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].

  • 2
    $\begingroup$ Regarding problem 1: Looking at the relevant OEIS entries, it looks like the real and imaginary parts are strictly, rapidly, increasing in magnitude, so I'd be surprised if it is not hard to show that they are never zero again after $n=3$. oeis.org/A231530 oeis.org/A231531 $\endgroup$ Commented Dec 20, 2013 at 19:08
  • $\begingroup$ Related question regarding asymptotics: mathoverflow.net/questions/116336/… $\endgroup$ Commented Dec 20, 2013 at 19:09
  • 5
    $\begingroup$ Some heuristics and very rough eyeballing: As the absolute value of the product is $> n!$, the probability of landing less than $\frac12$ away from an axis on such a large circle is fairly low, $< \frac 4{2\pi n!}$. The sum of all these probabilities is $<\frac {2e}\pi$, so "less than two" solutions is about what we can expect. $\endgroup$ Commented Dec 20, 2013 at 19:21
  • 1
    $\begingroup$ This is the same proving that $\{0,1,3,4\}$ are the only natural solutions to the following equations: $$\Re{i\choose n}=0\qquad;\qquad\Im{i\choose n}=0\qquad;\qquad\Re{i+n\choose i}=0\qquad;\qquad\Im{i+n\choose i}=0.$$ $\endgroup$
    – Lucian
    Commented Dec 20, 2013 at 20:39
  • 1
    $\begingroup$ Little number theory notice: If $n\in \Bbb N$ is a solution to problem 1, then by conjugation we see that $\prod_{k=0}^n(i-k)=\pm \prod_{k=0}^n(i+k).$ Multiplying both sides by the left side gives $$\left( \prod_{k=0}^n(i-k) \right )^2 =\pm \prod_{k=0}^n(k^2+1).$$ So as $k^2+1$ is never divided by 4, we conclude that $n$ must be $$n\equiv 0,3 \mod 4$$. $\endgroup$
    – benh
    Commented Dec 21, 2013 at 2:20

5 Answers 5


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. (Edit: this statement is not true in general, see the comment by @mr_e_man.)

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 $\pi/4$

And the following is implied by both of these:

  1. $(1+1)(1+4) \cdots (1+n^2)$ is a perfect square.

As described in this MathOverflow post, in fact, all three are impossible for $n > 3$.

  • 4
    $\begingroup$ I posted the equivalent question over on MathOverflow and quickly got a very nice answer, citing a paper that proves n=3 is the only solution. See mathoverflow.net/questions/152908/…. $\endgroup$
    – KenWSmith
    Commented Dec 28, 2013 at 0:21
  • 3
    $\begingroup$ Why in the third paragraph $\frac{i(i-1)(i-2)…(i-n)}{k}$ is a Gaussian integer? $\endgroup$ Commented Aug 25, 2017 at 11:25
  • $\begingroup$ Surely you meant $2\pi/4$ rather than $\pi/4$? $\endgroup$
    – mr_e_man
    Commented Aug 6, 2020 at 22:41
  • 1
    $\begingroup$ I second the comment of @AlexRavsky. The Gaussian integer $z=3+4i$ has perfect square magnitude $k^2=|z|^2=25$, but $\frac zk=\frac35+\frac45i$ is not a Gaussian integer. $\endgroup$
    – mr_e_man
    Commented Aug 6, 2020 at 23:10

Not an answer, but an observation that is far too long for a comment.

It's a well known fact that $$z(z+1)\ldots (z+n-1) = \sum_{k=0}^{n}{n \brack k}z^k$$

Where ${n \brack k}$ denotes an unsigned Stirling number of the first kind. ${n \brack k}$ counts the number of permutations of $n$ elements that are the product of $k$ disjoint cycles.

