TITLE: On the parity of $[x^n]$
QUESTION [7 upvotes]: I am trying to find a problem which appeared years ago in the American Mathematical Monthly.  It went something like this:  There was a Putnam Competition question which asked to show that there is a number $x$ with the property that $[x^n]$ has the same parity as $n$ for all positive integers $n$. The square brackets indicate the floor function.  Demonstrate such a number. 
Does anyone know when this problem and solution appeared?

REPLY [9 votes]: Google turns up this "Mock Putnam Exam" from the U[niversity] of I[llinois], whose unattributed second problem asks to show that $[(\sqrt2+1)^n]$ has parity opposite to that of $n$ for each $n=1,2,3,\ldots\,$.  Perhaps that's what you remember?  (It should also be possible to make any infinite binary sequence arise as  $\{[x^n] \bmod 2\}_{n=1}^\infty$ for some $x$ by starting from $x$ large enough and using the sequence of conditions to define a sequence of approximations converging to $x$.)
[added later: an explicit example of $[x^n] \equiv n \bmod 2$ is the largest root $3.214319743377535\!\ldots$ of the cubic $x^3 - 3x^2 - x + 1$.]