# 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?

• See math.stackexchange.com/questions/533166/… your case is similar, though not exactly the same, but still $0\leq 2-\sqrt{3} < 1$ and you can take $(2-\sqrt{3})^{50} + (2+\sqrt{3})^{50}$ which is an integer. Nov 11, 2013 at 17:16
• At first I was thinking this was likely a case of floating point problems, but this is actually pretty neat. Nov 11, 2013 at 23:15
• Guys it was my preposition ...a general case Nov 13, 2013 at 5:08

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

• I don't quite see what $\frac1{2+\sqrt3}=2-\sqrt3$ is used for, other than to get $0<2-\sqrt3<\frac13$, which isn't too hard to see directly. Nov 11, 2013 at 17:50
• How about a generalization: Let $x=(a+\sqrt{b})^{n} + (a-\sqrt{b})^{n}$ with n going to infinity. Does the conjecture still hold true for $(a+\sqrt{b})^{n}$ to be almost an integer? Nov 11, 2013 at 17:55
• I think so, as long as $|a-\sqrt{b}| < 1$. Nov 11, 2013 at 17:59
• @imranfat Yes, a number $x$ for which $x^n$ gets close to an integer is called a Pisot number. The simple characterization, due Pisot, is that all algebraic conjugates have to be strictly less than $1$ in absolute value. The conjugate of $a+\sqrt{b}$ is exactly $a-\sqrt{b}$, but one can extend this result to numbers of the form $a+\sqrt[3]{b}$... Nov 11, 2013 at 18:24
• Now do $e^\pi-\pi = 19.99909997\ldots$, please!
– kba
Nov 12, 2013 at 0:38

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.

• Indeed, like the time I spent in high school raising $\varphi$ to a power to get near-integers and conjecturing they were all primes :p Nov 11, 2013 at 23:12

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 $y>1$.

• This is actually very nice.Well done Feb 15, 2015 at 10:28