<s>[INST] Find all the tuples of integers $(a, b, c)$ with $a>0>b>c$, where $a+b+c=0$ and $N=2017-a^3b-b^3c-c^3a$ is the perfect square of an integer Find all the tuples of integers $(a, b, c)$ with $a>0>b>c$, where $a+b+c=0$ and $N=2017-a^3b-b^3c-c^3a$ is the perfect square of an integer
I said that since $a+b+c=0$ then $c=-a-b$ and hence factoring $K=a^3b+b^3c+c^3a$ we have that $K=-(a^2+ab+b^2)^2$
This is where I got stuck. I can't work out how to finish off the solution. Could you please explain to me the full solution and how you intuitively thought of each step?
 [/INST] you have $$ 2017 + \left( a^2 + ab + b^2 \right)^2 = w^2 $$  or
$$ w^2 - \left( a^2 + ab + b^2 \right)^2 = 2017 $$
Difference of squares on the left, factors, meanwhile 2017 is prime. Thus
$$  w + a^2 + ab + b^2 = 2017 , $$
$$  w - a^2 - ab - b^2 = 1 , $$
or
$$ w = 1009$$
$$ a^2 + ab + b^2 = 1008 = 2^4 3^2 \cdot 7 = 7 \cdot 144 $$
You can check, this means $a,b$ are both divisible  by $12.$ Let $\alpha = \frac{a}{12}$ and $\beta = \frac{b}{12},$ then
$$  \alpha^2 + \alpha \beta + \beta^2 = 7$$
which has just a few integer solutions.

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