Post #1815104

@hallasurvivor Puzzle 1. Since ij=-ji etc., one has (a i + b j + ck)^2 = -(a^2 + b^2 + c^2). Puzzle 2. This is tricky and probably explained not precisely enough in the literature. The point is that while you can decide to evaluate a quaternionic polynomial f = sum q_n T^n at a quaternion a, setting f(a) = sum q_n a^n, this map f -> f(a) is not a ring morphism. So when you manage to divide a polynomial f which vanishes at a by the polynomial T-a, which you can do, you get f = g · (T-a), but you can't deduce that f(b)=g(b) · (b-a). In particular, there is no relation between the roots of f and those of g and the classic induction breaks down.