TITLE: Every prime number > 19 divides one plus the product of two smaller primes? 
QUESTION [57 upvotes]: This is a part of my answer to  this question I think it deserves to be treated separately. 
 Conjecture Let $A$ be the set of all primes from $2$ to $p>19$. Let $q$ be the next prime after $p$. Then $q$ divides $rs+1$ for some $r,s\in A$. 
I wonder if this conjecture is already known. I checked it for all $p<692,000$. 
A reformulation of the conjecture is this (motivated by  Gjergji Zaimi's comment). Let $p > 19$ be prime. Let $A$ be the set of primes $< p$ considered as a subset of the cyclic multiplicative group $\mathbb{Z}/p\mathbb{Z}^*$. Then the product $A\cdot A$ contains $-1$. 
It is interesting to know how large $A\cdot A$ is. This seems to be related to Freiman-type results of Green, Tao and others. Also as Timothy Foo pointed out, perhaps Vinogradov's method of trigonometric sums can apply.

REPLY [11 votes]: Perhaps this might be another perspective on this problem. In an answer to a question I had previously asked on Math Overflow, "A generalized Möbius function?", the following paper of Addison was cited (which I simply copy from that answer):
A Note on the Compositeness of Numbers, A. W. Addison, Proceedings of the American Mathematical Society, Vol. 8, No. 1 (Feb., 1957), pp. 151-154
In this paper, it is proven that if one divides the integers into $q$ classes according to whether $\Omega(n)$ is $0,1,\dots,q-1 \bmod q$, $q>2$ then the function $C_{q,i}(x)$ which counts integers less than $x$ with $\Omega(n) \equiv i \bmod q$ satisfies
$$
C_{q,i}(x) - \frac{x}{q} = \Omega_{\pm}\left(\frac{x}{(\log x)^r}\right)
$$
where $r = 1-\cos(2\pi/q)$. (Here the variables $q$ and $r$ are taken from the paper and mean different things from elsewhere in this page.)
Let's assume that the relative frequencies of $\Omega(n) \bmod q$ should not differ much whether we consider $n\in [1,p^2]$ or $n\in (p\mathbb{Z}-1)\bigcap [1,p^2]$. Then we take $x$ to be about $p^2$ and $q$ to be about 
$$
f(p)=\max_{n \in (p\mathbb{Z}-1)\bigcap [1,p^2]}\Omega(n).
$$ 
Then a necessary (but not sufficient) condition for the conjecture is that $C_{f(p),2}(p^2)\geq 1$. I'm not really sure if one can take $x$ and $q$ in this relative range, but if so, then one can't even get a necessary condition from this because, using $1-\cos x \approx x^2/2$ for small $x$, we have 
$$
f(p) \gg (\log p)^{1-\cos(2\pi/f(p))}.
$$