Question

Say we have three infinite sequences $\{a_i\},\{b_i\},\{c_i\}$ of natural numbers, satisfying the equations $$a_1+b_1=c_1,\dots, a_n+b_n=c_n,\dots $$ Assume further that $gcd(a_i,b_i,c_i)=1$ for each $i$ and that $(a_i,b_i,c_i)\neq (a_j,b_j,c_j)$ for all $i,j$. Now let's define $S$ as the set of primes $p$ which divide $a_ib_ic_i$ for at least one $i$. From the S-unit theorem we know that $S$ has to be infinite.

Now the question is: Can $S$ be sparse? This can be taken to mean Dirichlet density zero, for example.

I haven't thought much about this but there are reasons to believe the answer is yes, indeed if there are infinitely many Mersenne primes $q=2^p-1$ then the equations $1+q=2^p$ give such a sparse $S$. However, I am looking for an unconditional result.

Answer 1

Yes, in fact, you can make $S$ grow as slowly as you like. This follows, for example, from the fact that there exist 3-term arithmetic progressions of primes $(p,q,r)$ with $\min(p,q,r)$ arbitrarily large. For each such arithmetic progression, you can take $(a_i,b_i,c_i)=(p,r,2q)$. Now just choose these arithmetic progressions one at a time, making sure that the primes involved each time are much larger than any primes appearing earlier.

Answer 2

Alternatively, every prime dividing $2^{2^k}+1$ must be 1 mod $2^k$, hence must be at least $2^k$. So if you take your triples of the form $(1,2^{2^k},2^{2^k}+1)$ with the values of $k$ sufficiently sparse, you get what you want.

Answer 3

An even simpler answer: take triples of the form $(2,n!-1,n!+1)$ for a sufficiently sparse sequence of values of $n$.