MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## A problem on sums of arctangents of rationals

Let $S$ be a set of rational numbers.

For "special" sets $S$, we can ask if $\pi$ can be written as a $\mathbb{Q}$-linear or $\mathbb{Z}$-linear combination of elements from '$\{\tan^{-1}(x): x \in S\}$'.

In particular, let $\hat S$ be the closure of $S$ under both negation and the binary operation $p*q=\frac{(p+q)}{(1-pq)}$.

1. For any natural number $b \geq 2$, $S_b =\{ \frac{1}{b^k}: k \geq 1 \}$. Prove that $0 \notin \hat {S_b}$.

Note: $1 \notin \hat S$ since any rational number $P/Q$ in $\hat {S_b}$ in non-reduced form satisfies $(P,Q) \in \{(0,1), (0,-1),(1,0),(-1,0)\}$ (mod $b$).

2. $tan^{-1}(1/2) + tan^{-1}(1/3) = \pi/4$.

Let $b_1,b_2 \geq 2$ be two natural numbers. For which pairs does 1 belong to the closure of $S_{b_1} \cup S_{b_2}$? For which pairs does zero belong to the closure? Are there pairs for which $\pi$ can be written as a $\mathbb{Q}$-linear combination of the arctangents of their negative powers but not as $\mathbb{Z}$-linear combinations?

-
 This reads like a homework problem... nevertheless, as a hint notice your binary operation is in fact a well known trig identity (it should hopefully remind you of $\tan(p+q)$, though, depending on course material, you may need to derive the equivalent expression for arctans. – mathic Dec 3 2010 at 8:30 This is not a homework problem. In the post, I define the closure operation AFTER considering the problem of expressing $\pi$ as a sum of arctans in order to reduce this to a (hopefully) simpler problem about rational numbers. – Aravind Dec 3 2010 at 9:34 Aravind--there's a fancier way of looking at your note. You're adding one element--infinity--to Q and making the resulting set into a group, with r+(infinity)= -1/r. Now if you fix a prime p, the rationals whose ord at p is non-zero (together with the element infinity) form a subgroup under your group operation. So, as you say, if you "add together" various r whose ord at p is non-zero, you can't get 1. Thanks for an interesting problem. – paul Monsky Dec 3 2010 at 16:49

This is not an answer but an attempt to interpret your question in terms of the arithmetic of the field K=Q(i), where it seems much more natural.

The image in (Reals)/((pi)(integers)) of the set of u for which tan(u) is a rational, r/s, or "is 1/0" is an additive subgroup, G. Now u-->s+ir is a homomorphism from G to the multiplicative group L*/Q* where L is the field Q(i), and you're really studying L*/Q*. Your first question is basically this: If b>1 is an integer, are the elements (b^n)+i of L*/Q*, n=1,2,3..., multiplicatively independent? This looks very hard to me.

If instead of say the powers of 2, you look at the positive even integers, then there are relations. For example (2+i)^6, (8+i)^4, (12+i)^(-2) and (70+i)^2 multiply to an integer, and pi is 6 arc tan (1/2)+ 4 arc tan (1/8)- 2 arc tan (1/12) + 2 arc tan (1/70). A more tractable problem might be to show that for any l there are such relations expressing pi as a Z-linear combo of arc tan (1/k) with l dividing each k.

-
 Thanks for your answer, Paul. Looking at the problem in terms of multiplication of Gaussian integers does make it look clearer. Now, norm considerations can be used to rule out numbers like 1+i as the product of numbers in lZ+i, for an integer l>1. By the way, I am curious about how you obtained your example of numbers which multiply to an integer. – Aravind Dec 6 2010 at 8:54 Aravind--There are only 3 primes, 5,13 and 29 dividing the norms of the 4 elements 2+i,8+i,12+i and 70+i. So there has to be a multiplicative relation in L*/Q*, and by factoring these 4 elements into irreducibles, one quickly writes it down. I could more simply have used the 3 elements 2+i,8+i and 18+i and the 2 primes 5 and 13. – paul Monsky Dec 7 2010 at 16:22

In addition to Paul's comment let me point out that the structure of the underlying group was investigated in

• Sam Northshield, Associativity of the secant method, Amer. Math. Mon. 109 (2002), 246-257,

and that the relation with the arithmetic of ${\mathbb Z}[i]$ was observed by

• John Todd, A problem on arc tangent relations, Amer. Math. Mon. 56 (1949), 517-528

The associativity of the secant method is, by the way, a consequence of the usual geometric interpretation of the group law on conics.

-