So I was working on one of the exercise question for school and I came across this question and I wanted to know if someone can help me in providing a proof for this question. So the question is:
Prove if $p_1, p_2,\dots , p_n$ are distinct prime numbers where $p_1 = 2$ and $n > 1$ then $p_1p_2p_3 \dots p_n + 1$ can be written in the form $4k + 3$ for some $k \in \mathbb N$.
Pose $P = p_1\dots p_n +1$. Then, as $p_1 = 2$, you know that $P$ is odd, thus either in the form $4k+1$ or $4k+3$.
Suppose it is of the form $4k+1$. Then $P-1 = p_1\dots p_n = 4k$. Hence, $p_2\dots p_n = 2k$, but each $p_i$, $i\geq 2$ is odd. That's a contradiction.