 mersenneforum.org Maybe some Ms can be further limited to factors 10m1 = +-1
 Register FAQ Search Today's Posts Mark Forums Read 2013-03-10, 03:39 #1 Andrew   Feb 2013 111002 Posts Maybe some Ms can be further limited to factors 10m1 = +-1 I was looking at numbers of the form "x*(x+3) + 1" - They all end in either 5,1, or 9. I can also say that for any x, at least one of its factors/divisors must have already been in a previous, or lesser, "x*(x+3) + 1." This shows when you see that "(x+1)^2 + x" > equivalent to > x*(x+3) + 1 (X+1)^2 + X - (a*(a+3) + 1) = (K-a)^2 + (2a + 3)*(K-a) We see that some K-a must divide some lesser number of the same form as X, and hence, one of its divisors must have already been a divisor in an "x*(x+3) +1)" Given that as the case, it can be inferred that none of the factors ever end in 7 or 3. I can see that Mersenne numbers are divisors/factors in a some of these. 2^5,13,17,25... - 1. So if you test Mersennes that also are divisors for an x*(x+3) + 1, then that's a few more factors you can check out of your testing. Unless you already knew this, or I am wrong. Probably a bad first explanation. I'll come back.   2013-03-10, 15:36   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts Quote:
 Originally Posted by Andrew I was looking at numbers of the form "x*(x+3) + 1" .
in other words x^2+3x+1

Quote:
 They all end in either 5,1, or 9.
if x is replaced with x mod 10 we get the following:

0^2+3*0+1 = 1
1^2+3*1+1 = 5
2^2+3*2+1 = 11
3^2*3*3+1 = 19
4^2+3*4+1 = 29
5^2+3*5+1 = 41
6^2+3*6+1 = 55
7^2+3*7+1 = 71
8^2+3*8+1 = 89
9^2+3*9+1 = 109

Quote:
 I can also say that for any x, at least one of its factors/divisors must have already been in a previous, or lesser, "x*(x+3) + 1."
of course 1 is a divisor of every number

as to the rest I can barely follow right now.   2013-03-12, 21:38   #3
Andrew

Feb 2013

22×7 Posts Quote:
 Originally Posted by science_man_88 in other words x^2+3x+1
Yeah, but I like the other form as it says to me "We're multiplying two numbers whose difference is three, and adding one."
Quote:
 if x is replaced with x mod 10 we get the following: 0^2+3*0+1 = 1 1^2+3*1+1 = 5 2^2+3*2+1 = 11 3^2*3*3+1 = 19 4^2+3*4+1 = 29 5^2+3*5+1 = 41 6^2+3*6+1 = 55 7^2+3*7+1 = 71 8^2+3*8+1 = 89 9^2+3*9+1 = 109 of course 1 is a divisor of every number as to the rest I can barely follow right now.
Well here's the guess I made, firstly:
----------------------------------------------
Every divisor, and prime factor, of a number of such form ends in 1, 5, or 9.

I tried to reword this as "Every prime factor is congruent to +-1 mod 10, and 5, mod 10." <- I think this is at least better.
----------------------------------------------
After the guess I tried to figure out why this had to be the case, so:

It appears that at least one divisor of this kind of number must divide a lesser number, also of the aforementioned form. Eg.:

Taking for example 11*14 + 1 = 155, then starting from 12, 12^2 + 12 = 156. [In the first case, 12 if it were to be a divisor, would have to divide one and it doesnt.] 11^2 + 3*11 is 1 less than 155. 10^2 + 5*10 = 150. 10 Does not divide 5. 9^2 + 7*9 = 144. If 9 were an integer divisor it would have to divide 11.

So in here :

11 !/! 1
10 !/! 5
9 !/! 11

…down through:
5^2 + 25*5 + {(11-5)*(11-2) + 1}
5 does divide 55, so 5 is a divisor. This is not to say 5 * 11 is the number 155.

So if you will go along, this pattern goes on. The X selected will always be equal to

(X+1)^2 + X
and
(X+1)^2 + X = (X-A)^2 + (2A + 3)*(X-A) + (A^2 + 3A + 1)
Where (X-A) is the potential divisor.
-----------------------------------------------
I say that this must mean that:

1. At least one of the divisors divides some lesser number of the form.

2. Since all of these numbers end in 1,5,9, then the other divisor is either
A: Prime, and must also end in 1,5,9
B: Composite, in which case its prime components divide some lesser number of the form, and hence each also ends in 1,5,9.

If that's not strong enough, I've looked at a few hundred of these and seen no evidence to contradict, so there’s probably another good reason.

So I am thinking if this all is right, then here’s one example:

31 divides 155 (and more of these) and is also a Mersenne number, prime in this case. There are other Mersenne numbers that divide these.

Say you wanted to test if 2^25 – 1 is prime. I know it doesn’t have a prime exponent. If you wanted to test this, and knew it also divided one of these numbers, then you’ve already taken away about half of the factors you’d have to test.

I haven’t considered whether it would be more work to check this, than it would be to check all those factors you would have gotten rid of.

In any case, one last thing, the Mersenne numbers that divide and x*(x+3) + 1 all seem to have an exponent of the form 4n+1.

