<s>[INST] Find all $n$ such that $n \mid x_n$ where $x_n = x_{n-1} + \lfloor n^2 / 4 \rfloor$, $x_0 = 0$ $X_n$ is sequence such that $x_n=x_{n-1}+[\frac{n^2}{4}]$ and $x_0=0$. Find all positve integers $n$ for which $x_n$ is divisible by $n$. 
[X] means integer part.
 [/INST] $$x_{2k}=x_{2k-1}+k^2=x_{2k-2}+2k^2-k=2k^2-k+(2(k-1)^2-(k-1))+x_{2k-4}=\cdots=\sum_{n=1}^{k} 2n^2-\sum_{k=1}^{k} n=\frac{k(k+1)(2k+1)}{3}-\frac{k(k+1)}{2}=\frac{k(k+1)(4k+2-3)}{6}=\frac{k(k+1)(4k-1)}{6}$$
Also
$$a_{2k+1}=a_{2k}+k^2+k\\a_{2k+1}=\frac{k(k+1)(4k-1)}{6}+\frac{6k(k+1)}{6}=\frac{k(k+1)(4k+5)}{6}$$
Now splitting into two cases we can see that
$$2k+1\mid \frac{k(k+1)(4k+5)}{6}$$ Works only for $k=1$ since $\gcd(2k+1,k)=\gcd(2k+1,k+1)=1$ and $\gcd(4k+5,2k+1)\mid  3$ that implies that only possible solution is $2k+1=1$ or $2k+1=3$
$$2k\mid\frac{k(k+1)(4k-1)}{6}$$
Works for $k=12t+7$ and $k=12t+11$,this is because $$\frac{(k+1)(4k-1)}{12}\in\mathbb{N}$$
Putting it all together for $n=3$ and $n=24t+14,n=24t+22$ we have that $n\mid a_n$
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