NMO Selection Tests for the Balkan and International Mathematical Olympiads

Clearly, $f=(X^{n_1}+X^{n_2}+\cdots +X^{n_p}-p)/(X^d-1)$ 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 $$X^{n_1}+X^{n_2}+\cdots +X^{n_p}-p=(X^d-1)f.(\ast )$$ Suppose, if possible, that $f$ has a root $z$ in the closed unit disc. Then $$p=|z^{n_1}+z^{n_2}+\cdots +z^{n_p}|\le |z{|}^{n_1}+|z{|}^{n_2}+\cdots +|z{|}^{n_p}\le p,$$ 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 $$p(w-1)=z^{n_1}+z^{n_2}+\cdots +z^{n_p}-p=0;$$ that is, $w=1$. Now write $d=n_1{t}_1+n_2{t}_2+\cdots +n_p{t}_p$, for some integer numbers $t_1,t_2,\dots ,t_p$, to get $$z^d=z^{n_1{t}_1+n_2{t}_2+\cdots +n_p{t}_p}=(z^{n_1}{)}^{t_1}(z^{n_2}{)}^{t_2}\dots (z^{n_p}{)}^{t_p}=1.$$ Finally, to reach a contradiction, evaluate the formal derivatives of both sides of $(\ast )$ at $z$. The left-hand side is $(n_1+n_2+\cdots +n_p)/z$, while the right-hand side vanishes.