TITLE: Proving there are only finitely many $n$-tuples of prime numbers such that $(p_{1}p_{2}...p_{n})|(p_{1}+k)(p_{2}+k)...(p_{n}+k)$
QUESTION [5 upvotes]: I have stumbled upon this claim: $\forall k\in \mathbb{N}$ there are only finitely many completely distinct $n$-tuples of prime numbers $p_{1},p_{2},...,p_{n}$ such that $(p_{1}+k)(p_{2}+k)...(p_{n}+k)$ is divisible by $p_{1}p_{2}...p_{n}$. There was no proof provided so I tried to have a go at this myself.
I tried expanding the terms but the results did not appear to help with the general case. I also attemped to split the problem into cases on the basis of parity of $n,k$ but without much success there either.
Is the right approach trying to directly express the form of such $n$-tuples and then enumerating or is there a neat method that allows us to enumerate right away?
I am grateful for any suggestions.

REPLY [2 votes]: Let $p_1,\ldots,p_n$ be such a tuple. Choose a sequence $q_0,q_1,\ldots\in\{p_1,\ldots,p_n\}$ such that $q_i|q_{i+1}+k$ for all $i$, and $q_0 = \max\{p_i\}$. Let $i_0$ be the smallest value of $i$ for which $q_{i}\neq q_{i+1}+k$. Write $\ell$ for the smallest prime not dividing $k$.
I claim $i_0< \ell$. If instead $i_0\geq\ell$, then since $k$ is coprime to $\ell$, one of the $\ell$ values $q_i=q_0-ik$ for $0\leq i\leq \ell-1$ would be divisible by $\ell$, hence equal to $\ell$, in which case $q_{i+1}=\ell-k\leq 1$ could not be prime.
Now, by assumption $q_{i_0} = q_0-i_0k$ is a proper divisor of $q_{i_0+1}+k$, so
$$
q_0 + k\geq q_{i_0+1} + k \geq 2(q_0-i_0k),
$$
which implies
$$
q_0\leq (2i_0+1)k <2\ell k.
$$
So we have a bound on $q_0=\max\{p_i\}$ depending only on $k$. This implies that for fixed $k$, there are only finitely-many tuples satisfying the given divisibility.