# I need help in solving this proof when $p_1$, $p_2, \dotsc, p_n$ are distinct prime numbers [duplicate]

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 $k \in \mathbb N$.

## marked as duplicate by rogerl, Matthew Towers, Clement C., Community♦Dec 2 '15 at 22:40

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## 1 Answer

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.

• can you please explain your answer a little bit more I mean like in English a little bit more – syed Dec 2 '15 at 16:54
• You know that there are 4 choices for a number : $4k$, $4k+1$, $4k+2$ and $4k+3$. For an odd number, you can eliminate $4k$ and $4k+2$. Suppose then that it is $4k+1$. Then $P-1$ is a multiple of 4, ie in the form $4k$. So, $\frac{P-1}{2}$ is in the form $2k$. But $\frac{P-1}2$ is the product of all your primes, except 2. As any prime greater than 2 is odd, the product cannot be even. Thus, there is a contradiction, and the only other option is $4k+3$, ie exactly what you want. – Kevin Quirin Dec 2 '15 at 16:58