# Prove a sequence of numbers is composite

I'm doing a problem that states: Let $p_1,p_2,p_3...(p_n+1)$ be the first $n + 1$ primes in order.

Prove every number between $p_1 * p_2 * .... * p_n + 1$ and $p_1 * p_2 * ... * p_n + (P_n+1) - 1$ is composite (not including the two numbers).

I'm stuck on the fact that all those numbers must be a product of primes. How can I continue?

• Well see for example that 2×3×5×7 + 2 is div. by 2, 2×3×5×7 + 3 is div. by 3, do you get the idea? – Sawarnik Mar 16 '17 at 3:15

## 1 Answer

Note how every number $N$ between $p_1\cdot p_2\ldots\cdot p_n+1$ and $p_1\cdot p_2\ldots\cdot p_n+p_{n+1}$ can be written in the form: $$N=p_1\cdot p_2\ldots\cdot p_n+k$$ with $k$ an integer and $2\le k\le p_{n+1}-1$. This means that $k$ is divisible by at least one of the primes in the product $p_1\cdot p_2\ldots\cdot p_n$ and since that prime also divides the product, that prime will divide $N$. However, it will not be equal to that prime (let's call it $q$) because $p_1\cdot p_2\ldots\cdot p_n\ge q$ (do you see why?) and $k\ge 2$.

So now we know that there is a prime $q$ such that $q$ divides $N$, but is not equal to $N$. We may conclude that $N$ is composite.