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Let $x = \cos(\pi/8) = \frac{1}{2} \sqrt{2+\sqrt{2}}$ and $y = \sin(\pi/8) = \frac{1}{2} \sqrt{2-\sqrt{2}}$. What is the lowest degree polynomial $p(x,y)$ with integer coefficients such that $p(x,y) = \frac{1}{\sqrt{2^n}}$?

(By degree I mean the joint degree in $x$ and $y$. For example, I am considering $p(x,y) = x^2 y$ to be degree 3.)

One can obtain $\frac{1}{\sqrt{2^n}}$ using a degree $2n$ polynomial, namely $p(x,y) = (x^2-y^2)^n$. I suspect this is optimal. I also have the feeling that this problem might be easily solved using some standard technique related to number fields but I don't know how to do it.

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    $\begingroup$ $xy=\frac{\sin(\pi/4)}{2}=1/\sqrt{2^3}$ so degree $\frac{2n}{3}$ suffices for $n$ a multiple of $3.$ Your remark that $x^2-y^2=1/\sqrt{2}$ shows a way to handle the other two cases. $\endgroup$ Oct 2, 2014 at 22:27
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    $\begingroup$ Since $\mathbb{Z}[2x]$ is the ring of integers in $K = \mathbb{Q}[x]$ and $2y \in \mathbb{Z}[2x]$, if $p(x,y) = \frac{1}{\sqrt{2^{n}}}$, then $2^{{\rm \deg}(p)} p(x,y) \in \mathbb{Z}[2x]$ and this implies that $\deg(p) \geq \lceil \frac{n}{2} \rceil$, since $\frac{1}{\sqrt{2}} \not\in \mathbb{Z}[2x]$. $\endgroup$ Oct 2, 2014 at 22:44
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    $\begingroup$ From the observation that $2^{\lceil \frac{3j}{2}\rceil + \lceil \frac{3}{2}\lfloor\frac{i-j}{2}\rfloor\rceil + (i-j \bmod{2})} x^iy^j\in \mathbb{Z}[2x]$ if $i\geq j$ (similarly for the reverse --- to see this write $x^iy^j = (xy)^j\cdot (x^2)^{\lfloor \frac{i-j}{2}\rfloor}\cdot x^{(i-j \bmod{2})}$ and use $xy = 2^{-3/2}, x^2 = 2^{-1} + 2^{-3/2}, y^2 = 2^{-1} - 2^{-3/2}$), and that $\lceil \frac{3j}{2}\rceil + \lceil \frac{3}{2}\lfloor\frac{i-j}{2}\rfloor\rceil + (i-j \bmod{2}) = \frac{3}{4}(i+j) + O(1)$, it follows that, by Jeremy Rouse's argument, $\frac{n}{2}\leq \frac{3}{4}\deg{p} + O(1)$. $\endgroup$
    – alpoge
    Oct 3, 2014 at 0:02
  • $\begingroup$ Thanks for all three comments! I would say that, collectively, they fully answer my question. $\endgroup$
    – StephenJ
    Oct 3, 2014 at 0:50

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