The following is a problem which looks likes elementary but I can not solve it. But I am very interesting know the answer to it. I also think it deserves a decent solution personally, so that I can not stopping thinking over it.
Let $p$ be an odd prime number and $Q$ be a power of $p$. Denote by $\mathbb{F}_Q$ the finite field with $Q$ elements and denote by $k$ the algebraic closure of $\mathbb{F}_Q$. Suppose $n>1$ is a positive integer and $A\in \mathbb{F}_Q$ with $A\notin \{0,+1,-1\}$. Consider the polynomial $f(x):=(x^{2n}-A)^2(x-1)^2+(1-A^2)(x^{4n}-1)(x-1)\in \mathbb{F}_Q[x]$. I hope to show that $f(x)$ is not a square of another polynomial in $k[x]$. In the context I got this question, $Q=q^2$, $a\in \mathbb{F}_{q^2}^*$ with $a^{q+1}\neq 1$ and $A=a^{(q+1)/2}$. You may assume these if you like.
In the special case that the characteristic $p$ divides $n$, it is true since each root $\alpha\neq 1$ of $f(x)$ in the field $k$ is simple. Surely it needs to check such a root indeed exists. In fact, my original question can be solved by solving this special case. But I am very curious about what will happen in the case $p$ does not divide $n$. I have tested for some small values of $n$ and wrote out the square roots of $f(x)$ in the ring $k[[x]]$ of formal power series. The square roots are never polynomials of degree $2n+1$ in all examples I tried. Note here $n>1$, since $f(x)$ could be a square if $n=1$. Furthermore, I guess any root $\alpha\neq 1$ of $f(x)$ is simple in the general case. So I tried to calculate the discriminant of the polynomial $f(x)/(x-1)^2$. But again I failed to show the discriminant is nonzero.
I am wondering what is the point I missed to notice? I will appreciate all kinds of solutions and suggestions about this particular question, or about the general and practical way to detrermine if a given polynomial over is a square of another polynomial.