Kurt Foster on Tue, 05 Dec 2023 18:41:41 +0100 |
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Re: number of ways of writing a nonnegative integer n as a sum of 3 squares (zero being allowed). |
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.