Prove that the number 2003 cannot be expressed as the sum of three cubes?

Question

Three integral cubes please.

Answer 1

Lemma:
Every integer cubed is equivalent to -1, 0 or 1 mod 9.
Let integer n = 9a + b for some integer a,
and b є {0,1,2,3,4,5,6,7,8}
Then n^3 = (9a+b)^3 = (9a)^3 + 3b(9a)^2 + 3b^2(9a) + b^3
= 9[81a^3 + 27a^2b + 3ab^2] + b^3
≡ b^3 mod 9
And {n, n^3, n^3 mod9} for all b are:
0, 0, 0
1, 1, 1
2, 8, -1
3, 27, 0
4, 64, 1
5, 125, -1
6, 216, 0
7, 343, 1
8, 512, -1

Therefore, the sum of three cubes can only be congruent to 0, ±1, ±2, or ±3 mod 9,
and can never be congruent to ±4 mod 9.

2003 ≡ -4 mod 9.

Answer 2

Well...

It happens to be a fact that all cubes are 0, +1 or -1, when expressed mod 9 . I guess this arises out of the arbitrary fact that our normal numbering system is base 10.

So when we add three cubes, we must get a number that is NOT 4 or 5, mod 9. Combinations of 0, -1, +1 in mod 9 arithmetic can give all other values {0, 1, 2, 3, 6, 7, 8} but never 4 or 5.

2003 is 5 mod 9

Therefore it cannot be expressed as the sum of 3 cubes.

(Note that it's much more difficult to reason this backwards: there are some quite small numbers that are not 4 or 5 mod 9 for which it is quite simply *unknown* if they can be expressed as a sum of 3 cubes. Anyhow, luckily that's not our problem!)