most recent 30 from http://mathoverflow.net 2013-05-20T18:32:11Z http://mathoverflow.net/feeds/question/59906 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59906/a-question-related-to-the-abc-conjecture Stanley Yao Xiao 2011-03-28T23:53:56Z 2011-03-29T01:48:33Z

<p>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</p> <p>$$\displaystyle c \leq C_\epsilon N(a,b,c)^{1 + \epsilon}.$$</p> <p>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?</p>