# 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 $x(1-t)$

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: $(2+\sqrt3)=(2+\sqrt[]{2^2-1})$ – Rajesh K Singh Aug 4 '12 at 4:36
NOTE: $(2+\sqrt[]{3})^6=((2+\sqrt[]{3})^3)^2=(26+\sqrt[]{26^2-1})^2=(1351+\sqrt[]{1351‌​^2-1})$ – Rajesh K Singh Aug 4 '12 at 5:43

## 1 Answer

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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And I was working all the numbers out by hand. I got as far as recognizing that the integer part of (2+sqrt(3))^6=1351+[75660/56] if I didn't make any mistakes anywhere... Checking in Python, I see I was off by one. I thought Newton's method gave under-approximations for square roots but I was wrong, it gives over-approximations – dspyz Aug 4 '12 at 5:03
$(2+\sqrt[]{5})^2$=$(9+\sqrt[]{9^2-1})$ – Rajesh K Singh Aug 4 '12 at 6:46
$(2+\sqrt[]{5})^3=(38+\sqrt[]{38^2+1})$ – Rajesh K Singh Aug 4 '12 at 6:53
Note that in this case we are alternately a bit below an integer and a bit above an integer. That's basically because $2-\sqrt{5}$ is negative. – André Nicolas Aug 4 '12 at 6:57
$(2+\sqrt[]{5})=\frac{-1}{2-\sqrt[]{5}}$ – Rajesh K Singh Aug 4 '12 at 6:59