Prove that there are infinitely many numbers that cannot be expressed as the sum of three cubes

Question

Prove that there are infinitely many numbers that cannot be expressed as the sum of three cubes. I thought this involved looking at cubes mod 7 but that doesn't work as they can be 0,+-1 so you can make any number mod7..... ok same thing different story for mod9.

Can this be solved using Fermat's little theorem? X^6is congruent to 1 mod 7 express the three cubes as (X^2)^3.

No actually this doesn't work as this is just the same as the method for working mod9 on the cubes, with out the negative for squared, and also I think that Fermat's little theorem assumes that x is not a multiple of 7.

Answer 1

Fix M and see the integers in the interval [0, M^3 - 1]

M^3 integers, as the sum of 3 cubes we can present not more when M (3 equal cubes) + M*(M-1) (2*a^3 + b^3 - sum for ab) + M(M-1)(M-2)/6 (3 distinct numbers)

M^3 - M - M(M-1) - M(M-1)(M-2)/6 -> +inf with M -> +inf

Answer 2

Any number $m\equiv\pm4\pmod9$ cannot be expressed as a sum of $3$ cubes:

• $n\equiv0\pmod9 \implies n^3\equiv0^3\equiv 0\pmod9$
• $n\equiv1\pmod9 \implies n^3\equiv1^3\equiv+1\pmod9$
• $n\equiv2\pmod9 \implies n^3\equiv2^3\equiv-1\pmod9$
• $n\equiv3\pmod9 \implies n^3\equiv3^3\equiv 0\pmod9$
• $n\equiv4\pmod9 \implies n^3\equiv4^3\equiv+1\pmod9$
• $n\equiv5\pmod9 \implies n^3\equiv5^3\equiv-1\pmod9$
• $n\equiv6\pmod9 \implies n^3\equiv6^3\equiv 0\pmod9$
• $n\equiv7\pmod9 \implies n^3\equiv7^3\equiv+1\pmod9$
• $n\equiv8\pmod9 \implies n^3\equiv8^3\equiv-1\pmod9$

No $3$ values chosen from $\{0,+1,-1\}$ will ever sum up to $\pm4\pmod9$.