n^2 = SumOfSquares for n=power of 2

Meeussen Wouter (bkarnd) wouter.meeussen at vandemoortele.com
Mon Nov 19 18:25:22 CET 2001

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

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  

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
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
	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

This email is confidential and intended solely for the use of the individual to whom it is addressed. 
If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited.
You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

More information about the SeqFan mailing list