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

Question

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?

Answer 1

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.