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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 $x^3 - 3x + 1$?

dxiv
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John Smith
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    @BillDubuque Sorry, but I don't see how that's a duplicate. The OP is well aware of the rational root theorem and wrote "I can check using the rational root test that there are no rational roots, and consequently no integer roots either". The question is precisely whether there is a different way to show there are no integer roots directly, and why "Lang seems to be going in the opposite direction". Voted to reopen. – dxiv Apr 24 '23 at 18:27
  • @dxiv By RRT the only possible integer roots are divisors of $f(0)=1$. Of course it is trivial to check these two divisors $,\pm1,$ are not roots of $f(x).,$ This is surely what Lang intended. We already have hundreds of answers showing how to use RRT (and similar methods). There is nothing at all novel in any of the answers below. – Bill Dubuque Apr 24 '23 at 18:38
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    @BillDubuque You are ignoring the fact that the OP is very specifically asking for a way other than RRT, and the posted answers do not use RRT (for the most part). Just because it's straightforward to answer it using RRT, and you think that's the best/easiest way, does not mean that RRT is the only valid answer, or that you should deny the poster the right to ask the question. – dxiv Apr 24 '23 at 18:45
  • No the question is "how is [Lang] checking that there are no integer roots?". Reading this section of Lang makes it clear that he is almost surely still using RRT just as in the first example. Lang is simply saying it more concisely since he has already given more details in the example in the prior paragraph. The OP has misread. The OP does not ask for proofs using other methods (but they too would all be dupes since this is a huge FAQ). Please don't fight site organization. It is becoming impossible for students to search the site due to rampant duplication. – Bill Dubuque Apr 24 '23 at 18:51
  • For the record, many comments above have been deleted by a moderator, and what remains does not adequately represent the prior discussion (which was influenced by a current meta discussion on this). – Bill Dubuque Apr 27 '23 at 16:25

5 Answers5

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

Dyako Soltanian
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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.

TheSilverDoe
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$x^3 - 3x + 1 = 0 \iff x(x^2-3) = -1\,$ 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 $x=\pm1 \implies x^2-3=-2$.

dxiv
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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 $x^3-3x+1$.

Prism
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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.

KCd
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