Let $ u=1-\sqrt{2}$ $ [u^{2n}]=0$ $ [u^{2n+1}]=-1$ Imply that $ [u^n]-n\equiv 0(\mod 2)$

Why $ [u^{2n}]=0$ and $ [u^{2n+1}]=-1$?

It's since $ -1<u<0$, so then $ 0<u^{2n}<1$ and $ -1 < u^{2n+1} < 0$ which means $ [u^{2n}]=0$ and $ [u^{2n+1}]=-1$.

Yes, in the litteral form of course, but $ 1-\sqrt{2}<0$. I thought he at least meant $ \sqrt{2}-1$ instead. Because the problem really isn't solved with this.

Peter wrote: Prove or disprove that there exists a positive real number $ u$ such that $ \lfloor u^n \rfloor - n$ is an even integer for all positive integer $ n$. Let $ x_0 = 2,\ x_1 = 3,\ x_n = 3x_{n - 1} + 2x_{n - 2}$ for $ n\geq 2$. It is easy to see that $ x_n$ is an odd positive integer for all $ n\geq 1$. It is also easy to obtain an explicit formula for $ x_n$: \[ x_n = \left(\frac {3 + \sqrt {17}}{2}\right)^n + \left(\frac {3 - \sqrt {17}}{2}\right)^n. \]Since the second summand in this formula is negative, for $ u = \frac {3 + \sqrt {17}}{2}$ and all $ k = 1, 2, \dots$ we have: $ \lfloor u^{2k}\rfloor = x_{2k} - 1$ is even; $ \lfloor u^{2k - 1}\rfloor = x_{2k - 1}$ is odd. Therefore, $ u = \frac {3 + \sqrt {17}}{2}$ gives affirmative answer to the original problem. P.S. The same solution I gave back in 2006 in Russian forum: http://dxdy.ru/post25766.html#p25766

$\frac{2a+1+\sqrt{(2a+1)^2+b}}{2}$ is a solution for all naturals $(a,b)$ with the condition $2a\geq b$. And also all quadratic irrationalities that satisfy the conditions of the problem are in this form.