On the “Reducibility” of Arctangents of Integers

Abstract

This is discussed a bit further in [1] with some numerical and theoretical evidence. In this note we point out the link of Conjecture 3 with the recent very deep work of Duke, Friedlander, and Iwaniec [2] on roots of quadratic congruences modulo primes and give some more support (heuristic and theoretical) for the conjecture, showing also that it is undoubtedly a very hard problem. The condition p | n + 1 that occurs in Lemma 1 means that n is a root of the quadratic congruence X + 1 = 0 (mod p). We recall Fermat’s classical theorem that this congruence has two solutions if p ≡ 1 (mod 4) (if ν is one, the other is −ν), none if p ≡ 3 (mod 4) and one if p = 2 (see, for instance, [7] for a proof). If ν = −1 (mod p), then the fractional part {ν/p} of ν/p, where ν is any integer congruent to ν modulo p, is well-defined in [0, 1]. According to the lemma, arctanm is irreducible if and only if there exists a prime p such that p | m+1 and p > 2m. This prime p is then unique, since p 6 q implies pq > 4m > m + 1. Thus one can write