets would be five times five, or 25 square metres, while the area of the room would be seven times seven, or 49 square metres. This area is almost twice 25 square metres, but not quite, so these values of n and p do not work.

Suppose, though, that the smallest possible values of n and p such that twice p times p is equal to n times n are available, and let the room and carpets have these dimensions.

We place these carpets on the floor of this room so that one corner of each carpet coincides with each of two opposite corners of the room. Since the area of both carpets together is equal to the area of the room, the area of the overlapping carpets must be equal to the area of the uncovered parts of the room. The overlapping part is a...