to show a special polynomial is not a square of another polynomial

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 pp be an odd prime number and QQ be a power of pp. Denote by FQ\mathbb{F}_Q the finite field with QQ elements and denote by kk the algebraic closure of FQ\mathbb{F}_Q. Suppose n>1n>1 is a positive integer and A∈FQA\in \mathbb{F}_Q with A∉{0,+1,−1}A\notin \{0,+1,-1\}. Consider the polynomial f(x):=(x2n−A)2(x−1)2+(1−A2)(x4n−1)(x−1)∈FQ[x]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)f(x) is not a square of another polynomial in k[x]k[x]. In the context I got this question, Q=q2Q=q^2, a∈F∗q2a\in \mathbb{F}_{q^2}^* with aq+1≠1a^{q+1}\neq 1 and A=a(q+1)/2A=a^{(q+1)/2}. You may assume these if you like.

In the special case that the characteristic pp divides nn, it is true since each root α≠1\alpha\neq 1 of f(x)f(x) in the field kk 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 pp does not divide nn. I have tested for some small values of nn and wrote out the square roots of f(x)f(x) in the ring k[[x]]k[[x]] of formal power series. The square roots are never polynomials of degree 2n+12n+1 in all examples I tried. Note here n>1n>1, since f(x)f(x) could be a square if n=1n=1. Furthermore, I guess any root α≠1\alpha\neq 1 of f(x)f(x) is simple in the general case. So I tried to calculate the discriminant of the polynomial f(x)/(x−1)2f(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.

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Are you familiar with computing the GCD of f(x)f (x) and its derivative to identify repeated factors?
– hardmath
2 days ago

  

 

@hardmath I think I am familiar with that. But I can not compute the gcd in this example. Surely you can solve the problem if you can show the degree of gcd is less than the half of the degree of f(x)f(x). Thank You!
– Joy-Joy
2 days ago

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Related: How do we check if a polynomial is a perfect square? and your similar Question from about three months ago.
– hardmath
yesterday

  

 

Unless (1+4n+A(−1+4n))=0(1 + 4 n + A (-1 + 4 n))=0 we have that x=−1x=-1 is a zero of multiplicity exactly two, and my answer to your earlier question works verbatim. So if AA is not in the prime field, the case is settled. This polynomial is palindromic as well 🙂
– Jyrki Lahtonen♦
yesterday

  

 

@Jyrki Lahtonen Thanks for your comments here and your former answer. But I think here in this case x=−1x=-1 is not a root of f(x)f(x). Isn’t it?
– Joy-Joy
yesterday

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