From: insight_love@hotmail.com (Lee Ho-Joo)
Subject: sums of five cubes
Date: 9 Jul 00 15:31:11 GMT
Newsgroups: sci.math.numberthy
Summary: [missing]

Dear, Number theorists

It is well-known that 6k=(k+1)^3 +(k-1)^3 -k^3 -k^3

and 6k+3=k^3 -(k-4)^3 +(2k-5)^3 -(2k-4)^3

I found the following identities.

6k+2 =k^3 +(k+1)^3 +(5k+1)^3 +(6k+2)^3 -(7k+2)^3

and 6k+4 =(6k+4)^3 +(k+1)^3 +(5k+4)^3 +k^3 -(7k+5)^3

Therefore, we easily get the following result.

If n=0, 2, 3, 4 (mod 6), then there exists A, B, C, D, E in Z  such that

n=A^3 +B^3 +C^3 +D^3 +E^3 and -2n <= A, B, C, D, E <= 2n .

Here, I have a question to you.

If n= +1 or -1 (mod 5), then there exists A, B, C, D, E in Z  such that

n=A^3 +B^3 +C^3 +D^3 +E^3 and -2n <= A, B, C, D, E <= 2n ??

And do you know anything about similar results?

Thanks in advance.

Sincerely yours,
Lee, Ho-Joo


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