A question related to the abc conjecture The abc conjecture asserts that whenever $a,b,c$ are pairwise coprime positive integers such that $a + b = c$ and $\epsilon > 0$, there exists a constant $C_\epsilon > 0$ (which depends on $\epsilon$ but not on $a,b,c$) such that if $N(a,b,c) = \displaystyle \prod_{p | abc} p$ is the radical of $a,b,c$, we have
$$\displaystyle c \leq C_\epsilon N(a,b,c)^{1 + \epsilon}.$$
Now, if we define $R(n)$ to be the number of ways of writing $n$ as the sum of two positive integers $a,b$ such that $a,b,n$ are pairwise coprime, then infinitely often (when $n$ is prime) we have $R(n) = n -1$. What if we defined $R_{\epsilon, C}(n)$ to be the number of ways of writing $n = a + b$ and $n > C N(a,b,n)^{1 + \epsilon}$? If the $abc$-conjecture is true then $R_\epsilon(n)/n$ should tend towards 0 (in fact, if the conjecture is true, then $R_\epsilon(n) = 0$ for all $n$ sufficiently large). Of course, this is a much weaker statement (one 'almost' version of $abc$ conjecture if you will) than the full conjecture. Is anything of this sort accomplished?
 A: For any $n$ the number of relatively prime $a,b<n$ such that
$n > ({\rm rad}(ab))^{1+\epsilon}$ is $o(n)$, indeed $o(n^{1-\epsilon'})$
for any $\epsilon' < \epsilon / (1+\epsilon)$.  Therefore this bound
is true a fortiori of the number of such $a,b$ for which $a+b=n$.
We use the following lemma:
For all $\delta > 0$ there exists $C_\delta$ such that for every $r,n$
the number of solutions of ${\rm rad}(x) = r$ with $x\leq n$
is at most $C_\delta n^\delta$.
Proof: We may assume $r$ squarefree, else the number of solutions is zero.
Then (even without the condition $x \leq n$) we compute
$$
\sum_{{\rm rad(x)} = r} x^{-\delta}
 = \prod_{p|r} \, (p^{-\delta} + p^{-2\delta} + p^{-3\delta} + p^{-4\delta} + \cdots)
 = \prod_{p|r} \frac{p^{-\delta}}{1 - p^{-\delta}}.
$$
All summands are positive, and each solution with $x \leq n$ contributes
at least $n^{-\delta}$, so the number of $x \leq n$ terms in the sum
is at most $n^\delta \prod_{p|r} p^{-\delta} / (1 - p^{-\delta})$.
But each factor in this product is less than $1$ except for the
finitely primes $p \leq 2^{1/\delta}$.  This proves the lemma with
$$
C_\delta = \prod_{p \leq 2^{1/\delta}} \frac{p^{-\delta}}{1 - p^{-\delta}}.
\qquad \Box
$$
Now, given $\epsilon$, the number of pairs $(r,r')$ such that
$n > (rr')^{1+\epsilon}$ is asymptotically proportional to
$n^{1/(1+\epsilon)} \log n$.  By the lemma, each one arises as
$({\rm rad}(a), {\rm rad}(b))$ at most $C_d^2 n^{2\delta}$ times
with $a,b<n$.  The $o(n^{1-\epsilon'})$ claim follows
because $\delta$ is an arbitrary positive number.
