Proving that $\lfloor a^2/2 \rfloor$ is even for all $a\in\mathbb{Z}$. $\forall a \in \mathbb Z, \lfloor a^2/2 \rfloor$ is even.
I am pretty sure this statement is true. The only suspicious cases to me are 0/2 and 1/2, but they also have even floors. How do I prove it though?
 A: Well everything except 0 divides 0 that's why zero is even. you could try cases when a is even its pretty easy when a is odd then u have an even number +1/2 for all odd numbers the floor won't affect that though it will send each 1/2 to 0. does that help?
A: Do it by cases. Either $a$ is even, or it’s odd.


*

*If $a$ is even, write $a=2n$; then $a^2/2=2n^2$ is certainly even. 

*If $a$ is odd, write $a=2n+1$; then $$\left\lfloor\frac{a^2}2\right\rfloor=\left\lfloor\frac{4n^2+4n+1}2\right\rfloor=2n^2+2n=2(n^2+n)$$ is also even.
Don’t shy away from case-by-case arguments; sometimes they’re the most straightforward way to prove a result.
A: Note that for every integer $a$, we have $a^2\equiv 0\,(mod\,4)\,
\lor a^2\equiv 1\,(mod\,4)$. If $a^2=4k$ (for some integer $k\geq 0$), then $\lfloor \frac{a^2}{2}\rfloor=\lfloor \frac{4k}{2}\rfloor=2k$. If $a^2=4k+1$, then $\lfloor \frac{a^2}{2}\rfloor=\lfloor \frac{4k+1}{2}\rfloor=\lfloor 2k+\frac{1}{4}\rfloor=2k$.
A: $\rm \begin{eqnarray}\rm Let\ \ a &=&\rm 2n\!+\!r\\ \rm for\  \ \ r&\in& \{0,1\}\ \end{eqnarray}\!\bigg\rbrace\:\Rightarrow\: \bigg\lfloor \dfrac{a^2}2\bigg\rfloor = \bigg\lfloor 2n^2\!+2nr+\dfrac{r^2}2\bigg\rfloor =\, 2n^2\!+\!2nr,\,\ $ by $\rm\ \ \dfrac{r^2}2 \in\bigg\{0,\,\dfrac{1}2\!\bigg\}$
