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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}$

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  • $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$
    – José Carlos Santos
    Commented Apr 17, 2023 at 10:04
  • $\begingroup$ gmpy2 with precision 500000 suggests that the fractional part of the number starts with the digit '9' repeated more than 2400 times. Not sure if this result can be trusted. $\endgroup$
    – Gribouillis
    Commented Apr 17, 2023 at 10:18
  • $\begingroup$ @Gribouillis it's true I think. And it has got something to do with $8^2 - (3\sqrt 7)^2 = 1$ $\endgroup$
    – D S
    Commented Apr 17, 2023 at 10:20
  • $\begingroup$ @Gribouillis Yes, it's correct if there was explanation it would be great $\endgroup$
    – Anar
    Commented Apr 17, 2023 at 10:22
  • $\begingroup$ Can you find your answer using this question and the accepted answer @Anar ? $\endgroup$
    – D S
    Commented Apr 17, 2023 at 10:29

1 Answer 1

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Methodology adopted from Thomas Andrews's answer here.

$(8+3\sqrt 7)^{2023} + (8-3\sqrt 7)^{2023}$ 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 $9$.

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.

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