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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'

jimjim
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Jose Garcia
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2 Answers2

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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”.

Tito Piezas III
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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.

John Gowers
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    Could something like that be concluded when the quadratic field has class number 2? $\hspace{1.61 in}$ –  Aug 19 '13 at 07:42
  • Complex multiplication and $j$ appear to explain why Heegner numbers make $e^{\pi\sqrt{n}}$ 'close' to being an integer, but out of curiosity does it also explain why/if they stand out among other naturals in this regard? Is there a "correct" "way of measuring" the deviation of $e^{\pi\sqrt{n}}$ from being an integer that provably makes Heegner numbers distinguished among naturals? – anon Aug 19 '13 at 12:17
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    They sort of aren't. For example, the 'best' (in the obvious sense) $m$ below $50$ is $37$: $e^{\pi\sqrt{37}}=199148647.9999780$, while the Heegner number $43$ only gives $e^{\pi\sqrt{43}}=884736743.999777$. Of course, $e^{\pi\sqrt{163}}$ is extremely close to an integer, and I would be surprised if there were any 'better' numbers of comparable magnitude. However, I would be more surprised to find out that there weren't any 'better' (albeit much larger) numbers, as my probabilistic argument demonstrates. – John Gowers Aug 19 '13 at 13:32
  • The special thing about the Heegner numbers is that there is a very good reason why they yield near-integers in this way, while other near-integers that arise like this may be coincidences (insofar as a coincidence ever exists in mathematics). In fact, there are many similar near-identities holding for non-Heegner numbers (see mathoverflow.net/questions/30787/why-is-exppisqrt232-an-almost-integer, for example). I suppose that with the Heegner numbers, you can argue that the fractional part drops off more quickly than for any other class, but there are only finitely many, so that's academic. – John Gowers Aug 19 '13 at 13:39
  • @john Gowers The number $37$ has the property that $2n^2+2n+19$ whose discriminant is $-4×37$ generates only primes until we hit the obvious multiple of $19$ (prime for $0\le n\le 17$). This indicates that a modified version of $\mathbb{Z}(\sqrt{-37})$ in which we add in the "body centers" of the rectangular lattice will have unique factorization (but not closure under multiplication, of course). How does this relate to $\exp(\pi\sqrt{37})$ being a near-integer? – Oscar Lanzi Mar 06 '19 at 17:57