All integers except possibly those of the form $9k \pm 4$ can be written as the sum of 4 cubes.
Are there any heuristics that suggest this is impossible in general, or is it the case that no one has found any identities?
2026-02-22 19:35:42.1771788942
Why $9k \pm 4$ can't be written as sum of four cubes
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(Updated):
If the number is of the form $9k\pm 4$, then it cannot be written as the sum of three cubes. Simply because any cube number has form $9s$ or $9s\pm 1$.
As for four cubes: maybe it is not proven that all numbers of the form $9k\pm 4$ can be written as the sum of four cubes.
It looks that for many values $k$ it is relatively easy to decompose numbers of the form $9k\pm 4$ to the sum of $4$ cubes.
I checked $k$ in the range $1\le k \le 10^4$, and result is:
if $1\le k \le 10000$, then equation $$9k-4=a^3+b^3+c^3+d^4\tag{1}$$ has integer solution $(k,a,b,c,d)$, where $\max\{|a|,|b|,|c|,|d|\}<300$.
Few relatively "hard" examples from this set:
$9\cdot 3112-4 = 28004 = (-175)^3 + (-214)^3 + (-226)^3 + 299^3$;
$9\cdot 9174-4 = 82562 = 104^3 + (-196)^3 + (-235)^3 + 269^3$;
$9\cdot 1264-4 = 11372 = 59^3 + 59^3 + 257^3 + (-259)^3$;
$9\cdot 1418-4 = 12758 = (-58)^3 + (-169)^3 + (-217)^3 + 248^3$;
$9\cdot 3105-4 = 27941 = (-100)^3 + (-106)^3 + (-232)^3 + 245^3$;
and equation $$9k+4=a^3+b^3+c^3+d^4\tag{2}$$ has integer solution $(k,a,b,c,d)$, where $\max\{|a|,|b|,|c|,|d|\}<520$.
Few relatively "hard" examples from this set:
$9\cdot 2474+4 = 22270 = 325^3 + 349^3 + 391^3 + (-515)^3$;
$9\cdot 6517+4 = 58657 = (-161)^3 + (-221)^3 + (-401)^3 + 430^3$;
$9\cdot 6608+4 = 59476 = (-101)^3 + 247^3 + 295^3 + (-341)^3$;
$9\cdot 7479+4 = 67315 = (-221)^3 + 250^3 + 307^3 + (-323)^3$;
$9\cdot 574+4 = 5170 = 22^3 + 67^3 + 319^3 + (-320)^3$;
Here are few first numbers of the form $9k-4$:
And few first numbers of the form $9k+4$: