# Integers $n$ such that $i(i+1)(i+2) \cdots (i+n)$ is real or pure imaginary

A couple of days ago I happened to come across [1], where the curious fact that $i(i-1)(i-2)(i-3)=-10$ appears ($i$ is the imaginary unit). This led me to the following question:

Problem 1: Is $3$ the only positive integer value of $n$ such that $i(i-1)(i-2) \cdots (i-n)$ is a real number or a pure imaginary number? If not, can we describe all such integer values of $n$?

Initial Analysis: We can view $i(i-1)(i-2) \cdots (i-n)$ as the result of applying a finite sequence of operations to $i,$ each of which involves a radial stretch from the origin and a rotation about the origin. Specifically, $i$ is moved radially outward by a factor of $\sqrt{1^2 + 1^2}\sqrt{1^2 + 2^2} \cdots \sqrt{1^2 + n^2}$ and rotated counterclockwise by the angle $\arctan 1 + \arctan 2 + \cdots + \arctan n.$ Since we don’t care about the magnitude of the result, only whether we land on the real axis or the imaginary axis, Problem 1 reduces to asking whether $3$ is the only positive integer value of $n$ such that $\arctan 1 + \arctan 2 + \cdots \arctan n$ is an integer multiple of $\frac{\pi}{4}.$

One can also check that $i(i+1)(i+2)(i+3) = -10.$ This is not a coincidence. Since $\arctan 1 + \arctan 2 + \arctan 3$ is an integer multiple of $\frac{\pi}{4},$ it follows that $\arctan (-1) + \arctan (-2) + \arctan (-3) = -\left(\arctan 1 + \arctan 2 + \arctan 3 \right)$ must also be an integer multiple of $\frac{\pi}{4}.$

Other than the products $(i-3)(i-2)(i-1)(i)$ and $(i)(i+1)(i+2)(i+3)$ (and two other products obtained by omitting the $i$ factor from these two), and products of the form $(i-k)(i-k+1) \cdots (i+k-1)(i+k)$ where $k$ can be any positive integer, I don’t know any product of the form $(i+m)(i+m+1) \cdots (i+n)$ for integers $m$ and $n$ with $m < n$ that equals a real number or a pure imaginary number.

Problem 2: Are $(i-3)(i-2)(i-1)(i)$ and $(i)(i+1)(i+2)(i+3)$ (and two other products obtained by omitting the the $i$ factor from these two), and products of the form $(i-k)(i-k+1) \cdots (i+k-1)(i+k)$ where $k$ can be any positive integer, the only pairs of integers $(m,n)$ with $m < n$ such that $(i+m)(i+m+1) \cdots (i+n)$ is a real number or a pure imaginary number? If not, can we describe all such pairs of integers $(m,n)$?

Of course, it is easy to come up problems having a broader scope, such as replacing “$i+m$” with an arbitrary complex number whose real and imaginary parts are integers and/or using factors that increment the imaginary part by $1$ (or increment both the real and imaginary parts by $1$) and/or using factors that increment the real part (or the imaginary part, or both the real part and the imaginary part) by a constant integer amount, etc.

I suspect the answers to these problems can be obtained from a careful analysis of Carl Størmer’s 1890s results involving Machin-like formulas, but my knowledge of French and of this field of mathematics is rather poor. For those interested, I suggest looking at Størmer [2]. I also suspect there is a more direct way to solve Problem 1, and perhaps also Problem 2.

Personally, I am only moderately interested in this issue, but I am posting it because I thought others in this group might find this something interesting to pursue. In particular, if there is not a fairly trivial way to solve Problem 1 and someone manages to find a solution that isn’t extremely difficult, I suspect that such a solution would make for an interesting Math-Monthly type paper.

[1] Charles-Ange Laisant (1841–1920), Remarque sur une équation différentielle linéaire [Remark on a linear differential equation], Bulletin de la Société mathématique de France 23 (1895), 62-63.

