Problem 1875

Let $p$ be a prime with $p \mid d$. We know that $p \mid a_{i}^{n}+a_{1} \cdot a_{2} \cdots a_{n}$ for all $i$. If $\exists i$ such that $p \mid a_{i}$, then $p \mid a_{1} \cdot a_{2} \cdots a_{n}$. Since $p \mid a_{j}^{n}+a_{1} \cdot a_{2} \cdots a_{n}$ for all $j$, we find $p \mid a_{j}$ for all $j$, which contradicts to the hypothesis.

Thus we consider the case $p \nmid a_{i}$ for all $i$, i.e. $p \nmid a_{1} \cdot a_{2} \cdots a_{n}$. We have the $n$ congruences
$$\begin{aligned} & a_{1}^{n} \equiv -a_{1} \cdot a_{2} \cdots a_{n} \pmod{p}, \\ & a_{2}^{n} \equiv -a_{1} \cdot a_{2} \cdots a_{n} \pmod{p}, \ldots, \\ & a_{n}^{n} \equiv -a_{1} \cdot a_{2} \cdots a_{n} \pmod{p} \end{aligned}$$
and by multiplying them, we obtain
$$\left(a_{1} \cdot a_{2} \cdots a_{n}\right)^{n} \equiv (-1)^{n} \cdot \left(a_{1} \cdot a_{2} \cdots a_{n}\right)^{n} \equiv -\left(a_{1} \cdot a_{2} \cdots a_{n}\right)^{n} \pmod{p}$$
which yields $p=2$.

This shows that $d$ must be a power of $2$. Suppose, now, that $4 \mid d$. Clearly, $a_{1}, a_{2}, \ldots, a_{n}$ must be all odd. We consider two cases

1. If $a_{1} \cdot a_{2} \cdots a_{n} \equiv 3 \pmod{4}$, then $\exists i$ such that $a_{i} \equiv 3 \pmod{4}$ then $a_{i}^{n}+a_{1} \cdot a_{2} \cdots a_{n} \equiv 3+3 \equiv 2 \pmod{4}$ which contradicts with $4 \mid d$.

2. If $a_{1} \cdot a_{2} \cdots a_{n} \equiv 1 \pmod{4}$, then $\exists i$ such that $a_{i} \equiv 1 \pmod{4}$ then $a_{i}^{n}+a_{1} \cdot a_{2} \cdots a_{n} \equiv 1+1 \equiv 2 \pmod{4}$ which also contradicts with $4 \mid d$. (In both cases, we made use of $3^{n} \equiv 3 \pmod{4}$.)

Therefore, $d=1$ or $d=2$. If $a_{1}, a_{2}, \ldots, a_{n}$ are all odd, then $d$ must be even, hence $d=2$. If some of $a_{1}, a_{2}, \ldots, a_{n}$ are even and some of them odd, then $d$ must be odd, hence $d=1$. As such, both of these values are realized.