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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?

user1510374
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    See http://math.stackexchange.com/questions/533166/my-conjecture-on-almost-integers/ 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. – Pablo Rotondo Nov 11 '13 at 17:16
  • At first I was thinking this was likely a case of floating point problems, but this is actually pretty neat. – AJMansfield Nov 11 '13 at 23:15
  • Guys it was my preposition ...a general case – Shivam Patel Nov 13 '13 at 05:08

3 Answers3

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

user85798
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    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. – Marc van Leeuwen Nov 11 '13 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? – imranfat Nov 11 '13 at 17:55
  • And how about complex number with an absolute value less than one? Just wondering – imranfat Nov 11 '13 at 17:58
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    I think so, as long as $|a-\sqrt{b}| < 1$. – user85798 Nov 11 '13 at 17:59
  • @imranfat What's more, the integers in question will be solutions to a Fibonacci-style recurrence relation with constant coefficients, of the form $x_{n+2} = 2ax_{n+1} - (a^2-b)x_n$, with suitable initial conditions. – Steven Stadnicki Nov 11 '13 at 18:24
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    @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 $a+\sqrt[3]{b}$... – N. S. Nov 11 '13 at 18:24
  • Seems like I could open a can of worms here on this one. Thanks for the replies – imranfat Nov 11 '13 at 18:26
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    Now do $e^\pi-\pi = 19.99909997\ldots$, please! – kba Nov 12 '13 at 00:38
  • I'm pretty sure that's a genuine coincidence! – user85798 Nov 12 '13 at 00:40
  • I want to give +10 for this!!! – N.S.JOHN May 13 '16 at 02:34
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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.

Community
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N. S.
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    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 – Thomas Nov 11 '13 at 23:12
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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 $y>1$.

newzad
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