TITLE: Divisibility bound on sums of numbers
QUESTION [5 upvotes]: Suppose $p$ and $q$ are two prime numbers. Let $N(p)$ be the set of natural numbers that are divisible only by primes $\leq p$. Define $N(q)$ similarly.
If $N(p,q)$ is the set of numbers that can be written as a sum of a number in $N(p)$ and a number in $N(q)$, what is the upper bound on the primes that can divide numbers in $N(p,q)$, if any?

REPLY [4 votes]: For every prime number $p$ we have $1,2^{2^n}\in N(p)$, and hence $2^{2^n}+1\in N(p,q)$ for any pair of prime numbers $(p,q)$. The number $F_n:=2^{2^n}+1$ is called the $n$-th Fermat number. It is currently unknown whether there are only finitely many Fermat primes or not, despite the fairly generous attention it receives.
It is known, however, that every prime factor $p$ of the Fermat number $F_n=2^{2^n}+1$ satisfies
$$p\equiv1\pmod{2^{n+2}}.$$
In particular every prime factor of $F_n$ satisfies $p>2^{n+2}$, and hence there exist arbitrarily large prime numbers dividing numbers of the form $2^{2^n}+1$. This shows that $N(p,q)$ is unbounded for all pairs of primes $(p,q)$.