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consider $x = (2+\sqrt[]{3})^6$, $x=[x]+t$, where $[x]$ is the integer part of $x$, and $t$ is the 'non integer' part of $x$. find the value of $x(1-t)$

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    NOTE: $(2+\sqrt[]{3})^6=((2+\sqrt[]{3})^3)^2=(26+\sqrt[]{26^2-1})^2=(1351+\sqrt[]{1351^2-1})$2012-08-04

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Note that $(2+\sqrt{3})^6+(2-\sqrt{3})^6$ is an integer, indeed an even integer. For imagine expanding each term, using the Binomial Theorem. The terms involving odd powers of $\sqrt{3}$ cancel.

We have $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$. So $(2-\sqrt{3})^6=\frac{1}{x}$, and $(2+\sqrt{3})^6+(2-\sqrt{3})^6=x+\frac{1}{x}.$

Since $2-\sqrt{3}$ is between $0$ and $0.3$, the number $(2-\sqrt{3})^6$ is positive but close to $0$. So $x$ is close to but below the integer $x+\frac{1}{x}$, and therefore $\lfloor x\rfloor=x+\frac{1}{x}-1.$

Also, $t=x-(x+\frac{1}{x}-1)=1-\frac{1}{x}$, so $1-t=\frac{1}{x}$. It follows that $x(1-t)=(x)(1/x)=1$.

Remark: The result is obviously structural. In particular, the exponent $6$ is irrelevant. The same argument works with, for example, $(3+2\sqrt{2})^{99}$, and in many analogous situations.

A crucial role was played by the conjugate $2-\sqrt{3}$ of the number $2+\sqrt{3}$. This sort of thing happens very frequently.

The fact that medium sized powers of $2+\sqrt{3}$ are almost integers (but just a little bit smaller than an integer) is at first a little startling. It may be enlightening to use the calculator to compute a few powers. For similar but not identical behaviour, compute some powers of $2+\sqrt{5}$.

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    Exactly. It is all connected with solutions of the Pell equation $x^2-dy^2=\pm 1$.2012-08-04