# Proving that the following equation does not have integer solutions

I want to prove that the following equation has no integer solutions $$a,b,c$$: $$-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc = 0$$ apart from the naive solution $$a=b=c=0$$.

The context, in case it helps, is the following: I am dealing with the field $$\mathbb{F}_p[x] / (x^3-x-1)$$ and this equation appeared when I tried to get a "formula" for the inverse of some element on this field. More precisely, if $$g(x) = a + bx +cx^2$$ is an element of the field and $$g(x)' = c + dx + e^2$$ is assumed to be the inverse of $$g(x)$$ I have tried to compute $$c,d$$ and $$e$$ as a function of $$a,b$$ and $$c$$. The result being: $$d = \frac{a²+2ac-bc-b²+c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}, \\ e = \frac{ba-c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}, \\ f = \frac{-b²+ca+c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}.$$

However, the expression has a denominator that happens to be the above equation. So, as $$\mathbb{F}_p[x] / (x^3-x-1) = \mathbb{F}_{p^3}$$ we know that every element on this field has to have a multiplicative inverse (except the $$0$$) and therefore the above equation cannot be $$0$$.

How can I be more precise in the proof?

• $x^3-x-1$ is not irreducible for all $p$. Plug in $x=n$ and let $p$ be any prime factor of $n^3-n-1$. Then the polynomial has a zero, and cannot be irreducible. May 23, 2022 at 18:55
• In other words, you are asking too much. And you should worry about the denominator being divisible by $p$. In this ring $p=0$, so... May 23, 2022 at 18:57
• @JyrkiLahtonen Yeah, you are right. I did not explicitly written that this polynomial is irreducible for the prime that I am using (otherwise, I would not be able to talk about field extensions). May 24, 2022 at 6:56

For a contradiction assume $$a, b, c$$ is a solution.

First of all, if $$a, b, c$$ have any common factor, we can factor it out so without loss generality assume $$a, b, c$$ have no common factor.

Now, reduce the equation $$\mod 2$$: \begin{align} -a^3 - b^3 -c^3 + ab^2 - ac^2 + bc^2 -2a^2c + 3abc &\equiv\\ a + b + c + ab + ac + bc + abc &\equiv \\ (1 + a)(1 + b)(1 + c) - 1 &\equiv 0 \mod 2 \end{align}
But this shows, $$a, b, c$$ are all even which is impossible because $$a, b, c$$ have no common factor.