Find all $n\in\mathbb N$, $n>3$ such that $p(n)=2n+16$, where $p(n)$ is the product of all the prime numbers less than $n$. 
Find all $n\in\mathbb N$, $n>3$ such that $p(n)=2n+16$, where $p(n)$ is the product of all the prime numbers less than $n$.

E.g. $p(7)=2\cdot3\cdot5$ (and $n=7$ is a solution).
Let $$p(n)=2\cdot3\cdot p_3\cdot p_4\cdot\ldots\cdot p_k$$Then $n\in(p_k;p_{k+1}]$.
And $$2n+16=2\cdot3\cdot p_3\cdot p_4\cdot\ldots\cdot p_k$$ $$n+8=3\cdot p_3\cdot p_4\cdot\ldots\cdot p_k$$
Therefore $$n\equiv 1\pmod 3$$
And we know that $n$ is odd. Hence the number is of the form $n=6l+1$ for some natural number $l$. So $$2l+3=p_3\cdot p_4\cdot\ldots\cdot p_k$$
I wonder if we could get to some kind of a restriction, i.e. $p(n)\le a$ for some natural number $a$. Or something else. I'd appreciate some ideas. Thanks.
 A: There are several ways to obtain the lower bound $n^2<p(n)$ for all $n\ge 8$. See for example here.  It follows that $2n+16>n^2$, which clearly excludes all $n\ge 8$. So it is enough to check the formula for $n=4,5,6,7,8$. Indeed, only for $n=7$ there is equality.
A: This answer is probably more what you are looking for.
As you already noticed, $3\mid n-1$. Suppose $n>13$, so $n-10$ is not prime.
Now $p(n-10)\mid p(n)=2(n-10)+36$. From here we see that every prime divisor $q\mid n-10$ satisfies $q\mid p(n-10)$ and $q\mid n-10$ so $q\mid36$. This means $n-10$ is of the form $2^a3^b$.
Because $4,9\nmid p(n)$ we have $a=0$ and $b=1$. Then $n=13$, which contradicts our assumption $n>13$.
So $n\leq13$, and it's not hard to check that only $7$ works.
A: Bertrand's postulate (also known as Chebyshev's theorem) states that for every integer $x>1$ there is a prime between $x$ and $2x$. Applying this for $\lfloor\frac n2\rfloor$ and $\lfloor\frac n4\rfloor$, we get an interesting lower bound on $p(n)$. I think you can continue, knowing this...
