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Let $n$ be a positive integer. What is the number of solutions to the equation $$8^n = a^3+b^3+c^3-3abc$$ with integers $a\geq b\geq c\geq 0$?

We have the factoring $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$ meaning that both factors must be powers of $2$.

  • 0
    If $a$ is odd, then.... – 2017-02-20
  • 2
    Related: http://math.stackexchange.com/questions/724990/integers-can-be-expressed-as-a3b3c3-3abc – 2017-02-20
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    Indeed, following @Martin R the answer the number of solutions is infinite since $3\not\mid 8^n$ for any $n$. – 2017-02-20
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    https://www.artofproblemsolving.com/community/c3046h1046862___ – 2017-02-20

2 Answers 2

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Note $$a^2+b^2+c^2-ab-ac-bc=(1/2)((a-b)^2+(a-c)^2+(b-c)^2)$$ Now suppose $a+b+c$ is even. Then all three variables are even, or else exactly one of them is even. If all three are even, then we can divide the original equation through by 8, getting a smaller solution. Repeat this enough times, and we get to a solution with exactly one even variable; let it be $a$. Then $a^2+b^2+c^2-ab-ac-bc$ is odd, but it's a power of 2, so it must be 1, so $(a-b)^2+(a-c)^2+(b-c)^2=2$. The only way to get three squares to sum to $2$ is for one of them to be zero and the others to be $1$, so $a-b=\pm1$, $a-c=\pm1$, $b=c$. Taking the plus sign, we get $3a-2=2^r$ for some $r$, so $a=(2^r+2)/3$, $b=c=(2^r-1)/3$. Then $$a^3+b^3+c^3-3abc=(1/27)((2^r+2)^3+2(2^r-1)^3-3(2^r+2)(2^r-1)^2)=2^r$$ so we need to take $r$ to be a multiple of $3$. For example, $r=6$ leads to $a=22$, $b=c=21$, $a^3+b^3+c^3-3abc=64=8^2$.

The reader may enjoy working through the case with the minus sign.

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Strangely enough, the solution is finite.

for the equation:

$$X^3+Y^3+Z^3-3XYZ=q=ab$$

If it is possible to decompose the coefficient as follows:

$4b=k^2+3t^2$

Then the solutions are of the form:

$$X=\frac{1}{6}(2a-3t\pm{k})$$

$$Y=\frac{1}{6}(2a+3t\pm{k})$$

$$Z=\frac{1}{3}(a\mp{k})$$