Team selection tests for GMO 2018

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 $$\begin{array}{rl}& a_1^n\equiv -a_1\cdot a_2\cdots a_n(\mathrm{mod}p),\\ & a_2^n\equiv -a_1\cdot a_2\cdots a_n(\mathrm{mod}p),\dots ,\\ & a_n^n\equiv -a_1\cdot a_2\cdots a_n(\mathrm{mod}p)\end{array}$$ and by multiplying them, we obtain $${\left(a_1\cdot a_2\cdots a_n\right)}^n\equiv (-1{)}^n\cdot {\left(a_1\cdot a_2\cdots a_n\right)}^n\equiv -{\left(a_1\cdot a_2\cdots a_n\right)}^n(\mathrm{mod}p)$$ which yields $p=2$.

This shows that $d$ must be a power of $2$. Suppose, now, that $4\mid d$. Clearly, $a_1,a_2,\dots ,a_n$ must be all odd. We consider two cases

1. If $a_1\cdot a_2\cdots a_n\equiv 3(\mathrm{mod}4)$, then $\exists i$ such that $a_i\equiv 3(\mathrm{mod}4)$ then $a_i^n+a_1\cdot a_2\cdots a_n\equiv 3+3\equiv 2(\mathrm{mod}4)$ which contradicts with $4\mid d$.

2. If $a_1\cdot a_2\cdots a_n\equiv 1(\mathrm{mod}4)$, then $\exists i$ such that $a_i\equiv 1(\mathrm{mod}4)$ then $a_i^n+a_1\cdot a_2\cdots a_n\equiv 1+1\equiv 2(\mathrm{mod}4)$ which also contradicts with $4\mid d$. (In both cases, we made use of $3^n\equiv 3(\mathrm{mod}4)$.)

Therefore, $d=1$ or $d=2$. If $a_1,a_2,\dots ,a_n$ are all odd, then $d$ must be even, hence $d=2$. If some of $a_1,a_2,\dots ,a_n$ are even and some of them odd, then $d$ must be odd, hence $d=1$. As such, both of these values are realized.