Exercise 2. Let $x$ be a real number such that $x^{3}=x+4$. Let $a, b, c, d$ be non-zero integers such that $a+b x^{2}+c x^{3}+d x^{4}=0$. Determine the value of $\frac{a c}{b d}$.
Solution to Exercise 2: By multiplying $x^{3}=x+4$ by $x$, we find that $x^{4}=x^{2}+4 x = x^{2}+4(x^{3}-4) = 4 x^{3}+x^{2}-16$, so $x^{4}-4 x^{3}-x^{2}+16=0$. For $(a, b, c, d)=(16,-1,-4,1)$, we have $\frac{a c}{b d}=64$.

(We prove that if $(a', b', c', d')$ is another quadruplet that works, then it is proportional to $(a, b, c, d)$. To do this, we first show that $X^{3}-X-4$ has no root in $\mathbb{Q}$, and thus is irreducible over $\mathbb{Z}$; then, $\left(0, a b'-a' b, a c'-a' c, a d'-a' d\right)$ also works, which provides a polynomial of degree at most 2 that annihilates $x$. This was not required in the statement, which implied that the ratio $a c /(b d)$ has a unique value.)