Show that $f=x^n+x+3$ is irreducible in $\mathbb Q[x]$ for all $n\geq 2$.

I am aware that this question has already been posted to the website but I just want to know if my attempt at proving it is correct. Here is my proof:

First suppose that $n$ is odd. Then $f$ certainly has at least one real root. Since $f'(x)$ is strictly positive for all $x\in\mathbb{R}$ this means $f$ is a strictly increasing function. Therefore it has exactly one real root $\alpha$. This root has multiplicity $1$ since $f'(\alpha)\neq 0$. We can write $f(x)=(x-\alpha)g(x)$ where $g$ is an irreducible element of $\mathbb{R}[x]$. In order to show $f$ is irreducible in $\mathbb{Q}[x]$, which by Gauss's lemma is equivalent to $f$ being irreducible in $\mathbb Z[x]$, it is enough to show that $\alpha \not\in \mathbb Z$. Since $f(-2)<0<f(-1)$ we can conclude that $\alpha \in (-2,-1)$ so $f$ is irreducible.

Now suppose $n$ is even. Since $f'=nx^{n-1}+1$ has exactly one root at $\beta= -(\sqrt[n-1]{n})^{-1}$ we can see that $f(\beta)$ is the global minimum of $f(x)$. Since $-1<\beta<0$ we have $f(\beta)=\beta^n+\beta+3>0-1+3=2$ so $f$ has no real roots which means it is irreducible in $\mathbb R[x]$ and therefore irreducible in $\mathbb Q[x]$.

  • 1
    $\begingroup$ Must a reducible polynomial have a linear factor? E.g. is $(x^2 + 1)(x^2 + 2) $ reducible in $\mathbb{Q}$? $\endgroup$
    – Calvin Lin
    Apr 1, 2020 at 3:23

1 Answer 1


This proof is not correct since it only establishes that $x^n+x+3$ has no root in $\mathbb{Z}$, which is equivalent to proving it has no linear factor (a factor of degree $1$). In order to prove irreducibility, you need to rule out also factors of higher degrees $2,3,\dots,\lfloor n/2 \rfloor $. For example, try your approach on $x^5+x+1$. Clearly derivative $5x^4+1$ is positive so it can have only one real root as well, yet $$x^5+x+1=(x^2+x+1)(x^3-x^2+1).$$

By the way, you can show non-existence of rational root by using rational root theorem. If $f(x)=x^n+x+3$ had a rational root $p/q$, then $p \mid 3$ and $q\mid 1$, so $p/q \in \{\pm 1,\pm 3\}$. Since clearly none of the $f(-1),f(1),f(3),f(-3)$ can be zero , we are done.

Just for completeness, see correct proofs in Show that $x^n + x + 3$ is irreducible for all $n \geq 2.$


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