Is there a polynomial $p$ such that $x^n$ divides $1+x-p^2$??

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I'm studying Ring Theory and I have a list of exercises to do, actually I'm trying to prove that for each nilpotent matrix $A \in F^{n^2}$, there is a matrix $N$ s.t $I_n+A = N^2$.

So, if we have for $n \geq 2$ a polynomial $p(x)$ s.t $(1+x - p^2) = x^n q(x)$, then applying $A$, we have $N= p(A)$.

May you help me with a tip about $p$??

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A particular solution is (writing formally the Taylor series) $N=(I+A)^{1/2}=I+\dfrac{1}{2}A-\dfrac{1}{8}A^2+\cdots$; note that the previous series is finite (because $A$ is nilpotent) and is really a square root.

We can use the same method for $I+A=N^k$ where $k$ is any positive or negative integer;