# n^2 = SumOfSquares for n=power of 2

Pantelimon Stanica stanica at strudel.aum.edu
Mon Nov 19 18:58:16 CET 2001

```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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