Thus $$i(i+1)\ldots(i+n-1) = \sum_{k=0}^{n}{n \brack k}i^k$$


$$\operatorname{Re}\left\{i(i+1)\ldots (i+n-1)\right\} = \sum_{k \equiv 0 \operatorname(mod) 4}{n \brack k} - \sum_{k \equiv 2 \operatorname(mod) 4} {n \brack k}$$


$$\operatorname{Im}\left\{i(i+1)\ldots (i+n-1)\right\} = \sum_{k \equiv 1 \operatorname(mod) 4}{n \brack k} - \sum_{k \equiv 3 \operatorname(mod) 4} {n \brack k}$$

Where our notation makes sense since ${n \brack k} = 0$ when $k < 0$ or $k > n$.

We then have the following combinatorial interpretation of your problem.

$i(i+1)\ldots (i+n-1)$ is purely real iff the number of permutations of $n$ elements with number of disjoint cycles congruent to 1 modulo 4 equals the number of permutations of $n$ elements with number of disjoint cycles congruent to 3 modulo 4. A similar statement can be made for the case when $i(i+1)\ldots (i+n-1)$ is purely imaginary.

Now let $$A_0(n) = \sum_{k \equiv 0 \operatorname(mod) 4}{n \brack k}$$ $$A_1(n) = \sum_{k \equiv 1 \operatorname(mod) 4}{n \brack k}$$ $$A_2(n) = \sum_{k \equiv 2 \operatorname(mod) 4}{n \brack k}$$ $$A_3(n) = \sum_{k \equiv 3 \operatorname(mod) 4}{n \brack k}$$

Unsigned Stirling numbers of the first kind satisfy the recurrence $${n \brack k} = (n-1){n-1 \brack k} +n{n-1 \brack k-1}$$

Plugging this into the equation for $A_0(n)$ gives us the recurrence $$A_0(n) = (n-1)\sum_{k \equiv 0 \operatorname(mod) 4}{n-1\brack k} + \sum_{k \equiv 0 \operatorname(mod) 4}{n-1 \brack k-1} =$$ $$(n-1)\sum_{k \equiv 0 \operatorname(mod) 4}{n-1 \brack k} + \sum_{k \equiv 3 \operatorname(mod) 4}{n-1 \brack k} =$$ $$(n-1)A_0(n-1) - A_3(n-1)$$

Plugging the recurrence into each equation gives us the system of recurrences $$A_0(n) = (n-1)A_0(n-1) - A_3(n-1)$$ $$A_1(n) = (n-1)A_1(n-1) - A_0(n-1)$$ $$A_2(n) = (n-1)A_2(n-1) - A_1(n-1)$$ $$A_3(n) = (n-1)A_3(n-1) - A_2(n-1)$$

with initial condition $$A_0(0) = 1\mbox{, } A_1(0) = A_2(0) = A_3(0) = 0$$

Letting $X(n) = A_{2}(n) - A_{0}(n)$ and $Y(n) = A_3(n) - A_1(n)$, we can derive the following system of recurrence for them by subtracting recurrences for the $A_{i}$'s.

$$X(n) = (n-1)X(n-1) - Y(n-1)$$ $$Y(n) = (n-1)Y(n-1) + X(n-1)$$

with initial condition $X(0) = -1$, $Y(0) = 0$.

If we can show that for $n \geq 5$, $X(n) \neq 0$ and $Y(n) \neq 0$ then we will have proven that $i(i+1)\ldots (i+n-1)$ cannot be purely real or purely imaginary for $n \geq 5$. This is where I'm stuck. I hope someone finds this direction helpful in solving this problem.

  • $\begingroup$ Stupid me. The recurrence is trivially derivable from the original formula. The permutations stuff might still be helpful though. $\endgroup$ Commented Dec 21, 2013 at 19:33

An element of $\Bbb Q[i]^\times$ is real or imaginary if and only if its image in the quotient $G = \Bbb Q[i]^\times / \langle i, \Bbb Q^\times \rangle$ is trivial.

