Showing that an equation has no solution in $\Bbb Z$

Question

Show that $x^3 + 10x^2 + 6x + 2 = 0$ has no solutions in $\Bbb Z$.

This seems rather trivial to do but I don't know how to rigorously show this is true. Having graphed this and attempted to factor,I see that it is indeed true. Could someone please explain how I would go about showing this rigorously?

Answer 1

Hint. $x\in\mathbb Z$ and $x^3 + 10x^2 + 6x + 2 = 0$ $\implies x\mid 2$

Answer 2

What is the context? Another way you could show this is rely on the fact the irreducibility over $\mathbb{Z}/(p)$ implies irreducibility in $\mathbb{Z}$ for a prime $p$, provided that $p$ does not divide the leading coefficient of your polynomial. Selecting $p = 3$, your polynomial reduces to $\tilde{f} = x^3 + x^2 + 2$. Since $\mathbb{Z}/(p)$ is relatively small, we can check each element to see if it is a root or not:

1. $\tilde{f}(0) = 0^3 + 0^2 + 2 \equiv 2 \mod 3$
2. $\tilde{f}(1) = 1^3 + 1^2 + 2 \equiv 1 \mod 3$
3. $\tilde{f}(2) = 2^3 + 2^2 + 2 \equiv 2 \mod 3$

Hence $\tilde{f}$ has no roots in $\mathbb{Z}/(p)$, which implies there are no solutions to the equation $x^3 +10x + 6x + 2 = 0$ in $\mathbb{Z}/(p)$ as well as $\mathbb{Z}$.