Distinct derivations of polynomial over finite field

31 Views Asked by At

I am a student studying algebra and cryptography.

I wonder below question is possible.

Can I make some polynomials $f(x)$ over finite field that all derivations $f^{(k)}(x)$ are distinct when x is fixed but picked randomly?

i.e. $\forall x \in F, \forall k \in \{ 0,...,n \}, \ where \ n \ is \ degree \ and \ F \ is \ finite \ field, \\ \exists f \ s.t. \ \forall f^{(k)}(x) \ are \ distinct \ when \ random \ x \ is \ fixed \ ?$

I think it will be impossible, but also wonder that is possible when the finite field or polynomial has some specific conditions...

If there exists some related works, please tell me..!!

1

There are 1 best solutions below

1
On BEST ANSWER

The choice of $x$ doesn't matter, because if $g(X) = f(X-x)$, $g$ is a polynomial of the same degree as $f$ and $g^{(k)}(x) = f^{(k)}(0)$. So let's choose $x = 0$ for convenience. Take any prime $p > n$. Then you can choose the coefficients $a_0, \ldots, a_n$ of $f(X) = a_0 + \ldots + a_n X^n$ so that $f^{(k)}(0) = k!\; a_k$ are congruent mod $p$ to any chosen $n+1$ values in $\{0,1,\ldots,p-1\}$.