Observe that for any $\epsilon > 0$ there are infinitely many triples of
$c^\epsilon$-[smooth](http://en.wikipedia.org/wiki/Smooth_number) coprime positive integers $a$, $b$ and $c$ such
that $a + b = c$. -- Considering triples of the form $(2^n-1,1,2^n)$
and the factorizations of the polynomials $x^n-1 \in \mathbb{Z}[x]$ into
cyclotomic polynomials, this holds since the set of quotients $n/\varphi(n)$
for positive integers $n$ is unbounded.


How much can this obvious observation be improved, i.e. how much can the
smoothness bound $c^\epsilon$ be lowered such that there are still infinitely
many such triples which satisfy that bound?


Or to be more concrete: is there an $\epsilon > 0$ such that there are
infinitely many triples of $e^{(\ln c)^{1-\epsilon}}$-smooth coprime
positive integers $a$, $b$ and $c$ satisfying $a + b = c$?
-- And if yes, which is the supremum of the set of values of $\epsilon$
for which this holds?



Smooth sums of coprime smooth integers
Balog and Sarkozy (Stud. Sci. Math. Hungarica 1984) showed that large $N$ may be written as $x+y+z$ where $x$, $y$, and $z$ are all $\exp(3\sqrt{\log N \log \log N})$ smooth. An analogous result applies to $a+b=c$, answering your question with a bound of the form 
$\exp((\log c)^{1/2+\epsilon})$. Much more is expected to be true. For a discussion of this problem see the paper of Lagarias and Soundararajan (Proc. London Math. Soc; <http://arxiv.org/abs/1102.4911> ) where it is shown on GRH that there are solutions to $a+b=c$ that are $(\log c)^{8+\epsilon}$ smooth.


Added: Recent impressive work of [Adam Harper](http://arxiv.org/abs/1408.1662) obtains unconditional results of about the same quality as Lagarias and Soundararajan -- that is, solutions to $a+b=c$ with $a$, $b$ and $c$ being $(\log c)^A$ smooth for some constant $A$.


