The current discussion[0] about the abc conjecture has a number of confused questions over what it's saying. This is a link to a presentation I found useful to say what the conjecture is about.
I may have misunderstood something, but I believe it is simple to prove that R cannot equal c.
Because c is relatively prime to both a and b, there exists some prime p which b is divisible by and c is not. Because b is divisible by p, R=rad(a,b,c) is also divisible by p. Since R is divisible by p and c is not, R!=c.
Comments
The current discussion[0] about the abc conjecture has a number of confused questions over what it's saying. This is a link to a presentation I found useful to say what the conjecture is about.
[0] http://news.ycombinator.com/item?id=4502856
I may have misunderstood something, but I believe it is simple to prove that R cannot equal c.
Because c is relatively prime to both a and b, there exists some prime p which b is divisible by and c is not. Because b is divisible by p, R=rad(a,b,c) is also divisible by p. Since R is divisible by p and c is not, R!=c.
The comments on the article make it clear that c!=R using reasoning effectively the same as yours, but yes, nice proof.