I am finding some problems with the following exercise.

Given a prime $p>2$ and the polinomial $f(x) = x^{p-2} + x + 2\in \mathbb{Z}/p\mathbb{Z}[x]$, I should prove that the root $\alpha = -1$ is simple for $f$.

It is explicitly asked to prove the above in the following way:

assume that the statement is false, i.e. the root is multiple for $f$ and prove that the same root is also multiple for the polynomial $g(x) = x^3 + 2x^2 -1\in \mathbb{Z}/p\mathbb{Z}[x]$ (which leads to a contradiction).

The way the second polynomial is linked to the first is the following: for all $0\ne\beta\in \mathbb{Z}/p\mathbb{Z}$ we have that $$f(\beta)=0 \iff \beta^2f(\beta) = 0 \iff \beta^3 + 2\beta^2 +\beta = 0$$

for Fermat's Little Theorem. Since one can easily show that $\alpha = -1$ is the unique root of $f(x)$, the polynomial $g$ appears because we can fix the constant term to be $-1$.

Now, I tried several ways following the suggested path. First of all, assuming that the root is multiple means that we can of course write $$f(x) = (x+1)^2h(x)$$ for a suitable polynomial $h$, since the mupltiplicity must be at least 2, and I should find a similar relation for $g(x)$, but I didn't find anything. I also tried other ways but they did not lead me anyway.

Consider that, in the context in which this exercise is given, the solution should not involve very sophisticated tools (more or less we are supposed to use basic tools like Fermat's Little Theorem, Euler's Theorem, Polynomial long division..).

  • 2
    $\begingroup$ Hint : Calculate the derivate of $f$ modulo $p$ $\endgroup$
    – Peter
    May 22 '18 at 14:22
  • $\begingroup$ No this is supposed to be solved without using derivatives $\endgroup$
    – Addc
    May 22 '18 at 14:38
  • $\begingroup$ I don't see how $\beta^2f(\beta) = 0 \iff \beta^3 + 2\beta^2 +\beta = 0$. This seems to be equivalent to $\beta^{p+2}=\beta$ for all nonzero $\beta\in\Bbb{Z}/p\Bbb{Z}$, which is certainly false. $\endgroup$
    – Servaes
    May 22 '18 at 14:43
  • $\begingroup$ Oh thanks for noticing. The polynomial f is indeed $f(x) = x^{p-2} + x + 2$, I made a mistake in writing the exercise down, I am sorry. I edited the post. $\endgroup$
    – Addc
    May 22 '18 at 14:47
  • $\begingroup$ The suggested solution seems completely nonsensical at every step. $\endgroup$
    – Servaes
    May 22 '18 at 15:00

The quotient $$ g(x)=\frac{f(x)}{x+1}=\frac{x^{p-2}+x+2}{x+1}=x^{p-3}-x^{p-4}+x^{p-5}\cdots -x+2 $$ satisfies $g(-1)=p-1\neq 0$ in $\Bbb{F}_p$, hence $-1$ is not a multiple root of $f$.

  • $\begingroup$ Yes I also solved exactly in this way, but I was trying to follow the suggested solution as well. I can not figure out how it could work. $\endgroup$
    – Addc
    May 22 '18 at 15:07

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