Prime numbers satisfying a congruence relation Let $p_1, p_2, \ldots, p_n$ prime numbers such that for each $i, 1 \le i \le n$, 
$$\prod\limits_{j \neq i} p_j  \equiv 1 \pmod{p_i}.$$ For example, $2,$ $3$ and $5$ satisfies these conditions. Then, it is true that one of $p_1, p_2, \ldots, p_n$ must be $2$?
 A: Just writing some interesting information related to the question here: OEIS has a list of solutions for this in A236434. The numbers made multiplying each list are called Giuga numbers. It seems that all known numbers are even (so there is no counterexample to your property), and if there is any, it has to have at least 14 different prime factors.
A: It is not what you are asking but you can make some conjecture on the density of solution, assuming independence and uniform distribution of congruences. Based on this you can do numerical experiments and compare with the truth.
Let $m=\prod_{j=1}^i p_j$ squarefree, if for some $q$ the set $\{ p_1,\ldots, p_i,q\}$ satisfies your condition then $q$ is the unique $a\in [1,m]$ such that $a \frac{m}{p_j}\equiv 1\bmod p_j$. Such a $a$ always exists, what is rare is that $a$ is prime and $m\equiv 1\bmod a$.
Thus we can estimate the probability that $m$ works as $\approx\frac1{\log m}\times \frac{\log m}m$ where $\frac1{\log m}$ is probability that $a$ is prime and $\frac{\log m}m\approx \Bbb{E}[\frac1a]$ is the probability that $m\equiv 1\bmod a$.
And hence the number of solutions with $m\le M$ is expected to be $$\approx \sum_{m\le M} \frac{|\mu(m)|}{m}\approx C \log M$$
