![]() |
|
|
#1 |
|
Nov 2004
32 Posts |
At some time ago i conducted a study for finding smaller bounds on the prime factors for integer composite numbers of type n=p*q, n>0
It is already known that 1<=p<=sqrt(n) and sqrt(n)<=q<=n. I was able to find smaller bounds by an empirical method (still remains to conduct more tests to validate). My question is: have these results pratical interest? or have these results cientific interest that justifies a publication in a jounal? |
|
|
|
|
|
#2 | |
|
Nov 2003
22·5·373 Posts |
Quote:
function of n. (2) Your claim of "smaller bounds", is therefore nonsense. Furthermore, one does not validate mathematics by "conducting tests". (3) You ask whether "these results" have any practical interest, but you have not presented any results.
|
|
|
|
|
|
|
#3 | |
|
Nov 2004
910 Posts |
Quote:
(2) There were many unproven and usefull theories until... someone proved it! (3) But is my intention to do it. I shall apply it to the infamous RSA challeng numbers, both factored and yet to be factored. I just want some feedback for the importance of establishing new bounds on factors, with or without a well founded mathematic theory behind. |
|
|
|
|
|
|
#4 | |
|
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
22·5·72·11 Posts |
Quote:
It is possible to prove, rigorously, that the known bounds are the best possible. An example, which you should be able to understand for yourself: let an integer N be the square of a prime p. That is, N=p*p. The existence of such N, and I have given you an explicit example, shows that if N=p*q where p>=q, then best possible upper bound for the smaller factor of N is sqrt(N) and the best possible lower bound on q is also sqrt(N) If, by some chance, you do not understand symbolic algebra and, therefore, the example given above, please explain how the smallest prime factor of 121 is less than sqrt(121). Remember that 1 is not prime and is a factor of every integer. Paul Last fiddled with by xilman on 2004-11-25 at 20:43 Reason: Correct a formatting typo |
|
|
|
|
|
|
#5 |
|
Dec 2003
Hopefully Near M48
110110111102 Posts |
xilman is right. But if n is a composite of some particular form (ex. (2^p) - 1
), you may be able to set better bounds.
Last fiddled with by jinydu on 2004-11-26 at 03:35 |
|
|
|
|
|
#6 | |
|
"Richard B. Woods"
Aug 2002
Wisconsin USA
22×3×641 Posts |
Quote:
If there are more than two prime factors, then you have a different case. But the original posting specified "n=p*q" (with the implication by context that p and q were both prime). Last fiddled with by cheesehead on 2004-11-28 at 09:48 |
|
|
|
|
![]() |
| Thread Tools | |
Similar Threads
|
||||
| Thread | Thread Starter | Forum | Replies | Last Post |
| What Bounds to choose, and what are Bounds | 144 | Information & Answers | 5 | 2017-03-15 13:36 |
| Question on P-1 bounds | NBtarheel_33 | Math | 1 | 2016-05-09 13:10 |
| Extending P-1 Bounds | TObject | Software | 4 | 2012-10-10 17:42 |
| ECM bounds | Unregistered | Information & Answers | 4 | 2011-07-17 20:18 |
| Progressive ECM bounds | Prime95 | Math | 14 | 2006-03-31 12:03 |