This answer pertains to Michael's comment above.

Let $c$ be $e^{\pi\sqrt{163}}$. It is well-known that $c$ is close to an integer. More precisely, $c$ is about $10^{-12}$ from an integer of size about $10^{17}$, so (squaring $n+\epsilon$) there's no reason for $c^2$ to be close to an integer (just from these facts alone). But it is, and here's a cheap reason why.

Set $q=e^{2\pi i\tau}$ with $\tau=(1+\sqrt{-163})/2$. Then we know from general theory that $j(\tau)=1/q+744+196884q+\ldots$ is an integer, and hence (because $q$ is tiny) that $-1/q$ (which is Ramanujan's constant) is also close to an integer.

But just take this result $1/q+744+196884q+\ldots\in\mathbb{Z}$ and square it. We get $1/q^2+1488/q+ 947304 + 335950912q +\ldots\in\mathbb{Z}$. Now we know $q$ is tiny, and we know $1488/q$ is close to being an integer (but not as close as $q$ is), hence $1/q^2$ is close to being an integer (but not as close as $1/q$ is).

Now rinse and repeat.

But unfortunately this cheap argument doesn't seem to show that $c^5$ is close to an integer, because the power series expansion of $j^5$ involves terms like $1410274829033621720q$, and $q$ is about $10^{-17}$ but this huge integer multiple of $q$ isn't much less than 1 any more so this cheap approach breaks down.

Maybe it's better to do the job properly, and think about what explicit class field theory says about $j(5\tau)$.

sqrt(163))^5 is within 10^(-5) of an integer. And this integer is not N^5 where N is the integer closest to exp(Pisqrt(163)). – Michael Lugo Nov 9 '09 at 17:08