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The $n$th metallic ratio $R_n$ is defined as,

$$R_n = \frac{n+\sqrt{n^2+4}}2$$

only the first ten of which listed by Wikipedia, with the first as $R_1=\phi=\frac{1+\sqrt{5}}2$. While making this recent question about the $17$-gon, I noticed something "odd".

I. Data

In the list of metallic means, there were odd powers of the golden ratio $\phi=\frac{1+\sqrt{5}}2$,

$$R_1 = \phi\;\\ R_4 = \phi^3\\ R_{11} = \phi^5\\ R_{29} = \phi^7$$

For the silver ratio $S = \frac{2+\sqrt{8}}2 = 1+\sqrt{2}$,

$$R_2 = S\;\\ R_{14} = S^3\\ R_{82} = S^5\\ R_{478} = S^7$$

For the bronze ratio $B=\frac{3+\sqrt{13}}2$,

$$R_3 = B\;\\ R_{36} = B^3\\ R_{396} = B^5\\ R_{4287} = B^7$$

and so on. The subscripts were the sequences,

\begin{align} x_1 &= 1, 4, 11, 29,\dots\\ x_2 &= 2, 14, 82, 478,\dots\\ x_3 &= 3, 36, 393, 4287,\dots\\ x_5 &= 5, 140, 3775, 101785,\dots \end{align}

namely A002878, A077444, A259131, respectively, and the $x$-values of the Pell-type equation,

$$x_n^2-(n^2+4)y_n^2=-4$$

Note: For $n=5$, I am familiar with the $5$th metallic ratio $R_5 = \frac{5+\sqrt{29}}2$ since it has a crucial role in Ramanujan's well-known 1/pi formula, but just noticed $x^2-29y^2 = -4$ has $x$-values all divisible by $5$.

II. Question

So is it true that given the $n$th metallic ratio,

$$R_n = \frac{n+\sqrt{n^2+4}}2$$

then its odd powers $k$,

$$(R_n)^{k} = R_m = \frac{m+\sqrt{m^2+4}}2$$

also belong to the sequence of metallic ratios for some integer $m$? For example, $(R_1)^3 = R_4,\, (R_1)^5 = R_{11}$, and so on. Moreover, why do only odd powers show this property?

Tito Piezas III
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1 Answers1

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Yes! This is true. Here's a proof.

For ease of notations, let $X=R_n$ and $Y=-1/R_n$. Then note that $X+Y=n$ and $XY=-1$ and hence $R_n$ is a root of $x^2-nx-1$. Now consider $$Z_m=X^m+Y^m$$ Since $X^mY^m=(XY)^m=(-1)^m$, letting $m$ be odd gives that $X^mY^m=-1$. Then $X^m$ is a root of $x^2-Z_mx-1$. If we can show that $Z_m$ is an integer, we are done! In fact, we will give an explicit way to find $Z_m$. Note that \begin{align*}Z_m&=X^m+Y^m\\&=(X+Y)(X^{m-1}+Y^{m-1})-XY(X^{m-2}+Y^{m-2})\\&=nZ_{m-1}+Z_{m-2}\end{align*} with the base cases $Z_0=X^0+Y^0=2$ and $Z_1=X+Y=n$.

Thus $R_n^m=R_{Z_m}$ where $m$ is odd and this finishes the proof.

HackR
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    Clever! No wonder it only works for odd powers. You let $z_1=X^m$ and $z_2 = Y^m$ and formed the quadratic $(z-X^m)(z-Y^m) = z^2-Z_mz-1 = 0$. I believe you made two typos though. It should be $R_n$ is a root of $x^2-n\color{blue}x-1=0$ and $Z_m$ is a root of $x^2-Z_m\color{blue}x-1=0$. – Tito Piezas III Oct 05 '24 at 08:02
  • Ah yes, silly typos. Thanks! – HackR Oct 05 '24 at 08:11
  • Perhaps I am not seeing something, but you write "Then $Z_m$ is a root of $x^2-Z_mx-1$. But how can this be? Said polynomial at $x=Z_m$ is $-1$ not $0$. What am I missing? – Zima Oct 05 '24 at 11:50
  • @Zima Well-spotted. It should say $X^m$ is a root. – J.G. Oct 05 '24 at 13:31
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    @Zima Thanks! I have corrected it. – HackR Oct 05 '24 at 13:39