Periodicity of fractional part of a sequence Let $u_n = \mathrm{frac}(a n^2)$, where $a$ is some real number and 
$\mathrm{frac}$ denotes the fractional part. 
Question 1)
Can $u_n$ be eventually periodic even if $a$ is irrational?
Question 2)
Is it true that whenever $a$ is rational, $u_n$ is eventually periodic?
(this is not homework; I think I have a solution for question 2 in the case where $a=p/q$ and $q$ is a product of prime numbers without any power greater than one, but I think it should be true in general)
 A: Proposition 1: If $u_n$ is "eventually periodic" (defined below), then $a$ is rational.  
Proof of Proposition 1:
Note that $\text{frac}(): \mathbb{R} \rightarrow (-1, 1)$ is a group homomorphism, the image of which is an abelian group, G, on $(-1, 1)$ whose operation is $+$ where we lop off the integral part.  Then we can view each element of $\{u_n\}$ as applying $\alpha := \text{frac}(a)$ to itself $n^2$ times.  That is, $\{u_n\} = \{\alpha, \alpha+\alpha+\alpha+\alpha = 4\alpha, 9\alpha, 16\alpha, \ldots\}$.  
Let $u_n$ be "eventually periodic".  That is, there exists some number $c > 0$ such that for some $N$, $\forall n \geq N$, $u_n = u_{n+c}$. This means that $G$ has a subgroup, $H$, that is cyclic of order $m := (N + c)^2 - N^2$ (since $u_{N + c} - u_N = 0 \in G$).  Therefore, $m \alpha = 0$.  Since $m \in \mathbb{Z}$ and $Ker(\text{frac}()) = \mathbb{Z}$, we have that $\exists z \in \mathbb{Z}$ such that $m a = z$, so $a \in \mathbb{Q}$, which completes the proof.  
Proposition 2: If $a$ is rational, then $u_n$ is "eventually periodic".  
Proof of Proposition 2:
Let $b, c \in \mathbb{Z}$ with $c \neq 0$ and $a = \frac{b}{c} \in \mathbb{Q}$.  If $b = 0$, then the proposition is trivially true.  If $b \neq 0$, then $G = Im(\text{frac}())$ has a cyclic subgroup $H$ of order $c$ generated by $\alpha$.  Then $\forall k \in \mathbb{N}$, $(n + kc)^2 a = (n^2 + 2nkc + k^2 c^2) \alpha = n^2 \alpha$, since the $c$'s send any $h \in H$ to the identity, $0$.  Therefore, $u_{n + c} = u_n$, so $\{u_n\}$ is eventually periodic, which completes the proof.  
Together, the two proofs show that $u_n$ is "eventually periodic" if and only if $a$ is rational.  
Interesting questions -- I enjoyed thinking through them.  Let me know what you think.  
