Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Clearly, $f = (X^{n_1}+X^{n_2}+\dots+X^{n_p}-p)/(X^d-1)$ is a polynomial with integral coefficients. If all $n_i$ are equal, then $f = p$, a constant polynomial; therefore assume at least two of the $n_i$ are distinct. By Gauss' Lemma, it is sufficient to prove $f$ irreducible in $\mathbb{Z}[X]$. We claim that the roots of $f$ all lie outside the closed unit disc in the complex plane. Assuming the claim, suppose, if possible, that $f = gh$ is a non-trivial factorization of $f$ in $\mathbb{Z}[X]$. Since $f(0) = p$, a prime, one of the numbers $|g(0)|$, $|h(0)|$ is 1. To reach a contradiction, notice that $g(0)$ and $h(0)$ are both products of roots of $f$, all of which lie outside the closed unit disc, so $|g(0)|$ and $|h(0)|$ are both greater than 1.

Back to the claim, write
$$X^{n_1} + X^{n_2} + \dots + X^{n_p} - p = (X^d - 1)f. \quad (*)$$
Suppose, if possible, that $f$ has a root $z$ in the closed unit disc. Then
$$p = |z^{n_1} + z^{n_2} + \dots + z^{n_p}| \le |z|^{n_1} + |z|^{n_2} + \dots + |z|^{n_p} \le p,$$
which forces $|z| = 1$ and $z^{n_k - n_\ell} > 0$ whatever $k$ and $\ell$. Consequently, the $z^{n_k}$ are all equal to some complex number $w$, so
$$p(w - 1) = z^{n_1} + z^{n_2} + \dots + z^{n_p} - p = 0;$$
that is, $w = 1$. Now write $d = n_1 t_1 + n_2 t_2 + \dots + n_p t_p$, for some integer numbers $t_1, t_2, \dots, t_p$, to get
$$z^d = z^{n_1 t_1 + n_2 t_2 + \dots + n_p t_p} = (z^{n_1})^{t_1} (z^{n_2})^{t_2} \dots (z^{n_p})^{t_p} = 1.$$
Finally, to reach a contradiction, evaluate the formal derivatives of both sides of $(*)$ at $z$. The left-hand side is $(n_1 + n_2 + \dots + n_p)/z$, while the right-hand side vanishes.