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,
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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