Prove that the roots $\in \Bbb R$ of $x^3+x+1=0$ aren't rational without RRT

Question

I need to prove that the roots $\in \Bbb R$ of $x^3+x+1=0$ aren't rational. Obviously, it's easy to use the rational root theorem to prove that there are not rational solutions to this equation, but i want a different approach. I saw a similar question here, but it didn't provide my "solution".

My try

$x^3+x+1=0$

$x^3+x=-1$

$x(x^2+1)=-1$

Here we have two options (product of two numbers to obtain a negative):

$1.$ $x \gt 0$ $∧$ $x^2+1 \lt 0$ (not possible in $\Bbb R)$

$2.$ $x \lt 0$ $∧$ $x^2+1 \gt 0$

So, we use $2.$ to prove that $x \ne 0$, then:

$x(x^2+1)=-1$

$x^2+1=\frac{-1}{x}$

$x^2=\frac{-1}{x}-1$

$x^2=-(\frac{x+1}{x})$

$x=\pm \sqrt {-(\frac{x+1}{x})}$

Then, $\frac{x+1}{x} \ge 0$ because $x \lt 0$ (the equality occurs when $x=-1$, but this doesn't satisfy the original polynomial equation) but here i'm missing the cases $-1 \lt x \lt 0$ (i don't know how to use this to prove that the root is irrational)

This implies that $2$ of the roots are complex, but there are $3$ roots to a third degree polynomial equation.

And here i'm stuck, because i don't know how to prove that the last solution is irrational.

Any hints?

Is there anyway to prove that the last root is irrational?

Is my proof good so far?

Answer 1

It suffices to show any rational root $w$ is an integer $\,n,\,$ by $\,1 = -n(n^2\!+\!1)\,\Rightarrow\, n\mid 1\,$ so $\,n = \pm1,\,$ contradiction. Suppose $\,w = c/d\in\Bbb Q.\,$ Note $d^2$ is a common denominator for all elements $r$ in the ring $\,R = \Bbb Z[w] = \{ a_o + a_1 w + a_2 w^2\ :\ a_i\in\Bbb Z\}.\,$ Thus $\,R\subseteq \Bbb Z/d^2,\,$ i.e. $\,r \in R\,\Rightarrow\,r = n/d^2\,$ for $\,n\in\Bbb Z.\,$ If $\,\color{#c00}{r\not\in\Bbb Z}\,$ then wlog we may assume $\,0 < r < 1\,$ by taking its fractional part - which lies in $\,R\,$ and is nonintegral iff $r$ is. Then $\,r\in \{ 1/d^2,\, 2/d^2,\ldots,(d^2\!-\!1)/d^2\}.\,$ If $r$ is the smallest element of $R$ in this set then $r^2$ is an even smaller such element, since $\,1 > r > r^2 > 0,\,$ contra minimality of $\,r.\,$ Therefore $\,\color{#c00}{r\in\Bbb Z}.$

Answer 2

If $x$ is real then $-1 = x(x^2+1) $ so $x = \frac{-1}{x^2+1} $ so $-1 < x < 0$.

If $x = -c/d$ with $(c, d) = 1$, then $\frac{c}{d} = \frac{1}{(c/d)^2+1} =\frac{d^2}{c^2+d^2} $ or $c(c^2+d^2) = d^3$.

If a prime $p$ divides $c$, then $p | d^3$ so $p | d$, which contradicts $(c, d) = 1$.

Therefore $c = 1$, so $1+d^2 = d^3$ or $1 =d^3-d^2 =d^2(d-1) $ which can not hold since it is false for $d = 1$ and $d^2(d-1) > 1$ for $d \ge 2$.

Therefore, there is no rational root.

Answer 3

Gotta hand it to Bill Dubuque.

Having said this:

Before I read Bill's enlightening answer, and in attempting to steer clear of RRT territory as much as possible, I argued as follows:

As we have seen, with

$x^3 + x + 1 = 0, \tag 1$

and

$x = \dfrac{p}{q}, \; p, q \in \Bbb Z, \; \gcd(p,q) = 1, \tag 2$

we have

$\dfrac{p^3}{q^3} + \dfrac{p}{q}+ 1 = 0; \tag 3$

which as has been seen leads directly to (upon multiplication by $q^3$)

$p^3 + pq^2 + q^3 = 0; \tag 4$

since $\gcd(p, q) = 1$ we may find $a, b \in \Bbb Z$ such that

$ap + bq = 1; \tag 5$

we multiply by $p^2$:

$ap^3 + bqp^2 = p^2; \tag 6$

from (4),

$q \mid p^3; \tag 7$

then from (6),

$q \mid p^2; \tag 8$

again from (5), this time multiplying by $p$,

$ap^2 + bpq = p; \tag 9$

thus

$q \mid p; \tag{10}$

so again by virtue of $\gcd(p, q) = 1$:

$q = \pm 1, \tag{11}$

then

$p^3 + p \pm 1 = 0, \tag{12}$

whence

$p(p^2 + 1) = \pm 1; \tag{13}$

well,

$p = -1, 0, 1 \tag{14}$

don't solve (13), and if

$\vert p \vert \ge 2, \tag{15}$

then

$p^2 + 1 \ge 5,\tag{16}$

which rules out $p$ as in (15). Thus (13) has no integer roots, and hence (1) has no rational roots.