On 8/2/06, Christian Boyer <cboyer@club-internet.fr> wrote:
With p,q,r consecutive prime integers, we may try to extend the problem.
For example, what are the solutions of: abs(p*q*r - n^3) < 20
Are 2* 3* 5 - 3^3 = 3 and 7*11*13 - 10^3 = 1 the only solutions???
Instead of using 20 or any other integer, same question with:
abs(p*q*r - n^3) < r
Or some other maximum?
Christian.
I searched for any solutions to this last inequality among consecutive primes, and there are none besides these two under r = 1,000,000, so seems like there are none. For the earlier inequality, just hunting for good values of abs(p*q*r - n^3) up to r = 100,000, I found the following in addition to your solutions: 3*5*7 = 105 is 20 away from 5^3 5*7*11 = 385 is 42 away from 7^3 13*17*19 = 4199 is 103 away from 16^3 11*13*17 = 2431 is 234 away from 13^3 So as you can see, all the good solutions happen right at the beginning. In fact the only ones where I see a distance less than 1000 from a cube are the above and 31*37*41 and 29*31*37 and 19*23*29 and 17*19*23 and 23*29*31 (which is 994 away). And indeed, under 100,000 the record-setters for farthest distance from a cube are the triples of primes starting with 99929, 99923, 99961, and 99971 ... pretty strong experimental evidence, anyway, that there aren't any more good solutions. [If my calculations are correct, that record, for 99929, is 10038882965 away from a cube!] This makes it look like the ratio (distance from cube)/p^2 might be a good measure of how close to a cube we've landed. After all, the distance between cubes is about 3p^2, so the maximum possible ratio here is about 1.5 or so (well, a little more, since I chose p as my denominator, perhaps unwisely). Now the best triple under 100,000 is 99817 * 99823 * 99829 which is "only" 3593628 from a cube. Seems to me that what this is showing is that (q-6) * q * (q+6) is pretty close to q^3 ... not so exciting after all, I suppose, since q-2, q, q+2 and q-4, q, q+4 can't all be prime. The best one not of that form is 99859 * 99871 * 99877 ... so still maybe not an interesting measure. Maybe to look for some exceptional ones, look at the ratio of the distance-to-cube divided by p instead of p^2 ... now the best one is 7*11*13 of course, and the second-best is 3*5*7, and indeed, it's all the little ones having the best ratios again ... well, up to about 30. Between 30 and 40 (again, measuring the ratio of distance-from-cube to p), we have 109*113*127 37*41*43 17*19*23 and then our old friend 99817 99823 99829 again. In the limit, q-6, q, q+6 will equal q^3 - 36q, so the ratio will equal 36q/(q-6) the way I have computed it -- I suppose I should have used q instead of p -- but anyway the limit there is 36, and apart from the first few terms I see none other under 100,000 where the ratio is less than 36. So indeed, it would appear that by any measure, all the "good" solutions are either with very small p,q,r or with q-6, q, q+6. --Joshua Zucker