# Argue that none of the $q_j$ are 3 and that each $q_j$ is different from each $p_i$ for $7(p_1\cdot p_2\dotsb p_s)+3=q_1\cdot q_2 \dotsb q_t$

Let $7(p_1\cdot p_2\dotsb p_s)+3=q_1\cdot q_2 \dotsb q_t$ where the $p_1,p_2,\dots,p_s$ and $q_1,q_2,\dots,q_t$ are primes and where none of the $p_i$ is $3$. Argue that none of the $q_j$ are 3 and that each $q_j$ is different from each $p_i$.

I'm not sure how to go about solving this. Therefore I will start off by taking the first 10 primes (excluding the number $3$ for some odd unknown reason to me) and making sure that when $7(\text{product of primes})+3=\text{a prime}$. If this is not true, then I would think this suggests that what was stated is false.

$$7(2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29\cdot31)+3=467974476973$$

However, $467974476973$ is not a prime, because $14519|467974476973$. Therefore, I clearly don't know what I'm talking about.

Perhaps, my strategy excluding the $3$ was incorrect.

$$7(2\cdot 3\cdot5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 29)+3=45287852613$$

Again, I am wrong, because $3|45287852613$, obviously.

In other words, I need some type of assistance regarding how to solve what is being asked.

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Hint $\$ Let $\rm\ P = 7\, p_1 p_2 \cdots p_s\$ and $\rm\ q = q_i.$

Since $\rm\, \ \ q\mid P\!+\!3,\quad q\mid P \iff q\mid 3 \iff 3\mid q,\:$ by $\rm\:q\:$ prime.

But $\rm\: 3\mid q\mid P\!+\!3\:\Rightarrow\:3\mid P,\:$ contra hypothesis.

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To show no $q$ is $3$, assume some $q$ is $3$, and show this implies some $p$ is $3$.

To show no $q$ is a $p$, assume some $q$ is a $p$, and show that said $q$ would then also have to be a factor of $3$ ---- then show $q$ can't be a factor of $3$.

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