``if the area of a rational right triangle were a square, then there would also be a smaller one with the same property, and so on, which is impossible.''OK, that wasn't so helpful. Anyway, we did everything on Thursday except 2b, i.e., suppose that is a solution in positive integers to the equation with minimal among all possible solutions. Show that there is a a smaller solution, hence deduce a contradiction.
Then
from which we obtain a smaller solution.
There are integers with such that
Since is a perfect square we can write either , or , for integers .
Then , so since are coprime we have , for integers . Since and , we have
which yields a smaller solution to .
Then , so since are coprime and have square product we have , for integers . Since and , we have
which yields a smaller solution to .
William Stein 2006-07-07