Since $\Bbb Z[i]$ has unique factorization, $(\Bbb Q[i]^\times,\times)$ is the direct product of :

  • the unit group $\{1,i,-1,-i\}$, which is annihilated in the quotient
  • $p^\Bbb Z$ for $p \equiv 3 \pmod 4$ , again annihilated
  • $\mathfrak p^\Bbb Z \overline{\mathfrak p}^\Bbb Z$ for $p \equiv 1 \pmod 4$ where $\mathfrak p \overline{\mathfrak p} = p$ is annihilated
  • $(1+i)^\Bbb Z$ where $(1+i)^2 = 2i$ is annihilated

Hence the quotient $G$ is isomorphic to $(\bigoplus_{p \equiv 1 \pmod 4} \Bbb Z) \oplus \Bbb Z/2\Bbb Z$, and we can look component by component at what $i(i+1)(i+2)\ldots(i+n)$ is.

We do so by looking at the prime decomposition of its norm. A norm divisible by $2$ means that we have a factor of $(1+i)$, and two of them cancel each other. Otherwise, a norm divisible by $p$ means that we have one of the two factors $\mathfrak p$ or $\overline{\mathfrak p}$. So $p$ comes in two "colors" that cancel each other. Both "colors" appear periodically with period $p$, and if we first see the prime $p$ at $i+k$, then the opposite color always appears at $i+(p-k)$.

Looking at the first few prime factorisations of $(1+k^2)$, and coloring the first occurence of a prime blue, we have :

$2,\quad \color{blue}{5},\quad 2.\color{red}{5},\quad \color{blue}{17}, \quad 2.\color{blue}{13}, \quad \color{blue}{37},\quad 2.\color{blue}{5^2}, \quad \color{red}{5.13},\quad 2.\color{blue}{41},\quad \color{blue}{101}, \quad 2.\color{blue}{61}, \quad \color{blue}{5.29},\quad 2.\color{red}{5.17},\quad \ldots$.

The product of the first three terms cancel completely, and it follows that $(i+1)(i+2)(i+3)$ is real or pure imaginary. Then, by $k=10$ we can't get rid of the $\color{blue}{101}$ before we get to $k=91$. By then, we will just have introduced some $\color{blue}{8101}$, which we can remove only at $k=8011$, and so on. So things are looking grim !

This suggests that we are never even close to have a complete cancellation again, but I don't see any simple argument proving it.

If we look at the product of the first $n$ terms and we find a prime $p>2n$, then if it occured at $k \le n$, it only occurs later at $p-k > 2n-k \ge n$, so it is not cancelled. Suppose we have a cancellation, and let us estimate the number of times a prime appears in the factorisation of $\prod(1+k^2)$.

Since about $2/p^d$ values of $k$ give $1+k^2$ divisible by $p^d$, we get something like $\sum_{d \ge 1} (2np^{-d} + O(1))$. If we want to have a complete cancellation, there is a maximum value of $d$ : at each level, one color can have at most $1$ term more than the other. If a level occurs in both colors, then $2n \ge p^d$, so that $d \le \log{2n}/\log p$. This gives us an upper limit on $d$ of $2\log{2n}/\log p$, and so the total number of times that $p$ appears is $N(n,p) = \sum_{d=1}^{2\log{2n}/\log p} (2np^{-d} + O(1)) = 2n/(p-1) + O(\log n / \log p)$.

Now, we look at the contribution of all the primes less than $2n$ : $L = \log (\prod_{p=2}^{2n-1}) p^{N(n,p)} 1_{prime}(p)) = \sum_{p=2}^{2n-1} N(n,p) \log p 1_{prime}(p)$.

Since number $p$ is prime with a square root of $-1$ about $1/{2\log p}$ of the time, we get $L \approx \int_2^{2n} N(n,p)\log p / (2 \log p) dp = \int_2^{2n} N(n,p)/2 = n \log(2n-1) + O(\log n Li(2n)) = n \log n + O(n)$

Meanwhile, we should have $L = \log \prod_{k=1}^n (1+k^2) = \sum_{k=1}^n \log(k^2+1) = 2\sum_{k=1}^n \log k + O(1) = 2n\log n + O(n)$

This shows that the contribution of primes $p \ge 2n$ is asymptotically half of the whole thing. So lots of them have to appear.

