Let $R$ be a ring with identity $1.$ We denote $\mathrm{UT}_n(R)$ by the unitriangular matrix group over $R$, that is the set of all matrices $(a_{ij})$ which $a_{ii}=1$ for all $i$ and $a_{ij}=0$ for all $i>j.$
Problem. Prove that if $1+1$ is invertible in $R$, then for every $c\in\mathrm{UT}_n(R)$ there exist a unique $b\in\mathrm{UT}_n(R)$ such that $b^2=c.$ Can $b$ be described?
For every $c\in\mathrm{UT}_n(R)$, I try by induction as below. This problem is true for $n=1.$ Let us assume this is true for all matrices of size less than $n$ for $n>1.$ Now, we can assume that $$c=\begin{pmatrix} 1&y\\0&z \end{pmatrix},$$ where $z\in\mathrm{UT}_{n-1}(R)$ and $y$ is a row by $n-1.$ By the induction hypothesis, there exists a $x_z\in\mathrm{UT}_{n-1}(R)$ such that $x_z^2=z.$ If $1+x_z$ is invertible, then we can choose $$b=\begin{pmatrix} 1&y(1+x_z)^{-1}\\0&x_z \end{pmatrix}.$$ Otherwise, I don't know. But if I use induction, I think that it is difficult to show that $b$ is unique. Moreover, it's even harder for me to describe $b$. If $R=\mathbb{C},$ then we can use the hypothesis of the invertiblity of $1+1$ and the concept of determinant in order to show that $1+x_z$ is invertible. But in general, I am not sure.