Square roots in ring of strictly upper triangular matrices Let $k$ be field, $M_n(k)$ the algebra of $n\times n$ matrices over $k$, and let $N\subset M_n(k)$ be the subring of strictly upper triangular matrices in $M_n(k)$.  Note that every element of $N$ is nilpotent.
Let $N^2=\{X^2|X\in N\}$, i.e. the set of matrices in $N$ with a square root in $N$.  My question is: can we describe $N^2$?
I conjecture that it is $N^2$ is the set of all matrices with zeroes below the second main diagonal, i.e.
$$
\begin{pmatrix}0 & 0 & * & \cdots & *
\\\vdots & \ddots & \ddots & \ddots &
\\ & & &\ddots & *
\\ \vdots & & &  \ddots & 0
\\ 0 & \cdots & & \cdots & 0
\end{pmatrix}
$$
where $*$ denotes any element of $k$.
Thoughts/ideas?  Any help is appreciated! 
 A: This is not true over any field. Consider a $5 \times 5$ upper triangular matrix
$$\begin{bmatrix} 0 & a & \ast & \ast & \ast \\ 0 & 0 & b & \ast & \ast \\ 0 & 0 & 0 & c & \ast \\ 0 & 0 & 0 & 0 & d \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$
When we square this we get
$$\begin{bmatrix} 0 & 0 & ab & \ast & \ast \\ 0 & 0 & 0 & bc & \ast \\ 0 & 0 & 0 & 0 & cd \\ 0 & 0 & 0 & 0 & 0 \\ 0& 0 & 0 & 0 & 0 \\ \end{bmatrix}$$
Note then if we want $bc = 0$ either $b = 0$ or $c = 0$. So the following matrix is not a square of an upper triangular matrix
$$\begin{bmatrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 &0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$
A: Edit: As NickR points out in the comments, the answer below is incorrect; in fact it describes the set of matrices which are squares of nilpotent matrices. I'll leave it here in case it can be salvaged.

For $l\geq0$ let $r_l(X)=\mathrm{rank}(X^l)-\mathrm{rank}(X^{l+1})$. Given $X\in N$, we have $X\in N^2$ iff $r_l(X)>r_{l+1}(X)$ whenever $r_l(X)$ is odd.
It is well-known that $r_l(X)$ is a non-increasing sequence which stabilises at $0$. First suppose $X=Y^2$ where $Y\in N$. Then
$$
  r_l(X)=r_{2l}(Y)+r_{2l+1}(Y).
$$
If $r_l(X)$ is odd then
$$
  r_{2l}(Y)>r_{2l+1}(Y)\geq r_{2l+2}(Y)\geq r_{2l+3}(Y)
$$
so $r_l(X)>r_{l+1}(X)$.
Conversely suppose $r_l(X)$ odd implies $r_l(X)>r_{l+1}(X)$. Note that Jordan decomposition holds for nilpotent matrices over any field, so we may choose elements $v_1,\ldots,v_m$ and integers $s_1\geq\ldots\geq s_m\geq1$ such that $X^{s_i}v_i=0$ and
$$
  B=\{X^jv_i\mid 1\leq i\leq m,\,0\leq j<s_i\}
$$
is a basis for $k^n$. In particular
$$
  r_l(X)=\max\{i\mid s_i>l\}.
$$
We may assume $m$ is even; if not, let $v_{m+1}=0$ and $s_{m+1}=0$. Now define $Y$ by
$$
  Y(X^jv_i)=\begin{cases}
    X^jv_{i+1}&\text{if }i\text{ odd},\\
    X^{j+1}v_{i-1}&\text{if }i\text{ even}
  \end{cases}
$$
for $0\leq j<s_i$. Note that if $b=X^jv_i\in B$ and $Y(b)\neq0$ then $Y(b)\in B$ and $Y^2(b)=X(b)$. Suppose $Y(b)=0$. If $i$ is even, this implies $j+1\geq s_{i-1}\geq s_i$, so $X(b)=0$. Finally suppose $i$ is odd. Then $s_i>j\geq s_{i+1}$. Thus $r_j(X)=i$ is odd, so $r_{j+1}(X)<i$ by assumption. That is, $s_i\leq j+1$, so $X(b)=X^{j+1}v_i=0$ as required.
