Why $\mathrm e^{\sqrt{27}\pi } $ is almost an integer?

Question

Is there any reason why the expression $\mathrm e^{\sqrt{27}\pi } $ is almost an integer ??

$ e^{\sqrt{27}\pi }=12288743.98 4 $

Is there an infinite set of numbers with integers 'a' and 'b' so

$ e^{\pi \sqrt{a}}= b+c $

and 'c' is a real number very close to 1 , for example c=0.99....

Does this method work with other numbers? For example given a (positive) real number d we have that $ d^{\pi \sqrt{a}} $ is an integer for certain values of 'a'

Answer 1

There is a set of integers $n$ called the Heegner numbers which give almost-integers when you take $e^{\pi\sqrt x}$.

More specifically, these numbers are square-free $n$ such that the imaginary quadratic field $\mathbb Q[\sqrt{-n}]$ has class number $1$. It can be shown from this property that $e^{\pi\sqrt{n}}$ becomes very close to an integer for large enough $n$.

Unfortunately, there are only finitely many Heegner numbers ($1,2, 3, 7, 11, 19, 43, 67, 163$), and only the last few are 'large enough'. $e^{\pi\sqrt1}$, for example, is $23.14$, hardly an 'almost-integer'. However, $e^{\pi\sqrt{163}}=262 537 412 640 768 743.99999999999925...$, so the numbers don't have to be that large to give numbers that are extremely close to integers.

$27$ is not a Heegner number, but you're naturally going to end up getting other numbers $m$ such that $e^{\pi\sqrt m}$ is close to being an integer. I would guess that the fractional part of $e^{\pi\sqrt m}$ is distributed more or less randomly: then, with probability $1$, and for any $\varepsilon>0$ there will be infinitely many integers $m$ such that $e^{\pi\sqrt m}$ is within $\varepsilon$ of an integer.

Answer 2

Yes, there is a reason. The clue should have been the famous $\color{blue}{744}$ hiding at the side of $x=12288\color{blue}{743.98}.$

There are 9 fundamental discriminants $d$ with class number $h(-d)=1$, but there are 4 non-fundamental ones. Hence, the j-function is also an integer for these, namely,

$$\begin{aligned} j(\sqrt{-3}) &= 2\cdot 30^3\\ j(\sqrt{-4}) &= 66^3\\ j(\sqrt{-7}) &= 255^3\\ j\big(\tfrac{1+\sqrt{-27}}{2}\big) &= -3\times 160^3\\ \end{aligned}$$

Thus, the last one explains your question on,

$$e^{\pi\sqrt{27}} \approx 12288743.984 \approx3\times160^3+743.984$$

You can calculate the j-function using this WolframAlpha command. There are of course other discriminants $d$. For example, the prime-generating polynomial,

$$P(n) =6n^2+6n+31$$

is prime for $n = 0\; \text{to}\; 28$. It has discriminant $d = b^2-4ac = -708 = -4\times177$. And a quick check with Class Numbers shows it has $h(-708) = 4.$ And we have,

$$e^{(\pi/6)\sqrt{708}}= 1060^2+9.999929\dots$$

The fact that the “excess” is close to 10 (and not 744) means the integer part does not involve the j-function, but a related function. There are many others. See also “Prime-Generating Polynomials”.