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Given $a^{3} + b^{3} + c^{3} - abc$, determine all the possible values of the expression where $A$, $B$, and $C$ are non-negative integers.

Note that this has been solved here

But I need help understanding why the proof generalizes when we let $f(A,A, A+1)$, $f(A, A, A-1)$ and $f(A, A+1, A-1)$.

The proof doesn't say anything about the following $f(A, A+j, A+k)$ where $k,j \in \mathbb{N}$.

But obviously it probably does as the proof is correct, I just can't seem to understand where. Any help understanding this would be great, thanks!

saulspatz
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1 Answers1

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When it considers $f(a,a,a\pm 1)$ and $f(a+1,a,a-1)$ it wants to show that you can get all numbers not divisible by $3$ or divisible by $9$. These examples are sufficient to get all those numbers. There is no need in this phase to consider any more values of the arguments.

The second phase is to show that numbers divisible by $3$ but not divisible by $9$ cannot be expressed. Here you have to consider other values for the arguments because (so far) it could be that there is some set that will express $3$, for example. It proves this by considering the numbers $\bmod 3$.

Ross Millikan
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  • So the second part of the proof is to show that besides numbers divisible by 9 or not divisible by 3, there is nothing else possible? – MathNoob420 Dec 18 '20 at 22:36
  • Also, when solving this proof, what is the intuition to stop only at $f(a,a,a+1), f(a,a,a-1)$, and $f(a,a+1, a-1)$? I guess if I were to solve this initially, I would consider more general approach than using a $+1$ but a variable instead. – MathNoob420 Dec 18 '20 at 22:39
  • For the first, yes, that is what is being proved. For the second, often one plays with lots of possibilities before realizing what is necessary for the proof at hand. Those other possibilities are usually not shown when writing up answers. One tries to reduce the proof to the minimum possible, particularly not showing dead ends that were explored. Those possibilities give all that is needed. – Ross Millikan Dec 19 '20 at 04:45