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Old 2017-06-28, 13:36   #183
science_man_88
 
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Jul 2009
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Quote:
Originally Posted by mahbel View Post
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 4-sq reps. The idea of my comment is to limit the search for factors to 4 sub expansions, that is the 1st 4-sq 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.

Last fiddled with by science_man_88 on 2017-06-28 at 13:50
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