# Why is $e^{\pi \sqrt{163}}$ almost an integer?

The fact that Ramanujan's Constant $e^{\pi \sqrt{163}}$ is almost an integer ($262 537 412 640 768 743.99999999999925...$) doesn't seem to be a coincidence, but has to do with the $163$ appearing in it. Can you explain why it's almost-but-not-quite an integer in layman's terms (I'm not a mathematician)?

-
mathoverflow.net/questions/30787 and mathoverflow.net/questions/4775 are relevant, but let's see somebody try to explain it without saying "modular". ;) –  Ｊ. Ｍ. Sep 13 '10 at 16:01
This is not really the kind of phenomenon with a layman's-terms explanation... –  Qiaochu Yuan Sep 13 '10 at 16:01
@Qiaochu, indeed! –  Mariano Suárez-Alvarez Sep 13 '10 at 16:02
I observed yet another interesting (IMHO) "almost an integer" related to exp(Pi*sqrt(163)) result, which is: sum((1/(exp(Pi*sqrt(163))^k))*(120/(8*k+1)-60/(8*k+4)-30/(8*k+5)-30/(8*k+6)), k = 0 ... infinity) = 94.000000000000000014789449792044364408558923807659819... Regards, Alexander R. Povolotsky –  Alex Jun 24 '12 at 15:12

I do not think "why" has a reasonable layman's-terms answer, but let me at least explain "how" with a simpler example of such a numerical coincidence. If one takes powers of the golden ratio $\phi = \frac{1 + \sqrt{5}}{2}$ it is not hard to see that they are close to integers. For example, $\phi^{20} = 15126.999934...$. One might ask an analogous question about why these numbers are close to integers. The answer is that

$$\phi^n + \varphi^n = L_n$$

where $L_n$ is the $n^{th}$ Lucas number (an integer), and where $\varphi = \frac{1 - \sqrt{5}}{2}$ has absolute value less than $1$. As $n$ gets larger, $\varphi^n$ gets smaller, so $\phi^n$ becomes a better and better approximation to the integer $L_n$.

A similar, but much more complicated, phenomenon is happening here. The reason $e^{\pi \sqrt{163}}$ is so close to an integer is that it, plus a small error term, gives a special formula for that integer. But "why" this formula exists is a rather long and complicated story (as Robin Chapman hints at) and I do not think there is any reasonable way to talk about it in layman's terms.

-
For a generalization of the above example, see en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number . –  Qiaochu Yuan Sep 13 '10 at 18:02
accepted as answer for explaining how some numbers approach integers, albeit with a completely different example. I understand Rumanujan's Constant is way beyond my understanding of mathematics. –  stevenvh Sep 14 '10 at 14:56
The above explanation is not correct. The approximation property of powers of the golden ratio are due to the fact that it is a Pisot number, i.e., it is > 1 and all its conjugates have absolute value < 1. While interesting, it has applications to Fourier analysis, the Salem-Zygmund theorem characterizing a class of sets of uniqueness, this is not at all what is going on here. –  ilan vardi Jul 17 '13 at 11:53
@ilan: I never claimed that it was. I claimed that it was similar (in that there is a formula for an integer which has a bulk term and a small error term). –  Qiaochu Yuan Jul 17 '13 at 19:08

This is quite a challenge to express in "layman's terms", but the reason is that $$j\left(\frac{1+\sqrt{-163}}{2}\right)$$ is an integer where $j$ is the $j$-function. When you substitute $(1+\sqrt{-163})/2$ into the $q$-expansion (see the wikipedia page) of $j$, all terms save the first two are small, and the first two equal $$-\exp(\pi\sqrt{163})+744.$$

The reason that this $j$-value is an integer is due to the quadratic field $\mathbb{Q}(\sqrt{-163})$ having class number one, or equivalently that all positive-definite integer binary quadratic forms of discriminant $-163$ are equivalent.

Added I'll try to explain the connection with binary quadratic forms. Consider a quadratic form $$Q(x,y)=ax^2+bxy+cy^2$$ with $a$, $b$ and $c$ integers. I'll only consider forms $Q$ which are primitive, so that $a$, $b$ and $c$ have no common factor $> 1$, and positive-definite, that is $a > 0$ and the discriminant $D=b^2-4ac < 0$. There is a notion of equivalence of quadratic forms, and two primitive positive-definite forms $Q$ and $Q'(x,y)=a'x^2+b'xy+c'y^2$ (necessarily also of discriminant $D$) are equivalent if and only if $$j\left(\frac{b+\sqrt{-D}}{2a}\right) =j\left(\frac{b'+\sqrt{-D}}{2a'}\right).$$ For each possible discriminant there are only finitely many equivalence classes. Thus we get a finite set of $j$-values for each discriminant, and the big theorem is that they are the solutions of a monic algebraic equation with integer coefficients. When there is only one class the equation has the form $x-k=0$ where $x$ is an integer, and the $j$-value must be an integer.

My recommended reference for this is David Cox's book Primes of the form $x^2+ny^2$. But these results appear towards the end of this 350-page book.

-
Related to this are the Heegner numbers: mathworld.wolfram.com/HeegnerNumber.html –  Ｊ. Ｍ. Sep 13 '10 at 16:15