[2] Fredrik Carl Mülertz Størmer [Störmer] (1874-1957), Sur l’application de la théorie des nombres entiers complexes à la solution en nombres rationnels $x_{1} \; x_{2} \; \dots \; x_{n} \; c_{1} \; c_{2} \; \dots \; c_{n}, \; k$ de l’équation: $c_{1} \text{arc tg}\, x_{1} + c_{2} \text{arc tg}\, x_{2} + \dots . + c_{n} \text{arc tg}\, x_{n} = k\frac{\pi}{4},$ [On an application of the theory of complex integers to the solution in rational numbers $x_{1} \; x_{2} \; \dots \; x_{n} \; c_{1} \; c_{2} \; \dots \; c_{n}, \; k$ of the equation: $c_{1} \arctan x_{1} + c_{2} \arctan x_{2} + \dots . + c_{n} \arctan x_{n} = k\frac{\pi}{4}$], Archiv for Mathematik og Naturvidenskab 19 #3 (1896), 95 + 1 (errata) pages.

UPDATE (30 December 2013): I have incorporated the comment benh made and I have made slight corrections to my Størmer paper citation, but otherwise I have left my original wording intact. I’m impressed with the variety of mathematical techniques brought up in the comments and solutions, especially the probabilistic analysis that Hagen von Eitzen gave.

I’m choosing KenWSmith’s answer because, more than anyone else, he is responsible for bringing to light a solution (to Problem 1): We observe that all products of the form $i(i+1)(i+2) \cdots (i+n)$ have the form $a+bi$ where both $a$ and $b$ are integers. Thus, if $a+bi$ is real or pure imaginary, we have $a=0$ or $b=0,$ and hence the modulus of $a+bi$ equals $b$ or $a,$ and hence the modulus of $a+bi$ will be an integer. On the other hand, the modulus of $a+bi$ also equals $\sqrt{1^2 + 1^2}\sqrt{1^2 + 2^2} \cdots \sqrt{1^2 + n^2}.$ Thus, Problem 1 is equivalent to finding all positive integer values of $n$ such that $\sqrt{1^2 + 1^2}\sqrt{1^2 + 2^2} \cdots \sqrt{1^2 + n^2}$ is an integer. Equivalently, find all positive integer values of $n$ such that $(1^2 + 1^2)(1^2 + 2^2) \cdots (1^2 + n^2)$ is the square of an integer.

KenWSmith then posted this last version in mathoverflow: When is the product (1+1)(1+4)…(1+n^2) a perfect square? On the same day Lucia supplied an answer by linking to a preprint version of a 2008 paper by Javier Cilleruelo [Journal of Number Theory 128 #8 (August 2008), 2488-2491], which was written solely to answer the question of whether $n=3$ is the only positive integer such that $(1^2 + 1^2)(1^2 + 2^2) \cdots (1^2 + n^2)$ is the square of an integer, which was conjectured and “partially verified” in the 2-month earlier paper by Amdeberhan/Medina/Moll [Journal of Number Theory 128 #6 (June 2008), 1807-1846].

Not an answer, but an equivalence…

As pointed out in the original question, $i(i-1)(i-2)…(i-n)$ landing on the real or imaginary axis implies $\arctan(1)+\arctan(2) + … + \arctan(n)$ is a multiple of $\pi/4$.
Since $i(i-1)(i-2)…(i-n)$ is a Gaussian integer, this also implies that its magnitude is an integer and so $(1+1)(1+4) \cdots (1+n^2)$ is a perfect square.

On the other hand, if $\arctan(1)+\arctan(2) + … + \arctan(n)$ is a multiple of $\pi/4$ then $i(i-1)(i-2)…(i-n)$ lands on the real or imaginary axis and so this statement about the sum of tangents is equivalent to the original question.

In a similar way, if $(1+1)(1+4) \cdots (1+n^2)$ is a perfect square then set $k := \sqrt{(1+1)(1+4) \cdots (1+n^2)}$ and note that the Gaussian integer $\frac{i(i-1)(i-2)…(i-n)}{k}$ has length 1. Since the only Guassian integers of length 1 are $\pm 1, \pm i$ then $i(i-1)(i-2)…(i-n)$ lies on the real or imaginary axis.

In other words, the following are equivalent.

1. $i(i-1)(i-2)…(i-n)$ lies on the real or imaginary axis,
2. $\arctan(1)+\arctan(2) + … + \arctan(n)$ is a multiple of $\pi/4$,
3. $(1+1)(1+4) \cdots (1+n^2)$ is a perfect square.