<s>[INST] 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?
 [/INST] 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 $y>1$.
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