Many descendants

Denote $x^{p+1}-x+1$ by $f_p(x),$ or simply $f(x)$ if $p$ is assumed fixed. We will consider zeroes of $f_p$ in the finite field ${\mathbb F}_p={\mathbb Z/pZ}$ and its finite extensions (alternatively, one can consider the zeroes in the algebraic closure of $\mathbb F_p$ ).

First of all, note that $0$ and $1$ are never zeroes of $f_p.$ The derivative of $f_p$ is $x^p-1=(x-1)^p$ (recall that we are working in characteristic $p,$ so $(a+b)^p=a^p+b^p$ ). So $f$ and $f'$ have no common roots, therefore $f$ is a product of distinct irreducible monic polynomials with coefficients in $\mathbb F_p.$

Lemma 1. If $p\equiv 2 \pmod 3,$ then $f_p(x)$ has no roots in $\mathbb F_p.$ If $p\equiv 1 \pmod 3,$ then $f_p(x)$ has exactly two roots in $\mathbb F_p.$

Proof. For $a$ in $\mathbb F_p,$ we have $a^p =a$ (in $\mathbb F_p$ ). So $f_p(a)=0$ if and only if $a^2-a+1=0.$ This is equivalent to writing that $a^3+1=0$ and $a\neq -1.$ Alternatively, $(-a)^3=1$ and $-a\neq 1.$ If $p=3k+2,$ then $(-a)^3=1$ and $(-a)^{p-1}=1$ imply $-a=\frac{(-a)^{p-1}}{(-a)^{3k}}=1,$ hence there are no roots.
One the other hand, if $p=3k+1,$ we use the theorem that the multiplicative group of $\mathbb F_p$ is cyclic (i.e., there exists a multiplicative generator modulo $p$ ). So there are exactly two residues modulo $p$ that satisfy the above condition. (Note: alternatively, one can use the quadratic reciprocity law: the discriminant of the equation $a^2-a+1=0$ is $-3$ and the equation has two or zero solutions depending on whether or not it is a quadratic residue).

Lemma 2. Every root of $f_p(x)$ that is not in $\mathbb F_p$ lies in the field of order $p^3.$ (That is, it is a root of an irreducible cubic polynomial with coefficients in $\mathbb F_p.$

Proof. Suppose $f_p(a)=0.$ Then $a^p=\frac{a-1}{a}.$ Therefore, $a^{p^2}=\left(\frac{a-1}{a}\right)^p=\frac{a^p-1}{a^p}=\frac{\frac{a-1}{a}-1}{\left(\frac{a-1}{a}\right)}=\frac{-1}{a-1}$ Continuing this, $a^{p^3}=\left(\frac{-1}{a}\right)^p=\frac{-1}{a^p}=\frac{-1}{\frac{a-1}{a} -1}=a$ The field with $p^3$ elements consists of all $a$ such that $a^{p^3}=a,$ so the lemma is proven.

The above two lemmas imply that for $p\equiv 2 \pmod 3$ the polynomial $f_p$ is a product of $\frac{p+1}{3}$ irreducible monic polynomials of degree $3.$ On the other hand, if $p\equiv 1 \pmod 3,$ then $f_p$ is a product of two linear polynomials and $\frac{p-1}{3}$ polynomials of degree $3$ . It is easy to check that for the primes $p<100$ the largest number of factors is for $p=97,$ which is $2+\frac{96}{3}=34.$