Quote:
Originally Posted by mahbel
I cannot do the calculations by hand.
Because we are dealing with a big number, the squares (a,b,c,d) themselves will have many 4sq reps. The idea of my comment is to limit the search for factors to 4 sub expansions, that is the 1st 4sq rep of a,b,c,d.

can't divide those big number's by 2 ? after all (2a)^2= 4(a^2) edit :to prove what you said, most of what you have to do is prove that, any of the squares, once expanded form a lower pair of a Pythagorean triple.