I thought of this question on Complex Numbers (perhaps related to Number Theory) a couple of months ago. It states that $$\prod\limits_{r=1}^{n}{\left( r+i \right)}$$ is purely imaginary if and only if $n=3$.
I believe that this is true but not sure how to prove it rigorously. My friend verified it for up to 9-digit numbers using a code but still haven't found any contradiction.
Motivation
I came across the Three Square Geometry Problem on Numberphile years ago and found it pretty fascinating that one could use Complex Numbers to solve this problem. In particular, using arguments, it is easy to show that $$\arg \left(( 1+i)(2+i)(3+i) \right)=\arg(1+i)+\arg(2+i)+\arg(3+i)=\frac{\pi}{2}.$$ Thereafter, I thought of generalising this problem to $n$ squares. Initially, I had the perception that the angle would converge but to my surprise, it diverges by the Integral Test. This is essentially because $$\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{{{\tan }^{-1}}\left( \frac{1}{i} \right)}$$ is a divergent series.
Recurrence Relation
Some of my friends tried to use a recurrence relation involving the addition formula of inverse tangent, that is $$\arctan u+\arctan v=\arctan \left( \frac{u+v}{1-uv} \right)$$ but their attempts were futile as there were no nice properties of the above formula.
Brute Force Expansion
This was my initial attempt but I felt it was quite difficult to manipulate the expression subsequently. $$\prod\limits_{r=1}^{n}{\left( r+i \right)}={{i}^{n}}+{{i}^{n-1}}\sum\limits_{\alpha =1}^{n}{\alpha }+{{x}^{n-2}}\sum\limits_{\alpha <\beta }{\alpha \beta }+{{i}^{n-3}}\sum\limits_{\alpha <\beta <\gamma }{\alpha \beta \gamma }+...+n!$$
Any leads would be appreciated. Thank you!
