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Show that area of Pythagorean triangle x,y,z cannot be square (U can use that a^4-b^4 = c^2 has no positive solution)

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the base and height of pythagorean triangle are 3d and 4d where d is a constant Area = 1/2 x 3d x 4d = 3d x 2d = 6d^2 this is never a square
Its not generalised though!!
It's a start... Let's say we have x^2 + y^2 = z^2 and xy = 2n^2 (triangle area)

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Okay so we take it as hypothesis and prove it a contradiction..I'll try
|dw:1350136054246:dw| will you get a quartic so as to use the given?
You need to multiply the first line of your first post with something...
Writing your first line in full: (x+y)^2 = x^2 + y^2 +2xy = z^2 + 4n^2
(x+y)^2 = x^2 + y^2 +2xy = z^2 + 4n^2 Your first line (x-y)^2 = x^2 + y^2 -2xy = z^2 -4n^2
that would yield extra non dependent solution.
*equation (solution)
We, are just trying to get a contradiction....
We are assuming a triangle that is square....
yeah ... three variables 3 non dependent equations ... it's possible to deduce from that.
If u multiply those two above, u get (x^2-y^2)^2 = z^4 - (2n)^4 which contradicts the given so no Pythagorean triangle area is square
yeah i see the trick!!
U almost had it, a bit more time, u would have had it, I think....
well .. I'm not too good with algebraic manipulations.
I shall put up a few more while it's stil quiet....
I usually takes quite lot more than necessary steps to do things.
I might go offline for 2hrs ... well I would enjoy if you tag me in few interesting problems if you have.

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