Let [z] mean the Greatest Integer less than or equal to z. Find a positive real number X, such that [X^n] is an even number whenever n is even, and [X^n] is an odd number whenever n is odd.
First of all, I believe that there is a typo in this problem. [X^n] cannot be an even number when n=0, since [X^0] = [1] = 1 for all positive real X. I think that the problem was intended to read,
"Find a positive real number X, such that [X^n] is an even number whenever n is odd, and [X^n] is an odd number whenever n is even."
(If I am wrong, Steve, please let me know.)
Using this assumption, we may take X= 1+sqrt(3). To see why this works, let Y= 1-sqrt(3), and consider the sequence {Z(1), Z(2), Z(3), ...} given by the relation
Z(n) = X^n + Y^n
We find:
Z(0) = 2
Z(1) = 2
Z(2) = 8
...
In fact, one can prove by induction that in general, Z(n) satisfies the recurrence Z(n+2) = 2*Z(n+1) + 2*Z(n). I won't prove this here since it's just simple induction.
Anyhow, it is clear from the recurrence that Z(n) is even for all n. Since Y(n) is a number of absolute value less than 1 that oscillates sign, and since Z(n) = X(n)+Y(n), we conclude that [X(n)] oscillates parity.
Solution
