Number of matrices $A \in M_n(\mathbb{F}_q)$ where $A^2 = 0$. What is the number of matrices $A \in M_n(\mathbb{F}_q)$ for which $A^2 = 0$ (as a function of $n$ and $q$)?
 A: I will let you write the details, this is just a sketch.
Even if we are not in an algebraically closed field, we can always conjugate those matrices to their canonical Jordan form (why?).
Then try to understand what are the possible canonical Jordan form verifying this condition (easy). 
Then compute the centralizer of each matrix obtained (this is possible, given any Jordan form to compute the centralizer), in particular you will have the cardinal of each.
Deduce from this the cardinality of each conjugacy class, the sum of all those cardinals will exactly give you  what you are looking for.
A: There are $p,r$ s.t. $A$ is similar over $\mathbb{F}_q$ to $U_{p,r}=diag(V_1,\cdots,V_p,0_r)$ where $V_i=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $2p+r=n$. Let $\alpha_{p,r}$ be the cardinality of the centralizer of $U_{p,r}$ in $GL_n(\mathbb{F}_q)$, that is $card(\{P\in GL_n;PU_{p,r}=U_{p,r}P\})$.
Note that $\{P\in M_n;PU_{p,r}=U_{p,r}P\}$ is a vector space of dimension $2p^2+2pr+r^2$; yet $\alpha_{p,r}$ is difficult to calculate; that is the sole difficulty of the problem and Clement did not rack his brains on the subject!
It is known that $card(GL_n)=(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$; then the cardinality of the conjugacy class of $U_{p,r}$ is $\beta_{p,r}=\dfrac{card(GL_n)}{\alpha_{p,r}}$. Finally, the required result is $\sum_{2p+r=n}\beta_{p,r}$.
Some results. For $n=2,3,4$, we obtain $q^2,q^4+q^3-q,q^8+q^6-q^2$. 
