Timeline for Mid-Square with all bits set
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 15, 2017 at 23:11 | comment | added | Kevin Buzzard | a/d is "close" to sqrt(bd) but at first glance it doesn't seem to be anywhere near close enough to say anything useful. Pell gives powerful things but we're doing Pell for bd and there are too many possibilities for b before we can get started, so I don't immediately see how to get it to give anything yet. There's a chance that the loop is the only way to go I guess. | |
| Jan 15, 2017 at 22:44 | comment | added | Gerhard Paseman | Let d=2^k. For m=2k even, the question boils down to finding c and b less than d such that bd^3 - c is a square of an integer . For such an a, a/d is close to sqrt(bd). Perhaps Pell equations or rational approximation of square roots might say what b are permissible. Gerhard "Looping Brain Instead Of Code" Paseman, 2017.01.15. | |
| Jan 15, 2017 at 22:15 | comment | added | Kevin Buzzard | @GerhardPaseman -- I don't understand the relevance of the equations with the cube in. Can you explain more? | |
| Jan 15, 2017 at 22:13 | comment | added | Gerhard Paseman | With reference to a^2 + c = bd^3, It seems bd must be very close to a square, and that Pell equations might say something about which n=2m have solutions. Gerhard "Perhaps This Will Inspire Someone" Paseman, 2017.01.15. | |
| Jan 15, 2017 at 20:36 | comment | added | Gerhard Paseman | Indeed. Except for my arithmetic mistake in my above comments I was thinking along these lines. More generally, one could look at (given d) certain constrained solutions of a^2 +c = bd^3, but I don't know enough theory to follow that up. Gerhard "Off To Practice Some Bit-shifting" Paseman, 2017.01.15. | |
| Jan 15, 2017 at 20:27 | history | answered | Kevin Buzzard | CC BY-SA 3.0 |