When is a given polynomial a square of another polynomial?

Question

I meet a problem in which I hope to show a special polynomial is not a square of another polynomial. More precisely, let's consider the polynomial
$$f(x):= 1-x+2bx^n-2bx^{n+1}-b^2x^{2n-1}+2b^2x^{2n}-b^2x^{2n+1}-2bx^{3n-1}+2bx^{3n}-x^{4n-1}+x^{4n}\in k[x],$$ where $k$ is a field of characteristic $p>0$, $n>2$ is an integer, and $b\in k$ with $b\neq 0,1,-1$. Indeed, in the context I meet, the field $k$ is just the finite field with $p$ elements, $n=(p^m+1)/2$ with $m\geq 1$. I just try to rephrase the question in a clean and a more general way, but you may use the further assumption in your proof. I have checked for a lot of examples and I found it is always not a square of another polynomial. I believe it is true in general but I fail to give a proof for it. Please be free to give me any kind of suggestions on this problem. Although I know how to check if a concrete polynomial is a square, I have no ideas about how to systematically determine whether a general polynomial is a square or not? If you know any theory related to it, please don't hesitate to tell me. Thanks a lot for your time!

Answer 1

Apparently $p>2$. An argument specific to this polynomial relies on the following observations:

1. $x=1$ is a zero with multiplicity exactly two.
2. $x=-1$ is not a zero of this polynomial.
3. $f(x)$ is palindromic. In other words $f(x)=x^{4n}f(1/x)$.

The claim then follows from these observations. Assume contrariwise that $f(x)$ would be the square of another polynomial. This means that any zero $\gamma$ of $f(x)$ in some extension field $K/k$ must have even multiplicity $2r$. By item 3, unless $\gamma=1/\gamma$, between them $\gamma^{\pm1}$ then account for $4r$ zeros counted with multiplicity.

But $\gamma=1/\gamma$ iff $\gamma=\pm1$, and items 1 and 2 show that these contribute exactly 2 zeros counted with multiplicity.

Thus the number of zeros (with multiplicity) is $\equiv2\pmod4$ contradicting the fact that the degree of $f(x)$ is a multiple of four.

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So we need to check those observations.

1. It is straightforward to check that $f(1)=f'(1)=0$ so $x=1$ is a zero multiplicity $\ge2$. Let's denote $P_k(x)=1+x+x^2+\cdots+x^k$. As $P_k(x)(1-x)^2=(1-x^{k+1})(1-x)$ we see that $$ g(x):=\frac{f(x)}{(x-1)^2}=P_{4n-2}(x)+2bx^nP_{2n-2}(x)-b^2x^{2n-1}. $$ Therefore $g(1)=4n-1+(2n-1)2b-b^2=1-b^2\neq0$, because $2n-1$ is a multiple of $p$ and $b\neq\pm1$.
2. We also see that $f(-1)=2(2\pm4b+2b^2)=4(b\pm1)^2$ where the sign depends on the parity of $n$.
3. This is clear by observation.