Let $f$ be a polynomial of degree 5 over $\mathbb F_p$ which has no multiple roots in $\overline{\mathbb F_p}$. ($p$ is prime and large enough. We can assume $p>100$).

Is it possible to find a such $f$ such that $$f^{(p^2-1)/2}(x)=s_0(x)+x^{p^2}s_1(x)+x^{2p^2}s_2(x)$$ where $s_0,s_1,s_2$ of degree $<p(p-1)$.

Note 1: For all such polynomials, there is always such $s_2$ because of the degrees of polynomials. So the problem is actually depending to $s_0$ and $s_1$.

Note 2: Computational, I couldn't find any such $f$.

The other version of this question (not same but this one gives that the other one exists)

Is it possible to find a such $f$ such that $$f^{(p-1)/2}(x)=s_0(x)+x^{p}s_1(x)+x^{2p}s_2(x)$$ where $s_0,s_1,s_2$ of degree $<p-3$.

Similar notes here too.

  • $\begingroup$ We have $f$. Then it is $(p^2-1)/2$-th power. Then the expansion of it. Can this expansion be of the form above with degree restriction. I hope it is clear now? $\endgroup$
    – vudu vucu
    May 15, 2017 at 13:32
  • $\begingroup$ The first version does not appear to be true for $f(x)=x^5-1$ and $p=101$ or $p=103$ (though it's much closer to being true in the latter case). Using Pari/GP I found that every $5^{th}$ coefficient between $101^2-101$ and $101^2-1$ was non-zero. $\endgroup$
    – sharding4
    May 25, 2017 at 23:45
  • $\begingroup$ @sharding4 Nice to know you have found a counterexample. It would be interesting to know if the first version is true for sufficiently large $p$ or not (or equivalently is there only a finite number of counterexamples). $\endgroup$
    – Alex Vong
    May 26, 2017 at 1:12
  • $\begingroup$ Actually, there was an answer here about the forms $x^5+a$ which was deleted. I also mentioned there that none of the $x^5+a$ satisfies the conditions. $\endgroup$
    – vudu vucu
    Jun 3, 2017 at 22:15


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