(2+sqrt(3))^50 close to being an integer

[Ciaran]

Q:Explain this phenomenon by finding a sequence of integers a_i, defined by a recursion relation similar to the Fibonacci sequence, such that a_n = (2 + √
3)^n + (2 −√3)^n.
Make sure to explain why constructing such a sequence proves (2 + √3)^50 is close to an integer, though you needn’t analyze precisely how close it is.

I managed to come up with a proof of why the number is so close to an integer but I have not been able to explain it in the way the question is asking me. Can somebody please help?

 

[MarkFL]

To find the recursion having the given closed form, we should look for the minimal polynomial having the roots:

\(\displaystyle 2\pm\sqrt{3}\)

So, we could write:

\(\displaystyle x=2\pm\sqrt{3}\)

\(\displaystyle x-2=\pm\sqrt{3}\)

\(\displaystyle (x-2)^2=3\)

\(\displaystyle x^2-4x+4=3\)

\(\displaystyle x^2-4x+1=0\)

So, using this as your characteristic equation, and the initial values of the given closed form, can you construct the recursion?

 

[Ciaran]

Thanks for your reply! This was actually as far as I got, because I was told to look at the Golden Ratios and how they are the solutions to x^2= x+1, and apply this to 2 +- sqrt3

 

[MarkFL]

Okay, using the characteristic equation we found, along with the fact that:

\(\displaystyle a_0=2,\,a_1=4\)

we may state:

\(\displaystyle a_{n+1}=4a_{n}-a_{n-1}\)

Does this make sense?

 

[Ciaran]

Yes, this makes sense! It's analogous to the one for the Fibonacci sequence

 

[MarkFL]

Yes, the method for finding the closed form of a linear recurrence is to raise each of the roots of the corresponding characteristic equation by a power of $n$ and multiply each by a parameter which you can determine from the given initial conditions.

So, knowing that $a_n$ is an integer for all $n$ (which we can see from the recurrence), then we know that \(\displaystyle \left(2+\sqrt{3}\right)^n\) will differ from an integer by the value of \(\displaystyle \left(2-\sqrt{3}\right)^n\). Since $2-\sqrt{3}<1$, we know that as $n$ increases, this value will approach zero.

 

[Ciaran]

Thank you very much indeed! It was the forming of the recurrence relation that got me; you made it very clear indeed.