2 Aug
2006
2 Aug
'06
12:40 p.m.
Almost certainly n=10 is the only case where n^3 + 1 is the product of three consecutive primes. As others have noted, n^3+1 = (n+1)(n^2-n+1). As noted, the middle prime must be n+1. Taking p = n+1, let p-i be the preceding prime, and p+j be the following prime. Then this equation can be rewritten as p(p-i)(p+j) = p(p^2-3p+3), which simplifies to (j - i) p - i j = -3p + 3, or p = (i j + 3) / (j - i + 3). Since the denominator is at least 1 (j and i are both even, so it is not zero), this means that p is (at least) approximately the square of (either of) the prime gaps around it. While it is unproven, it is almost certainly true that prime gaps are asymptotically considerably less than this. Franklin T. Adams-Watters