How do I check that $x^3 - 3x + 1$ has no roots in $\mathbb{Z}$?

Question

Serge Lang in Algebra (pg 270) says that $x^3 - 3x + 1$ has no roots in $\mathbb{Z}$ and hence no roots in $\mathbb{Q}$.

I can check using the rational root test that there are no rational roots, and consequently no integer roots either.

Lang seems to be going in the opposite direction. He seems to be using Gauss's lemma and that's fine but how is he checking that there are no integer roots?

There is another polynomial on the same page $x^3 - x + 1$ that he checks directly to be irreducible using the rational root test. Why is he not doing the same for $x^3 - 3x + 1$?

Answer 1

Hint: $x^3 - 3x + 1$ is an odd number for any integer $x$.

Answer 2

Let $f(x)=x^3-3x+1$.

One has $f(-2)=-1$, $f(0)=1$, $f(1)=-1$ and $f(2)=3$. So the three real roots of $f$ lay respectively in the intervals $(-2,0)$, $(0,1)$, and $(1,2)$. The only integer in these intervals is $-1$, but $f(-1)\neq 0$, so $f$ has no integer root.

Answer 3

$x^3 - 3x + 1 = 0 \iff x(x^2-3) = -1\,$ so if $x$ is an integer then both $x$ and $x^2-3$ would have to be divisors of $-1$, so both need to be $\pm1$. That's not possible since $x=\pm1 \implies x^2-3=-2$.

Answer 4

Suppose $x^3-3x+1$ had some integer root, say $x=m$ where $m\in\mathbb{Z}$. This means that $x-m$ would divide $x^3-3x+1$ in the polynomial ring $\mathbb{Z}[x]$. Thus, $x^3-3x+1 = g(x)\cdot (x-m)$ where $g(x)\in\mathbb{Z}[x]$ (not just in $\mathbb{Q}[x]$).

We can write $g(x) = ax^2+bx+c$ for some $a, b, c\in\mathbb{Z}$. You can check that $a=1$ by comparing the coefficient of $x^3$. By comparing the constant terms on both sides of $x^3-3x+1 = g(x)\cdot (x-m)$, you can conclude that $1 = - m \cdot c$. Since $m$ and $c$ are both integers, what can we conclude about $m$? There should only be two possibilities for $m$!

Now, check that each of those possibilities do not lead to a root of $x^3-3x+1$.

Answer 5

For a monic polynomial in $\mathbf Z[x]$, a rational root must be an integer dividing the constant term. So for $x^3 - 3x + 1$ and $x^3 - x + 1$, the only possible rational roots are $\pm 1$, and neither works. Thus there is no rational root for either polynomial.

Another approach is to use the reduction mod $p$ test: both $x^3 - 3x + 1$ and $x^3 - x + 1$ are congruent to $x^3 + x + 1 \bmod 2$, which is an irreducible polynomial in $(\mathbf Z/(2))[x]$ because it is cubic mod $2$ without a root in $\mathbf Z/(2)$ (try $0$ and $1$).

For $f(x) = x^3-3x+1$, another approach to irreducibility is to see that $f(x-1) = x^3 - 3x^2 + 3$ (that's not $x^3 - 3x + 3$!), so $f(x-1)$ is irreducible over $\mathbf Q$ by Eisenstein's criterion for the prime $3$. Thus $f(x-1)$, so also $f(x)$, is irreducible over $\mathbf Q$. This won't work for $x^3 - x + 1$: for no integer $c$ is $(x+c)^3-(x+c) + 1$ an Eisenstein polynomial.