5
$\begingroup$

Let $S_{n,k}$ be the set of all numbers that can be written as the product of $n$ odd primes plus $2k$. Is there integers $n>1$ and $k>1$ such that $S_{n,k}$ contains finite number of primes?

  • 0
    Are you asking if $S_{n,k}$ is non-empty for some $n$ and $k$ or do you want to know if $S_{n,k}$ is finite for some $n$ and $k$? (You say "contains finite number of primes" which is a bit unclear to me...)2011-10-07

1 Answers 1

3

Assuming Schinzel's Hypothesis H, these sets are always infinite:

Step 1: Let $t$ be the product of $n-1$ primes such that $\gcd(t,2k)=1$.

Step 2: Let $f(x)=x$ and $g(x)=tx+2k$. Define $Q(x)=x(tx+2k)$.

Hypothesis H asserts that if $Q(x)$ has no fixed prime divisor $q$, then $f(x)$ and $g(x)$ are simultaneously prime an infinite number of times.

Step 3: If $x=1$ then $q$ divides $Q(x)=t+2k$, and thus $q$ is odd. If $x=2$, then $q$ divides $Q(x)=2(2t+2k)$, and thus $q$ divides $t+k$.

Step 4: Since $q$ divides both $t+2k$ and $t+k$, we find $q$ divides both $k$ and $t$, contradicting that $\gcd(t,2k)=1$.

Hence $Q(x)$ has no fixed prime divisor, and Hypothesis H applies. Thus, since $g(x)$ has the desired form, we conclude that $|S_{k,n}|=\infty$.