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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.

Bill Dubuque
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grigori
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2 Answers2

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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 $\pm4\pmod9$.

barak manos
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  • It seems that only the multiples of $9$ work as a modulus for this argument! – lhf Nov 08 '16 at 15:21
  • the multiples of 9 work because if you multiply the results above by any integer, then as 9 isn't prime they will all be distinct or there will be some repeats mod9, so there will be less or equal to 4 different results mod 9 – grigori Nov 08 '16 at 15:28
  • Correct me if I'm wrong, but couldn't you simplify this and leave out the $8$s? Just ${0, \pm1}$. – Colm Bhandal Nov 09 '16 at 09:56
  • @ColmBhandal: Hmmmmm... I was about to say no, but with $3$ terms... Yes You Can (although, I admit that saying it today is a somewhat ironic). – barak manos Nov 09 '16 at 09:58
  • @ColmBhandal: Improved accordingly, thanks for pointing it out :) – barak manos Nov 09 '16 at 10:06
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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

kotomord
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