Solve $x^3-3y^3 + 6y^2 - 16x + 8 = 0$

Question

I was given the following equation: $$x^3-3y^3 + 6y^2 - 16x + 8 = 0$$ with $x, y \in \mathbb{Z}$.

I need to solve it in $\mathbb{Z}^2$. I have already the answer: The equation is not solvable in $\mathbb{Z}^2$.

I have a problem to understand the approach behind the answer, which states that in order to solve the equation in $\mathbb{Z}^2$ it is sufficient to solve it in $\mathbb{Z}/3\mathbb{Z}.$

Could anyone explain the idea behind this approach?

Answer 1

The idea is to say that if the polynomial can be equal to zero, it must definitely be able to represent some number which is a multiple of $3$ because $0=0\cdot 3$.

But if we are looking for a multiple of $3$ as a value, we can throw out any multiple of $3$ we like - which is essentially what we mean by working modulo $3$.

Of course we could find that the polynomial could be a multiple of $3$, but that wouldn't mean it had a solution.

Throwing out multiples of $3$ here is good to try because it gets rid of the terms involving $y$ and strategic adjustment by multiples of $3$ leaves $x^3-x-1$. Now little Fermat or the fact that $x^3-x=(x-1)\cdot x\cdot (x+1)$ tells us that $x^3-x$ is always a multiple of $3$ so we can delete that too and end up with $-1$ which is never a multiple of $3$. Since our polynomial represents only numbers which leave a remainder of $-1$ when divided by $3$, it can never represent a multiple of $3$ and can therefore never take the value zero.

Answer 2

if $x^3-3y^3 + 6y^2 - 16x + 8 = 0$ has a solution $x, y \in \mathbb{Z}$ then it has a solution modulo 3

$x^3 = x \pmod 3$

$x -x + 2 \ne 0\pmod 3$

Answer 3

Hint: Consider the equation modulo $3$. Then we have $x^3+2x+2=0$, which is irreducible over $\mathbb{F}_3$. Hence there is no solution.