# Why is $(2+\sqrt{3})^{50}$ so close to an integer?

I just worked out $(2+\sqrt{3})^{50}$ on my computer and got the answer

$39571031999226139563162735373.999999999999999999999999999974728\cdots$

Why is this so close to an integer?

#### Solutions Collecting From Web of "Why is $(2+\sqrt{3})^{50}$ so close to an integer?"

Let $x=(2+\sqrt{3})^{50} + (2-\sqrt{3})^{50}$

$x$ is clearly an integer, since all terms with $\sqrt{3}$ cancel.

Notice that $0<2-\sqrt{3}<1$, so $0<(2-\sqrt{3})^{50}\ll1$

So $(2+\sqrt{3})^{50} =x-(2-\sqrt{3})^{50}\approx x$.

Numbers which have this property are called Pisot Vijayaraghavan numbers.

Pisot proved that $x$ has the property that for a “large” $n$ the numbers $x^n$ get very close to integers, if and only if, all the the algebraic conjugates of $x$ satisfy $|x’| <1$.

In this case, the only algebraic conjugate of $2+\sqrt{3}$ is $2-\sqrt{3}$, and since $0< 2-\sqrt{3} <1$ it follows that $2+\sqrt{3}$ is a Pisot number.

P.S. The Golden mean $\frac{1+\sqrt{5}}{2}$ also has this property, which leads to some interesting properties for Fibonacci.

Let $$x=2+\sqrt { 3 }, \\$$then$${ x }^{ 2 }=7+4\sqrt { 3 }, \\ 4x=8+4\sqrt { 3 } ,$$ therefore $${ x }^{ 2 }+1=4x,\\ x+\frac { 1 }{ x } =4.$$ Take the square of both side and repeat the process $${ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ 4 }^{ 2 }-2\\ { x }^{ 4 }+\frac { 1 }{ { x }^{ 4 } } ={ \left( { 4 }^{ 2 }-2 \right) }^{ 2 }-2\\ { x }^{ 8 }+\frac { 1 }{ { x }^{ 8 } } ={ \left( { \left( { 4 }^{ 2 }-2 \right) }^{ 2 }-2 \right) }^{ 2 }-2$$ You can see that $$\lim _{ n\rightarrow \infty }{ \frac { 1 }{ { x }^{ { 2 }^{ n } } } } =0,$$ because $x=2+\sqrt { 3 }>1.$ Although you can intuitively say $\frac { 1 }{ { x }^{ 50 } } \approx0,$ you can verify it by saying $x=y^{50/2^n}$, where $y>1$.