Show that a specific $w$ cannot be the root of an quadratic with integer coefficients. Let $w$ be the only real root of $x^3-x-1=0$. Show that
 $w$  cannot satisfy the quadratic $ax^2 + bx + c$ ,where $a,b,c\in \Bbb Z$.
I have written 
$$w^3=w+1$$
but I can't go any further than this. Thank you.
 A: Note that, by Euclidien division
$$x^3-x-1=Q(x)(ax^2+bx+c)+x \left(\frac{b^2}{a^2}-\frac{c}{a}-1\right)+\frac{b c}{a^2}-1$$
So, if $w$ is a root of both $x^3-x-1=0$ and $ax^2+bx+c=0~$, then we would have
$$
w \left(\frac{b^2}{a^2}-\frac{c}{a}-1\right)+\frac{b c}{a^2}-1=0
$$
So if $a^2+a c-b^2\ne0~$ then
$$
w=\frac{b c-a^2}{a^2+a c-b^2}\in\mathbb{Q}
$$
which is absurd since $x^3-x-1=0~$ has no rational solutions.
Now what happens if $a^2+a c-b^2 =0~$?
In this cas we must also have 
$a^2=bc$. Without loss of generality we may suppose that $\gcd(a,b,c)=1$. From 
$a^2=bc$ we conclude that $\gcd(b,c)=1$, because otherwise any common prime divisor would also divide $a$.
From $\gcd(b,c)=1$ and $a^2=bc$ we conclude that 
$$b=\epsilon\beta^2, c=\epsilon\gamma^2, a=\epsilon'\beta\gamma,\qquad\hbox{with $\gcd(\beta,\gamma)=1,~(\epsilon,\epsilon)'\in\{+1,-1\}^2$}$$
Replacing in $a^2+a c-b^2 =0$ we get $\gamma^2(\beta+\epsilon\epsilon'\gamma)=\beta^3$, so $\gamma|\beta^3$, but $\gcd(\beta,\gamma)=1$ so,
$\gamma=1$ and consequently $ \beta+\epsilon\epsilon'=\beta^3$. This implies that $\beta $ is an integer solution of  $x^3-x-1=0$ or $x^3-x+1=0$ which is clearly absurd. So, this case cannot happen.
A: $p(x)=x^3-x-1$ is an irreducible polynomial over $\mathbb{Q}$, since it is irreducible over $\mathbb{F}_2$ (since it is a third-degree polynomial, proving that it is irreducible over a finite field is equivalent to proving that it has no roots in the field. But $p(x)$ is a trinomial, so $p(0)=p(1)=1$ in $\mathbb{F}_2$) . This gives that any root of $p(x)$ is an algebraic number of degree $3$ over $\mathbb{Q}$, so it cannot be the root of any quadratic polynomial with integer coefficients.
A: Suppose that $w \in \mathbb{R}$ is a root of $x^{3} - x - 1 = 0$ and $ax^{2} + bx + c = 0$ with $a, b, c \in \mathbb{Q}$ and $a \neq 0 \neq c$. It is easy to establish that the only real root $w$ of $x^{3} - x - 1 = 0$ is irrational. Also Note that in this case $ax^{2} + bx + c = 0$ has two roots and these must be unequal otherwise $w$ will be rational. If both these roots are also the roots of $x^{3} - x - 1 = 0$ then it follows that $ax^{2} + bx + c$ divides $x^{3} - x - 1$ and thus the remaining factor must be linear with rational coefficients leading to a rational root of $x^{3} - x - 1 = 0$ which is not possible. It follows that both $x^{3} - x - 1$ and $ax^{2} + bx + c$ have only one common factor of type $(x - w)$. It therefore follows that GCD of these polynomials $ax^{2} + bx + c$ and $x^{3} - x - 1$ is a linear polynomial. The GCD of these two polynomials however must have rational coefficients which will imply that $w$ is rational.
The above contradiction shows that there can't be any polynomial $ax^{2} + bx + c$ with rational coefficients having $w$ as its root. The proof shows that if an irrational number $\alpha$ is the only real root of a cubic polynomial $P(x)$ with rational coefficients, then $P(x)$ is irreducible over rationals.
