# 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

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.

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Thank you very much! –  Stanley Yao Xiao Aug 3 '14 at 0:50

## protected by Felipe VolochAug 2 '14 at 21:22

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