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From: alainverghote on 27 Jun 2010 12:22

Good evening

We might propose as a diophantine equality of form
x1^2+x2^2+x3^2+x4^2 = y^2
this one:
(2(a+b))^2+(2(a-b))^2+(2a^2+2b^2-1)^2+(2(a^2+b^2)(a^2+b^2+1))^2
=
(2(a^2+b^2)(a^2+b^2+1)+1)^2

Alain
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