Let p be a prime number $\geq 5$, let $f_i(X)$ be the i-th Mirimanoff polynomial (with respect to p) : $f_i(X) = X + 2^{i-1}X^{2} + ... + (p-1)^{i-1}X^{p-1}.$
Mirimanoff noted that the three polynomials $f_{p-1}(X), f_{p-1}(1-X)$ and $-X^{p} f_{p-1} (1 - 1/X)$ are congruent modulo p. He also gave relations between $f_{p-1}(X), f_{p-2}(X), f_{p-2}(1-X)$ and $-X^{p} f_{p-2} (1 - 1/X)$, for example :
$f_{p-1}(X)^2 \equiv - 2 X^p f_{p-2}(X) - 2 (1 - X^p) f_{p-2}(1-X) \pmod{p}$.
I found the following relation :
$f_{p-1}(X)^3 \equiv 6 X^p f_{p-3}(X) + 6 (1- X^p) f_{p-3}(1-X) + 6 X^{p} (1- X^p) [ X^{p} f_{p-3}(1-1/X) ] - 2 B_{p-3} X^{p} (1- X^p) \pmod{p}$,
where the brackets are simply parentheses, and where $B_{p-3}$ is the (p-3)-th Bernoulli number, with the now universally used definition of these numbers. Since $B_{p-3}$ is a p-integral number, the above congruence makes sense.
This congruence can be checked by showing that the derivatives of both members are congruent mod p (using the congruence $X f'_{i}(X) \equiv f_{i+1}(X) \pmod{p}$ ) and treating the coefficients of $X^p$ and $X^{2p}$ separately (and using relations between Bernoulli numbers and sums of powers).
Do you know if there exists an "analogous" relation where $f_{p-4}(X)$ and perhaps $f_{p-4}(1-X)$ or $X^{p} f_{p-4}(1-1/X) ]$ appear ? Thanks in advance.
P.S. I put this question on the xkcd forum, but it remained unanswered. So, I put it here and I will say it on the xkcd forum.