Numbers a^3+b^3+c^3-3abc

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#1 h Dec 25, 2018, 3:11 AM • 1 Y

Y by Adventure10

A number is special if it can’t be written as $a^3+b^3+c^3-3abc$ where $a,b,c \in \mathbb{Z}$ . How many special numbers are there from $1$ to $2013$

This post has been edited 1 time. Last edited by justkeeptrying, Dec 25, 2018, 3:13 AM
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#2 h Dec 25, 2018, 3:22 AM • 3 Y

Y by RudraRockstar, Adventure10, Mango247

Take $a=b=c-1$ . You get $a^3+b^3+c^3-3abc=3a+1$ . Take $a=b-1=c-1$ . You get $a^3+b^3+c^3-3abc=3a+2$ . Take $c-1=b=a+1$ to get $a^3+b^3+c^3-3abc=9b$ . We show that every number which is divisible by 3 but not by 9 is special. Suppose $3x$ is not special, i.e. $a^3+b^3+c^3-3abc=3x$ . Taking this mod 3, we get that $a, b, c$ are either the same mod 3 or $1,0,-1$ mod 3. Consider the factorisation $a^3+b^3+c^3-3abc = 1/2 (a+b+c)((a-b)^2 + (b-c)^2+(c-a)^2)$ . In case 2, the first and second factor both are divisible by 3, and in case 1, the first factor is divisible by 3 and the second is divisible by 9. So $x$ is divisible by 3.
Thus all that remains is to count such numbers, which is trivial.

This post has been edited 2 times. Last edited by WizardMath, Dec 25, 2018, 3:24 AM
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