MHB Solve Integer & Inequality: $x=(x-1)^3$ for $N$

[anemone]

Let $x$ be a real number such that $x=(x-1)^3$. Show that there exists an integer $N$ such that $-2^{1000}<x^{2021}-N<2^{-1000}$.

 

[Opalg]

Let $x$ be a real number such that $x=(x-1)^3$. Show that there exists an integer $N$ such that $-2^{{\color{red}-}1000}<x^{2021}-N<2^{-1000}$.

Spoiler

We want to find $N$ such that $|x^{2021} - N| < 2^{-1000}$.

Apart from the real root $x\approx 2.3247$, the cubic equation $z = (z-1)^3$ has a pair of conjugate complex solutions, say $\alpha$ and $\overline{\alpha}$. Write the equation as $z^3 - 3z^2 + 2z - 1=0$ to see that $x\alpha\overline{\alpha} = 1$. Since $x>2$ it follows that $|\alpha|^2 = \alpha\overline{\alpha} < \frac12$ and so $|\alpha|<2^{-1/2}$.

For $n\geqslant 1$ let $p_n$ be the sum of the $n$th powers of the roots: $p_n = x^n+\alpha^n + {\overline{\alpha}}^n$. By Newton's identities, $p_1 = 3$, $p_2 = 5$, $p_3 = 12$ and for $n\geqslant4$ $p_n =3 p_{n-1} - 2p_{n-2} + p_{n-3}$. By an easy induction argument, $p_n$ is an integer for all $n$.

Since $p_n = x^n+\alpha^n + {\overline{\alpha}}^n$, it follows that $|x^n-p_n| = |\alpha^n + \overline{\alpha}^n| \leqslant 2|\alpha|^n <2^{1- (n/2)}$.

Now let $N = p_{2021}$ to see that $|x^{2021} - N| < 2^{-1009.5} < 2^{-1000}$.