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Possibly a very straightforward question, but:

Question. For which complex numbers $\alpha$ and $\beta$ is it true that $\alpha^n+\beta^n$ is always an integer for all $n=1,2,3\ldots$?

For example, $$\alpha = \frac{1+i\sqrt{7}}{2}, \beta = \frac{1-i\sqrt{7}}{2}$$ have this relationship.

A couple of remarks. Firstly, a way of finding such $\alpha$ and $\beta$ pairs show's up in Silverman's book "The Arithmetic of Elliptic Curves." In particular:

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Secondly, something similar seems to occur in connection with the Fibonacci numbers. Following this line of thought, perhaps a better question would be: for which complex numbers $\alpha$ and $\beta$ does there exist a complex number $k$ such that $$\frac{\alpha^n+\beta^n}{k}$$ is always an integer?

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goblin GONE
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  • It reminds me somehow on a book called "Algebraische Zahlentheorie" (Algebraic Number Theory), which is opened with a question and whose impressive given proof theoretical answer involves something what you ask. Maybe a take a look in this book later.. – hänsel Dec 22 '17 at 10:34
  • See https://math.stackexchange.com/questions/936479/proving-that-frac-phi4001-phi200-is-an-integer and https://math.stackexchange.com/questions/2572494/prove-frac1i-sqrt724-frac1-i-sqrt724-1 – lab bhattacharjee Dec 22 '17 at 11:00
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    $T^2-aT+q$ is a monic quadratic polynomial $\in \mathbb{Z}[T]$. – reuns Dec 22 '17 at 11:29

1 Answers1

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This is certainly true if $\alpha$ and $\beta$ are conjugate quadratic integers because $\alpha^n+\beta^n$ is a symmetric function of $\alpha$ and $\beta$ and so is an integer polynomial expression in $\alpha+\beta$ and $\alpha\beta$.

Conversely, if $\alpha+\beta$ and $\alpha^2+\beta^2$ are integers, so is $2\alpha\beta$. Therefore, $\alpha$ and $\beta$ are roots of a polynomial $x^2+ax+\frac{b}{2}$ with $a,b\in\mathbb Z$, and so are definitely conjugate quadratic numbers, though perhaps not necessarily quadratic integers.

Now, by the same argument, $\alpha^2+\beta^2$ and $\alpha^4+\beta^4$ are integers implies $2\alpha^2\beta^2$ is an integer, that is, $2(\frac{b}{2})^2=\frac{b^2}{2}$ is an integer. Therefore, $b$ is even.

Bottom line: $\alpha^n+\beta^n$ is an integer for all $n$ iff $\alpha$ and $\beta$ are conjugate quadratic integers.

lhf
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    Put $u_n=\alpha^n+\beta^n$. You have $u_{n+1}=-au_n-\frac b2 u_{n-1}$ so if $b$ is odd the $2$-content of the $u_n$ will decrease - integers can persist for arbitrarily long periods, but not for ever. – Mark Bennet Dec 22 '17 at 11:09
  • @MarkBennet, what does $2$-content mean in this context? – goblin GONE Dec 23 '17 at 00:41
  • Can you clarify what you mean by the terms 'quadratic number' and 'quadratic integer'? I think a quadratic integer is a solution in $\mathbb{C}$ to an equation of the form $x^2+ax+b$ where $a$ and $b$ are ordinary integers, but I'm not sure what quadratic numbers are. – goblin GONE Dec 23 '17 at 00:58
  • The $2$-content of $u_n$ is the highest power of $2$ which divides $u_n$ – Mark Bennet Dec 23 '17 at 08:14
  • @MarkBennet, thanks. It was surprisingly hard to Google for! – goblin GONE Dec 26 '17 at 10:24