# Can we differentiate Fermat's little theorem?

From Fermat's little theorem and factor theorem, for any $$x \in \mathbb{Z}/p\mathbb{Z}$$, $$x(x-1)(x-2)\cdots(x-p+1)\equiv x^p-x$$ is satisfied. If we take derivative of this, we get $$\sum_{i=0}^{p-1}\prod_{j=0,i\neq j}^{p-1}(x-j)\equiv -1.$$ This formula can be derived by another way. \begin{align}&\sum_{i=0}^{p-1}\prod_{j=0,i\neq j}^{p-1}(x-j)\\ \equiv&\prod_{j=0,j\neq x}^{p-1}(x-j) +\sum_{i=0,i\neq x}^{p-1}\frac{1}{x-i}\prod_{j=0}^{p-1}(x-j) \\ \equiv&\prod_{j=1}^{p-1}j+\sum_{i=1}^{p-1}i\prod_{j=0}^{p-1}(x-j)\\ \equiv& -1 \end{align}

I prefer former way (because it's simpler!). However, I don't know why I can apply derivative although it's mod p. Can we prove it's ok to apply derivative?

• You can formally define the derivative of polynomials in $F[x]$, and then prove that things like the product rule, etc, are satisfied. Commented Nov 10, 2019 at 23:03
• @littleO : I think the difficult point is how to get from a point wise equality ($P(x) \equiv Q(x)$ for all $x \in \mathbb{Z}$) to a polynomial equality ($P \equiv Q$). Commented Nov 10, 2019 at 23:20

Namely, let us denote $$P \equiv Q \pmod p$$ if all the coefficients of $$P-Q$$ are multiples of $$p$$ (we'll call that polynomial equality mod $$p$$). Notice that polynomial equality implies point wise equality :
$$P \equiv Q \pmod p \implies \forall x \in \mathbb{Z}, P(x) \equiv Q(x) \pmod p$$
But beware the converse isn't true in general (take for example $$P=X^p-X$$ and $$Q=0$$). Now we prove that the polynomial equality can be differentiated (while point wise equality cannot, using the same counter example) : indeed, if we assume $$P \equiv Q \pmod p$$, then we can write $$P(X)-Q(X) = p R(X)$$ with $$R \in \mathbb{Z}[X]$$. And by differentiating, we get $$P'(X) - Q'(X) = p R'(X)$$, so $$P' \equiv Q' \pmod p$$.
Now let's denote $$P = X(X-1)(X-2)\ldots (X-p+1)$$ and $$Q = X^p -X$$. And let's prove that $$P \equiv Q \pmod p$$, which will allow us to differentiate. Let us consider $$R = P-Q$$ and notice that $$\deg R = p-1$$ because the terms of degree $$p$$ of $$P$$ and $$Q$$ cancel out. Now remember that $$k = \mathbb{Z}/p\mathbb{Z}$$ is a field, and that $$R(a) = 0$$ for all $$a \in k$$. In other words, the polynomial $$R \in k_{p-1}[X]$$ has $$p$$ roots in $$k$$, so it must be $$0$$. Which means $$P-Q \equiv 0 \pmod p$$.