Why is $(\sqrt{2}+\sqrt{3})^{2008}$ so close to an integer?

Question

Using 5000-digit precision in PARI/GP, I discovered that the fractional part of $(\sqrt{2}+\sqrt{3})^{2008}$ is extremely small, less than $10^{-999}$. Is there a simple explanation for this fact ?

This looks like a Pisot number issue (similar questions have already been studied on MSE, see for example Why is $(2+\sqrt{3})^{50}$ so close to an integer?), but it’s a more complicated situation.

Related : Show that $(\sqrt{2} + \sqrt{3})^{2009}$ is rounded to an even number.

Answer 1

Your intuition is correct: $$ (\sqrt{2} + \sqrt{3})^{2008} = (5 + 2 \sqrt{6})^{1004} $$ and $5 + 2\sqrt{6}$ is a Pisot integer. Further, Newton's identities imply that $$ (5 + 2\sqrt{6})^n + (5 - 2\sqrt{6})^n \in \Bbb{Z} $$ for every positive integer $n$, thus the distance between $(5 + 2\sqrt{6})^{1004}$ and the closest integer is at most $$ (5 - 2\sqrt{6})^{1004} \approx 2.6743 \cdot 10^{-1000} $$

Answer 2

That is because

$$(\sqrt{3}+\sqrt{2})^{2m} + (\sqrt{3}-\sqrt{2})^{2m}$$

is an integer, namely

$$\begin{align} (\sqrt{3}+\sqrt{2})^{2m} + (\sqrt{3}-\sqrt{2})^{2m} = 2\sum_{k=0}^{m} \binom{2m}{2k} 3^{m-k}2^k, \end{align}$$

and $(\sqrt{3}-\sqrt{2})^{2m} <\frac{1}{3^{2m}}$.