n^2 = SumOfSquares for n=power of 2

[Meeussen Wouter (bkarnd)]

d=2 :: n^2=(2^k)^2= a^2+b^2 ; non-equivalent:: b>=a>=0
 	 only one solution : 0^2+(2^k)^2,  
WHY?

d=3 :: n^2=(2^k)^2= a^2+b^2+c^2 ; non-equivalent:: c>=b>=a>=0
 	 only one solution : 0^2+0^2+(2^k)^2  
WHY?

d=4 ::        (2^k)^2= a^2+b^2+c^2+d^2 ; non-equivalent :: d>=c>=b>=a>=0
	always exactly 2 solutions :
	0^2+0^2+0^2+(2^k)^2  and
(2^(k-2))^2+(2^(k-2))^2+(2^(k-2))^2+(2^(k-2))^2
and no others, WHY?


d=5 ::    interesting number of solutions {1, 2, 3, 7, 27, 147, 963, 6947}
given by
	(4^k - j^2)+ SumOfSquares_of_j^2_in_4_terms  , for j=0 upto
Floor[2^(k-2)]
	but with quite some multiple counting.

the WHY's above are the crux of my question in this mail.

Oh, b.t.w., I know (of) the theorem that says that each integer can be
written as the sum of four squares. Was it Gauss' ?


Wouter Meeussen
tel  +32 (0)51 332 124
fax +32 (0)51 332 175
mail: wouter.meeussen at vandemoortele.com




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