n^2 = SumOfSquares for n=power of 2

[Pantelimon Stanica]

I forgot to mention, but for d=4, the answer is given by
Jacobi's Four Square Theorem:
The number of representations of an integer as the sum of four squares is
equal to eight times the sum of all its divisors which are not divisible
by 4. 

Pante Stanica
Auburn Univ. Montgomery

On Mon, 19 Nov 2001, Pantelimon Stanica wrote:

> 
> As far as I know, it was Lagrange who claims the theorem you
> mention at the end.
> 
> Pante Stanica
> Auburn Univ. Montgomery
> 
> On Mon, 19 Nov 2001, Meeussen Wouter (bkarnd) wrote:
> 
> > 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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