Finding the 23th digit after decimal point of $(8+3\sqrt{7})^{2023}$

Question

What is the $23th$ digit after decimal point of $(8+3\sqrt{7})^{2023}$ ?

I tried to solve it with help of complex numbers but it doesn't seem to go any further.

=$(8-3\sqrt{7}i^2)^{2023}$

=$((2\sqrt{2}-\sqrt{3\sqrt{7}}i)(2\sqrt{2}+\sqrt{3\sqrt{7}}i))^{2023}$

=$(2\sqrt{2}-\sqrt{3\sqrt{7}}i)^{2023}$$(2\sqrt{2}+\sqrt{3\sqrt{7}}i)^{2023}$

Answer 1

Methodology adopted from Thomas Andrews's answer here.

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$(8+3\sqrt 7)^{2023} + (8-3\sqrt 7)^{2023}$ is an integer, and it can be easily proven using the binomial theorem. Note that $$0<8-3\sqrt 7 = \frac 1{8+3\sqrt7}< \frac1{8+7} = \frac1 {15} \approx 0.0666667$$ So, $$(8-3\sqrt 7)^{2023} < \left(\frac{0.666667}{10}\right)^{2023} = \frac{0.666667^{2023}}{10^{2023}} < \frac{1}{10^{2023}}$$ We get that $(8-3\sqrt 7)^{2023}$ must have at least $2023$ zeroes in its decimal expansion.

$$(8+3\sqrt 7)^{2023} + 0.\overbrace{\text{0000000}\ldots}^{\text{at least 2023 0's}} = \text{integer}$$ Thus $$\boxed{(8+3\sqrt 7)^{2023} = \text{number.}\underbrace{\text{9999999}\ldots}_{\text{at least 2023 9's}}}$$ The $23$rd digit after the decimal point is certainly $9$.

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As Jean-Claude Arbaut mentioned in the comments, $(8+3\sqrt7)$ is a Pisot–Vijayaraghavan number. Its minimal polynomial is $x^2 - 16x +1$ since it is the monic polynomial of lowest degree that $(8+3\sqrt7)$ is root of. The other root, lies strictly inside the complex unit circle (or its absolute value is less than $1$), which makes $(8+3\sqrt7)$ a Pisot–Vijayaraghavan number. Quoting from the Wikipedia article:

When $x$ is a Pisot number, the $n$th powers of the other conjugates tend to $0$ as $n$ tends to infinity. Since the sum is an integer, the distance from $x^n$ to the nearest integer tends to $0$ at an exponential rate.

In fact, the number on the linked answer by Thomas is also a PV number.