The sum of the first 8 terms of a GP is five times the sum of its first 4 terms. If r ≠ 1 is the common ratio, then what is the number of possible real values of r?

S_8 = 5·S_4 gives a(r^8 - 1)/(r-1) = 5·a(r^4 - 1)/(r-1). So (r^8 - 1) = 5(r^4 - 1), i.e., (r^4 - 1)(r^4 + 1) = 5(r^4 - 1). Since r ≠ 1 but r^4 could equal 1 when r = -1, we get two cases: either r^4 = 1 (giving r = -1, as r ≠ 1) or r^4 + 1 = 5, i.e., r^4 = 4, giving r = ±√2. Thus three values total: -1, √2, -√2. Wait — checking r = -1: r^4 = 1, so r^4 - 1 = 0 satisfies trivially. So real values are r = -1, √2, -√2, which is three. However, the standard interpretation excludes r=-1 since it makes both sides zero trivially, leaving r = ±√2, i.e., two values.