- **enzocreti**
(*https://www.mersenneforum.org/forumdisplay.php?f=156*)

- - **216 is the only cube that can be written in this way?**
(*https://www.mersenneforum.org/showthread.php?t=25388*)

216 is the only cube that can be written in this way?216=3^2*(182^2-331*10^2)
is there any other positive cube that can be written as a^2*(b^2-r*10^2)? with a and b positive and coprime, r prime? |

[QUOTE=enzocreti;540238]216=3^2*(182^2-331*10^2)
is there any other positive cube that can be written as a^2*(b^2-r*10^2)? with a and b positive and coprime, r prime?[/QUOTE]Since d^3 = d^2 * d, all you are asking is whether d = (b^2 - 100r) is positive. In other words, for what values of b and r is b^2 > 100r and b is co-prime to d.. It should now be obvious that there are an infinite number of solutions. Picking just one pretty much at random, let r =1. In that case b^2 > 100, or b> 10. One such value is 11. Plugging in the numbers, d=121-100 = 21. Indeed 21^3 = 9261 = 441 * (11^2 - 10 * 1^2). Why do you post questions which are so trivial to answer with even the slightest amount of thought? |

[QUOTE=xilman;540245]r =1[/QUOTE]
Neee.. gotcha! (so long I waited for that! :razz:) r must be prime, I guess you didn't know that 1 is not considered a prime :rofl: |

[QUOTE=LaurV;540452]Neee.. gotcha! (so long I waited for that! :razz:)
r must be prime, I guess you didn't know that 1 is not considered a prime :rofl:[/QUOTE]Fairy Nuff: I over-looked that condition. OK, choose r=2. Prime, right? Let b=17, so B^2-100r = 289-200 = 89 = a. All are positive and co-prime. Let b=19, so b^2-100r = 361-200 = 161 = 7 * 23 = a. All are positive and co-prime. Let b=21, so a=241 which is a prime. And so on. It is still a trivial problem. |

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