# Why is $(2+\sqrt{3})^n$ nearly an integer?

Funny thing. Type $(2 + \sqrt{3})^{20}$ into Wolfram Alpha. (Or, if you’re really lazy, click this link.). It’s 274,758,382,273.999999999996 or so.

The higher the power you pick, the closer $(2 + \sqrt{3})^n$ gets to an integer value – although it never quite gets there, because $\sqrt{3}$ is irrational. So, how come?

The key is to think about $(2 - \sqrt{3})^n$.

If you expand both of these expressions binomially, you get:

$(2 + \sqrt{3})^n = 2^n + n \times 2^{n-1} \times \sqrt{3} + \frac{n(n-1)}{2}\times 2^{n-2}\times 3 + …$ $(2 - \sqrt{3})^n = 2^n - n \times 2^{n-1} \times \sqrt{3} + \frac{n(n-1)}{2}\times 2^{n-2}\times 3 - …$

If you add these together and call the result $S$. The odd terms – the ones with $\sqrt{3}$s in – all vanish, which means $S$ is an integer. How does that help? Well, $(2-\sqrt{3})$ is about 0.268, and when you take powers ((powers greater than 1, at least)) of a number between 0 and 1, it gets smaller. In particular, $(2-\sqrt{3})^{20}$ is very small, about $3.6 \times 10^{-12}$.

This means that $S - (2-\sqrt{3})^n$ is a tiny amount smaller than an integer – but, because of how $S$ is formed, that’s the same as $(2+\sqrt{3})^n$!

Edited 2015-07-10 to fix a footnote.