SELECTION and TRAINING SESSION

Denote the set of all residues modulo $p$ by ${\mathbb{F}}_p$. Fermat's little theorem implies that the roots of the polynomial $x^{p-1}-1$ are all nonzero elements of ${\mathbb{F}}_p$. Since $x^{p-1}-1=(x^d-1)(x^{d(m-1)}+x^{d(m-2)}+\cdots +x^d+1)$ where $p-1=md$, the polynomial $x^d-1$ divides $x^{p-1}-1$. In this equality the sum of the degrees of the factors equals the number of roots, hence $x^d-1$ has $d$ roots. Denote these roots $a_1,a_2,\dots ,a_d$.

Consider all polynomials of the form $x^d-C$. If such polynomial has a root $b$ then it has $d$ roots $a_{1}b,a_{2}b,\dots ,a_{d}b$ and cannot have more roots than its degree, so it has exactly $d$ roots. On the other hand, each $b\in {\mathbb{F}}_p\setminus \{0\}$ is a root of such polynomial, namely $x^d-b^d$. Therefore ${\mathbb{F}}_p\setminus \{0\}$ is the disjoint union of $m$ sets, consisting of $d$ roots of some polynomial $x^d-C$. Such partition satisfies the problem condition since by Vieta's theorem the sum of the roots and the sum of the pairwise products of the roots of $x^d-C$ are congruent to zero modulo $p$, whence the sum of their squares is likewise congruent to zero.