Does there exist a positive real number α such that [α^{n}] - n is even for all integers n > 0?

**Solution**

Answer: yes.

The idea is that an interval length 1 is acceptable for α^{n}. This corresponds to an interval length α for α^{n+1}. So if α is sufficiently large (in fact > 3), this will always contain an acceptable subinterval of length 1 for α^{n+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 I_{1} = [5, 5.9]. Then any point of I_{1} has integral part 5, which is odd. This corresponds to the interval I_{1}' = [25, 34.8] for α^{2}. This contains the interval [26, 26.9] every point of which has integral part 26, even. It corresponds to I_{2} = [26^{1/2}, 26.9^{1/2}] which is a subinterval of I_{1} (it is roughly [5.10, 5.19]. It corresponds to I_{2}' = [26^{3/2}, 26.9^{3/2}], or roughly [132.6, 139.5] for α^{3}. This has the subinterval [133, 133.9] every point of which has integral part 133, odd. It corresponds to I_{3} = [133^{1/3}, 133.9^{1/3}] (roughly [5.10, 5.12] ), which is a subinterval of I_{2}. This process can be continued indefinitely, because I_{n}' 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 I_{n}'.

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