Re: number of ways of writing a nonnegative integer n as a sum of 3 squares (zero being allowed).

[Kurt Foster <drsardonicus@earthlink.net>]

On Dec 4, 2023, at 1:03 AM, Thomas D. Dean wrote:
> If I consider each of a,b,c to be distinct even though the are not  
> numberically:
> N [a,b,c]
> 0 [0,0,0] -> 6 OEIS says 1 What about ordered tripes?
> 1 [0,0,1] -> 6 OEIS says 6 OK
> 2 [0,1,1] -> 6 OEIS says 12 - How do they get this?
> 3 [1,1,1] -> 6 OEIS says 8 - ?
> 4 [0,0,2] -> 6 OEIS says 6 OK
> 5 [0,1,2] -> 6 OEIS says 24??
> 6 [1,1,2] -> 6 OEIS says 24??
> 7 none
> 8 [0,2,2] -> 6 OEIS says 12
> ...
> 25 [0,0,5] [0,3,4] -> 12 OEIS says 30?
> 26 [0,1,5] [1,3,4] -> 12 OEIS says 72?
>
> What am I missing? Can someone explain this to me?

The "standard" count is of the number of triples (a, b, c) with a, b,  
and c integers for which a^2 + b^2 + c^2 = N.  Consider these as  
points on the sphere x^2 + y^2 + z^2 = N.

What you're missing is minus signs.  If none of a, b, or c is 0, there  
are 8 ways to choose the signs.  If one of them is 0, there are 4  
ways.  If two of them are 0, there are 2 ways.

So if 0 < a < b < c, , N = a^2 + b^2 + c^2, the triple (a, b, c)  
determines 8*3! = 8*6 = 48 points on x^2 + y^2 + z^2 = N.  If 0 = a <  
b< c, they determine 4*3! = 24 points.

Thus for N = 26, (0, 1, 5) determines 24 points and (1, 3, 4)  
determines 48 points.  And 24 + 48 = 72.

The other cases (e.g. 0 < a < b = c, etc) are left as exercises.