# 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?

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You wrote the conjecture in a sloppy way by introducing $a$, $b$, $c$, and $\varepsilon$ all before bringing in the constant $C_\varepsilon$. Since the constant depends only on $\varepsilon$, it would be much better if you wrote it as: for any $\varepsilon > 0$ there is a constant $C_\varepsilon > 0$ such that for all pairwise coprime positive integers $a$, $b$, $c$ satisfying $a + b = c$, we have [inequality here]. –  KConrad Mar 29 '11 at 1:34
I suggest trying Rankin's method: $$R_{\epsilon,C}(n) = \sum_{a+b=n, \hskip3pt n > CN(a,b,n)^{1+\epsilon}} 1 \le \sum_{a+b=n} \frac n{CN(a,b,n)^{1+\epsilon}}.$$ You then want lower bounds on $N(a,b,n)$, at least on average over $a$. Sorting the right-hand side by $g = {}$gcd$(a,n)$ is one way to proceed. This idea made less immediate progress than I was hoping, but maybe there's some value to it. –  Greg Martin Mar 29 '11 at 18:51
Thanks Professor Martin! –  Stanley Yao Xiao Mar 29 '11 at 23:11