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From: alainverghote on 17 Jun 2010 06:13
Bonjour,

It seems easy to generate 'squares' equalities.

Just starting with positive integer terms:
2^2+3^2 = 13 = 1*13=(7-6)(7+6)
So 2^2+3^2+6^2 = 7^2 (1)

1^2+4^2+7^2 = 66 (2)
=1*66=2*33=3*22=6*11
no product of same parity

3^2+5^2+9^2+11^2 = 236
= 2*118 = (60-58)(60+58)
So 3^2+5^2+9^2+11^2 +58^2 = 60^2 (3)

For all sum(xi^2) <> 2 mod.4 cf (1) ,(3)
we may find at least one difference of two squares
and an equality sum(xi^2) = y^2 ,xi ,y positive integer.

Alain
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