Does there exist a positive real number α such that [αn] - n is even for all integers n > 0?
Solution
Answer: yes.
The idea is that an interval length 1 is acceptable for αn. This corresponds to an interval length α for αn+1. So if α is sufficiently large (in fact > 3), this will always contain an acceptable subinterval of length 1 for αn+1. Translated back to intervals for α itself this gives to a sequence of nested intervals, which must have non-zero intersection.
So, being slightly more precise, define I1 = [5, 5.9]. Then any point of I1 has integral part 5, which is odd. This corresponds to the interval I1' = [25, 34.8] for α2. This contains the interval [26, 26.9] every point of which has integral part 26, even. It corresponds to I2 = [261/2, 26.91/2] which is a subinterval of I1 (it is roughly [5.10, 5.19]. It corresponds to I2' = [263/2, 26.93/2], or roughly [132.6, 139.5] for α3. This has the subinterval [133, 133.9] every point of which has integral part 133, odd. It corresponds to I3 = [1331/3, 133.91/3] (roughly [5.10, 5.12] ), which is a subinterval of I2. This process can be continued indefinitely, because In' always has length 0.9 α > 4 and hence contains at least 3 consecutive integers. We pick one of the two smaller, so that it is even or odd as required. This gives us an acceptable subinterval [m, m+0.9] of In'.
Note that we could have started with [3, 4) and used half-open intervals of length 1, but then the argument is slightly more complicated because an intersection of non-closed intervals is not guaranteed to be non-empty.
© John Scholes
jscholes@kalva.demon.co.uk
16 Jan 2001