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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!

user26857
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Joy-Joy
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  • One way is to show that it takes on a value that is not a square of another value (i.e., suppose that $f(x) = g(x)^2$ for every $x$. Then for some value $a$, you must have $f(a) = g(a)^2$. But if $f(a)$ is, for instance, a prime, then this cannot be true. – John Hughes Jul 25 '16 at 23:45
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    You are over a field. If your polynomial is a square, take its formal derivative, then the gcd of that and the original polynomial. You should be able to find reading on use of the formal derivative for this purpose. – Will Jagy Jul 26 '16 at 00:08
  • Examining the first and last terms, if $f(x)=g(x)^2$, then $g(x)$ must look like $x^{2n} - \frac12x^{2n-1} + \cdots - \frac12x + 1$ (not implying any pattern in the dots, but those must be the first and last terms). Now expand out $g(x)^2$ using this form, and look at the coefficients of $x^2$ and $x^{4n-2}$. What does that tell you about $n$ and $b$? – Greg Martin Jul 26 '16 at 00:13
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    There has been much work on polynomials whose squares (and/or powers) are "sparse", i.e. have fewer terms than the original polynomial. The results may well apply to your case. To find papers search on keywords: sparse polynomial (square OR power) Davenport, Schinzel. – Bill Dubuque Jul 26 '16 at 00:17
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    Simul-posted to MO, http://mathoverflow.net/questions/245085/when-is-a-given-polynomial-a-square-of-another-polynomial – without notification to either site. Please don't do that! – Gerry Myerson Jul 26 '16 at 07:28

1 Answers1

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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.

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.
Jyrki Lahtonen
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  • Initially I fumbled this by miscalculating that $x=1$ would be a simple zero. Proceed with caution. – Jyrki Lahtonen Jul 26 '16 at 08:51
  • It is so beautiful and it seems that you use all the assumption in such a clever way. To be honest, I noted all the observation you mentioned as above, but I failed to notice the multiplicities of its roots in the algebraic closure. Thank you so much for this clever idea. – Joy-Joy Jul 26 '16 at 15:20
  • I focused to much on the symmetry of this polynomial. As you may noted, if otherwise $f(x)$ is a square of a polynomial $g(x)$, then the coefficients of $g(x)$ must be determined by the coefficients of terms $x^i$ in $f(x)$ for $0\leq i\leq n$. That is to say, only one choice of the coefficients for other terms in $f(x)$ can make it a square. That seems impossible especially when $n$ is big. I tried to get a contradiction from the line. But I didn't succeed even if I did some computation in this direction. – Joy-Joy Jul 26 '16 at 15:26
  • This argument (if/when it checks out) is so ad hoc that I'm inclined to think the root cause is hidden in the way the polynomial was constructed. Can you shed some light to that? @Joy-Joy – Jyrki Lahtonen Jul 26 '16 at 15:30
  • As for your curiosity about the origin of this problem, I begin with some problem on permutation polynomial. Then I meet a lot of problem about the irreducibility of polynomial over finite fields. In one quite special case I get this special polynomial after a few steps of reduction. Personally, I think this kind of polynomials are quite interesting for their own reason. As you know, I can not write all the details here. But if you are interesting, we may discuss by email in more details. Anyway, thanks again for your clever proof. – Joy-Joy Jul 26 '16 at 15:34