  • $\begingroup$ Attempting to think through the lines suggested by mercio … since the product i(i-1)(i-2)…(1-n) is a Gaussian integer then if it is real or pure imaginary then it is, in fact, a rational integer or a rational integer times i. So the magnitudes are very important. We need a product of square roots $\sqrt{2}\sqrt{5}\sqrt{10}\sqrt{17}\cdots\sqrt{1+k^2}$ that is integral. Equivalently $\prod (1+k^2)$ has to be a perfect square. When 17 occurs in the product, it must have a mate further in the list. Same with 37. Etc. (Like everyone else, I find this hard to believe… but I don't see a proof.) $\endgroup$
    – KenWSmith
    Commented Dec 25, 2013 at 1:30

This is a detail for @AlbertSteppi 's answer, and also suggests, that the conclusion of increasing real&imaginary components is a misconception. Here is a plot of the argument (the angle around the origin in the complex plane) of the resulting complex number, where $f(n)=i(i+1)(i+2)...(i+n)$ displayed for $n=1 \ldots 128$

enter image description here

This does not mean that I really expect there is an integer $n$ at the $0$ or $\pm 1 \pi$ args of the function-values, but the image suggests, that we might have infinitely many revolutions around the origin which makes the growth of the absolute values of the components of $f(n)$ an irrelevant effect.

  • $\begingroup$ I'm not sure I follow you. That the ratio of the imaginary part to the real part can, I suspect, be found arbitrarily close to zero is interesting, but doesn't suggest that hitting zero exactly is possible, at least not to me. For instance, with $n=82$ (at the second crossing in your plot) the ratio of real to imaginary is $\approx 480.7$, so the argument is very small, but the imaginary part is greater than $10^{122}$, so very, very far from real. Perhaps I'm missing something? Cheers! $\endgroup$ Commented Dec 31, 2013 at 19:13
  • $\begingroup$ @Matthew: (I've been out for two days) Well, I didn't want to suggest that "hitting exact zero is possible"; I just wanted to contradict the impression, that the real and imaginary parts simply grow away from zero and by this a perfect imaginary/real number for $f(n)$ were unlikely/impossible. The fact of rotation around the origin only says that this growth is no more an argument and we need something else to suggest, that the real/imaginary axis is never met. $\endgroup$ Commented Jan 1, 2014 at 20:01
  • $\begingroup$ If we can bound the imaginary part away from zero, then the number is not real. That's all that is needed: if the imaginary part is never zero, the number is never real. The argument may get arbitrarily close to zero, but we may still conclude the number is never real. The growth of the absolute values of the components is far from irrelevant here. Cheers! $\endgroup$ Commented Jan 2, 2014 at 4:38

Find all $n$ so that $z=i(i+1)(i+2)(i+3)...(i+n)$ is real or pure imaginary.



$z\bar{z}=\Pi_{k=0}^n(k^2+1)=R^2$ .

$R\in N$


$R^2=\Pi_{k=0}^n k^2(1+1/k^2)$

$2\ln R=\sum_{k=0}^n2\ln k+\sum_{k=0}^n \ln (1+1/k^2)$

$\ln|1+x|\approx x-x^2/2$

$\ln R <\sum_{k=0}^n\ln k +\pi^2/12$


$k!\le n!<R, (k^2+1)|R^2$

$z^2=\Pi_{k=0}^n [(k^2-1)+2ik]$

$2\ln z=\sum_{k=0}^n\ln [(k^2-1)+2ik]=\sum \ln(k^2+1)+i\sum \arctan(\frac{2k}{k^2-1})$

$\ln z=\ln R +(i/2)\sum \arctan (\frac{2k}{k^2-1}) $

$z=\pm R\implies \ln z =\ln R + \pi i\implies (i/2)\sum \arctan(\frac{2k}{k^2-1})=m\pi i\implies \sum \arctan (\frac{2k}{k^2-1})=2m\pi$

$z=\pm iR\implies \ln z = \ln R \pm \pi i/2 \implies \sum \arctan(\frac{2k}{k^2-1})=(2m+1)\pi$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .