Daniel's triple challenge

The equation becomes $(a^3-b)^2 + (b^3-c)^2 + (c^3-a)^2 \; = \; 0$ so we deduce that $b = a^3$ , $c = b^3$ and $a = c^3$ . Thus $a^{27} = a$ , and hence either $a$ is $0$ (in which case $a=b=c=0$ or else $a$ is one of the twenty-six $26$ th roots of unity (in which case $b=a^3$ and $c=a^9$ . There are $27$ different values for $a$ , which then determine the values for $b$ and $c$ . Thus there are $27$ ordered triples.

First, we move all terms to one side and expand: $a^6+a^2+b^6+b^2+c^6+c^2-2a^3b-2b^3c-2c^3a=0$ Now, seeing the rational condition, we might wonder what is has to do with the problem, and with a little bit of looking at our equation, it isn't hard to see the factorization: $(a^3-b)^2+(b^3-c)^2+(c^3-a)^2=0$ Now, since all three squares are rational and positive, they must all be 0. We have three equations: $a=c^3$ $c=b^3$ $b=a^3$ Clearly, one solution is $(a,b,c)=(0,0,0)$ . We can substitute: $a=c^3=(b^3)^3=((a^3)^3)^3=a^{27}$ If $a\neq 0$ , we can divide by $a$ , giving: $a^{26}=1$ Therefore, $a$ is a 26th root of unity, and similarly, so are $b$ and $c$ . Let $a=e^{i\dfrac{n\pi}{13}}$ Therefore, using our equations: $b=e^{i\dfrac{3n\pi}{13}}$ $c=e^{i\dfrac{9n\pi}{13}}$ $a=e^{i\dfrac{27n\pi}{13}}$ Since $e^{ix}$ is periodic with period $2\pi$ : $a=e^{i\dfrac{27n\pi}{13}}=e^{i\dfrac{n\pi}{13}}$ Therefore, every 26th root of unity satisfies our equation, and gives one triple $(a,b,c)$ , and so the answer is $\boxed{27}$ .

Noticing that the squares of the given rational numbers resemble the parts in the equation, the equation can be rewritten as:

$a^6-2a^3b+b^2 + b^6-2b^3c+c^2 + c^6-2c^3a+a^2 =0$

$(a^3-b)^2+(b^3-c)^2+(c^3-a)^2=0$

As squares of rational number are always greater than or equal to $0$ , it follows that;

$a^3=b$ , $b^3=c$ and $c^3=a$ .

Substituting in eachother gives us that $a^{27}=a$ , so either $a=0$ or $a=e^{\frac{2n \pi i}{26}}$ ( $0 \le n \le 25$ ).

So there are $27$ ordered triplets satisfying the given.