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Compute the least positive integer $n$ for which $(-\sqrt{2}+i\sqrt{6})^n$ will be an integer, where $i$ is the imaginary unit.

I did the binomial expansion and just plugged in numbers for $n$ starting from $1$ to see any pattern. I coudn't find any pattern but I eventually solved the problem to be $n=6$, but is there any easier more practical approach to this problem?

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I've added LaTeX formatting to your question; did I interpret your meaning correctly? – Zev Chonoles Dec 1 '11 at 6:09
Yes it is perfectly portrayed. Thanks Zev. :) – Kelly Rocks Dec 1 '11 at 6:17
up vote 4 down vote accepted

$$ \begin{align} (-\sqrt{2} + i \sqrt{6})^n & = \left(\sqrt{2} \left(-1 + i \sqrt{3} \right)\right)^n & = 2^{n/2} \left(-1 + i \sqrt{3} \right)^n\\ & = 2^{n/2} \left( 2 \left( - \frac12 + i \frac{\sqrt{3}}{2} \right) \right)^n & = 2^{n/2} 2^n \left( - \frac12 + i \frac{\sqrt{3}}{2} \right)^n \\ & = 2^{3n/2} \left( - \frac12 + i \frac{\sqrt{3}}{2} \right)^n & = 2^{3n/2} \left(\cos \left(\frac{2 \pi}{3} \right) + i \sin \left(\frac{2 \pi}{3} \right)\right)^n \end{align} $$ $$(-\sqrt{2} + i \sqrt{6})^n = 2^{3n/2} \left(\cos \left(\frac{2 n\pi}{3} \right) + i \sin \left(\frac{2 n\pi}{3} \right)\right)$$ For $2^{3n/2}$ to be an integer, we need $\frac{3n}{2} \in \mathbb{Z}$. We also need $\sin \left(\frac{2 n\pi}{3} \right) = 0$. This gives us $ \frac{2n}{3} \in \mathbb{Z}$.

Hence, $n = \frac{3k_1}{2} = \frac{2k_2}{3}$ where $k_1,k_2 \in \mathbb{Z}$. Hence, $n = 6k$ where $k \in \mathbb{Z}$.

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This is my favorite explanation out of the six I recieved for this questiow on. But what I dont understand how did you get from radical 2 to 2^3n/2? – Kelly Rocks Dec 1 '11 at 6:50
@Kelly: $(\sqrt2)^n=2^{n/2}$. Sivaram proceeded to remove a factor of $2$ from $-1+i\sqrt 3$: $-1+i\sqrt 3=2\left(\frac{-1+i\sqrt 3}{2}\right)$. That $2$, raised to $n$, gives $2^n$. Finally, one applies the laws of exponents: $(2^{n/2})(2^n)=2^{n+n/2}=2^{3n/2}$. – J. M. Dec 1 '11 at 7:22

A somewhat more systematic way to do this problem is to write the number you're taking powers of in polar form, as

$$ (-\sqrt{2} + i \sqrt{6}) = \sqrt{8} {-1 + i \sqrt{3} \over 2} = \sqrt{8} \left( {\cos {2\pi \over 3}} + i {\sin {2\pi \over 3}} \right) = \sqrt{8} e^{-2i \pi/3}$$.

Therefore $$ (-\sqrt{2} + i \sqrt{6})^n = 8^{n/2} e^{-2i \pi/3 \times n}. $$ For this to be an integer, we must have that $8^{n/2}$ is an integer and that $2 \pi/3 \times n$ is a multiple of $\pi$ (therefore making the whole thing a real number). The first condition is satisfied if $n$ is even; the second is satisfied if $n$ is a multiple of 3. In particular $n = 6$ is the smallest $n$ that will work, and

$$ (-\sqrt{2} + i \sqrt{6})^6 = 8^3 e^{-2i\pi} = 8^3.$$

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You have swapped cos and sin: $e^{i\theta} = \cos \theta + i\sin\theta$. – TonyK Dec 1 '11 at 8:20
You're right! I've fixed it. – Michael Lugo Dec 1 '11 at 14:17

By the time you got to $n=3$ and got something like $16\sqrt2$ I would have thought it would be clear that $n=6$ was going to work, and that $n=4$ and $n=5$ couldn't lead to real, much less integer, values.

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Let $z=-\sqrt{2}+i\sqrt{6}$. Observe that $|z|=2\sqrt{2}$, so $z$ can be written as $2\sqrt{2}e^{i\theta}$ for some value of $\theta$. What is that value (called the argument)? Once you find the argument of $z$, computing powers like $z^n$ is much easier, in fact, $z^n=|z|^ne^{in\theta}$. If you're familiar with polar notation for complex numbers, it should become clear what values of $n$ will yield an integer value.

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Do notice that $z = \sqrt2 + i \sqrt6 = \frac{1}{\sqrt2} (\frac{1}{2} + i \frac{\sqrt 3}{2}) = \frac{1}{\sqrt 2}\operatorname{cis}(\frac{\pi}{3})$. Therefore, $$z^3 = (\tfrac{1}{\sqrt2})^3\operatorname{cis}(\pi) = \frac{1}{\sqrt8}(\cos(\pi) + i \sin (\pi)) = \frac{-1}{\sqrt 8}.$$ And you can easily check that for every $n<3$, $z$ is still a complex number. Therefore, $z^6$ is an integer.

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How do you know that $\frac{1}{2}+i\frac{\sqrt{3}}{2}=\operatorname{cis}(\frac{\pi}{3})$? – Kelly Rocks Dec 1 '11 at 6:37
@Kelly: You kind of try to remember the sines and cosines of common angles. Remember the 30-60-90 triangle? – J. M. Dec 1 '11 at 7:19

HINT $\ $ If $\rm\ x,\:y,\:xy\:$ all $\not\in \mathbb Q\:$ and $\rm\:x^2\!,\ y^3\in\mathbb Q\:$ then $\rm\:(xy)^n\in \mathbb Q\ \iff\ 6\ |\ n\ \ $ (put $\rm\:x = 2\sqrt{2},\ y = \zeta_3$).

NOTE $\ $ In $\:\mathbb C/\mathbb Q\:$ this amounts to $\rm\:xy\:$ generates $\rm\:\langle x,y\ |\ x^2 = y^3 = 1\rangle\ \cong\: \mathbb Z/2 \times \mathbb Z/3\ \cong\mathbb Z/6$

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