Prove or disprove that every integer can be expressed as either the sum of 3 cubes, or the sum of 3 cubes + 4.
'In the mathematics of sums of powers, it is an open problem to characterize the numbers that can be expressed as a sum of three cubes of integers, allowing both positive and negative cubes in the sum. An obvious necessary condition for n to equal such a sum is that n cannot equal 4 or 5 modulo 9, because the cubes modulo 9 are 0, 1, and -1, and no three of these numbers can sum to 4 or 5 modulo 9. It is unknown whether this necessary condition is sufficient.'
If the condition is sufficient, then the solution follows at once, because 4 could simply be added to the proximate solution for (n-4), with value 0 or 1, mod9, as required. However, the existence of such a solution is itself an 'open question' as Wikipedia confirms.
A further possibility is that not all moduli are created equal. Thus for numbers of the form (9k+1) all solutions for n less than 100 have a minimal solution of small size {0,0,1}, {1,1,2}, {-2,0,3}, {0,1,3} etc. The same can be said for most solutions for (9k+2) {0,1,1}{-2,-2,3}, {1,1,3}, {-3,1,4} etc. Unfortunately, 9*8+2 = 74 has the minimal solution {-284650292555885, 66229832190556, 283450105697727}, only discovered in 2016, which appears to rule out this approach.
Edited on June 1, 2019, 2:24 am
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Posted by broll
on 2019-06-01 01:43:59 |
Maybe