Last fiddled with by Andrew on 2013-03-12 at 22:10 Reason: I made an error: Not 35*5, it is 25*5   2013-03-12, 23:13   #4
R.D. Silverman

Nov 2003

22·5·373 Posts Quote:
 Originally Posted by Andrew Yeah, but I like the other form as it says to me "We're multiplying two numbers whose difference is three, and adding one." Well here's the guess I made, firstly: ---------------------------------------------- Every divisor, and prime factor, of a number of such form ends in 1, 5, or 9. I tried to reword this as "Every prime factor is congruent to +-1 mod 10, and 5, mod 10." <- I think this is at least better. ---------------------------------------------- After the guess I tried to figure out why this had to be the case
(1) You should learn some elementary number theory. We can recommend
some excellent books.

(2) As for your "discussion", I give two hints:

(A) Discriminant

(3) Stop prattling. Go learn some math. The stuff you are trying to
discuss could be given as simple exercizes in any introductory course
in number theory.   2013-03-13, 02:45 #5 LaurV Romulan Interpreter   Jun 2011 Thailand 34·113 Posts I don't want to encourage it, but let them play, we are in misc math thread and they are young... for some people this is how they learn. As long as they are aware of the fact that what they do is exercising and learning, and don't get the idea that they invented a new type of wheel, (a squared wheel?! ), then everything is ok... Last fiddled with by LaurV on 2013-03-13 at 02:46   2013-03-13, 11:13   #6
literka

Mar 2010

26·3 Posts Quote:
 Originally Posted by LaurV I don't want to encourage it, but let them play, we are in misc math thread and they are young... for some people this is how they learn. As long as they are aware of the fact that what they do is exercising and learning, and don't get the idea that they invented a new type of wheel, (a squared wheel?! ), then everything is ok... You didn't invent new type of wheel either. So, don't talk down.   2013-03-13, 11:36   #7
R.D. Silverman

Nov 2003

164448 Posts Quote:
 Originally Posted by literka You didn't invent new type of wheel either. So, don't talk down.
The people in this forum seem to have a strong aversion to actually
reading and learning mathematics. This is despite claiming to
like mathematics.

Everytime I suggest that people should read and/or study, people in this
newsgroup push back. I suggest that as a group you need to acquire some
intellectual maturity

There is a reason why elementary number theory books were
written. Unless one has a lot of mathematical maturity and training
it is almost impossible to learn mathematics via the types of 'dabbling' and
numerology we see here. It is clear that the OP does not have the
mathematical sophistication to learn by "dabbling". He will never gain
an understanding of what he is doing by following his current path.   2013-03-13, 14:15   #8
Mr. P-1

Jun 2003

7·167 Posts Quote:
 Originally Posted by LaurV ...a squared wheel?!    2013-03-13, 16:13   #9
Andrew

Feb 2013

348 Posts Quote:
 Originally Posted by R.D. Silverman (1) You should learn some elementary number theory. We can recommend some excellent books. (2) As for your "discussion", I give two hints: (A) Discriminant (B) Quadratic Residue (3) Stop prattling. Go learn some math. The stuff you are trying to discuss could be given as simple exercizes in any introductory course in number theory.
You reminded me of my old math/econ professor, except prattling isn't something I could see him saying as he uses American English. He made fun of me once for accidentally using the word 'fancy.' I liked that guy. No one else did as far as I could tell. Idk if he liked me.

_______

Anyway, yes, I have been 'playing.' I 'rediscovered' a bunch of other crap in this way, but I didn't say anything about that. Here I spoke up because this seemed like something relatively obscure while at the same time relatively useless.

I thought it might be like showing you all you lost a dime in the couch, but I guess you've already scrounged underneath that cushion. Anyway, I've seen lamer questions asked about what is the form of prime factors of Mersenne numbers.   2013-03-13, 16:45   #10
R.D. Silverman

Nov 2003

22·5·373 Posts Quote:
 Originally Posted by Andrew You reminded me of my old math/econ professor, except prattling isn't something I could see him saying as he uses American English. He made fun of me once for accidentally using the word 'fancy.' I liked that guy. No one else did as far as I could tell. Idk if he liked me. _______ Anyway, yes, I have been 'playing.' I 'rediscovered' a bunch of other crap in this way, but I didn't say anything about that. Here I spoke up because this seemed like something relatively obscure while at the same time relatively useless.
A classic example of Dunning and Kruger. You are so totally ignorant about
this subject that you aren't even aware of what it is that you don't know.
It is neither obscure nor useless. It permeates all of computational number
theory.

What you have been discussing is not "relatively obscure".
It is, in fact, an easy homework exercize for any first course in number
theory.

If you are interested in this subject, go READ.

If you are not willing to read, and want to remain willfully ignorant, then
go away. It would be impossible for you to discuss this subject intelligently
without first learning the basic "tools of the trade".

Last fiddled with by R.D. Silverman on 2013-03-13 at 16:46 Reason: pagination   2013-03-13, 19:08   #11
only_human

"Gang aft agley"
Sep 2002

2×1,877 Posts Quote:
 Originally Posted by R.D. Silverman If you are not willing to read, and want to remain willfully ignorant, then go away. It would be impossible for you to discuss this subject intelligently without first learning the basic "tools of the trade".
Bob, Chill. Chili Palmer, Elmore Leonard, Leonard Nimoy, Vulcan Mind Meld, Jedi Mind Trick, Jedi Mind Meld, These are not the exit directives you are looking for.
Quote:
 McCoy: C'mon, Spock, it's me, McCoy. You really have gone where no man's gone before. Can't you tell me what it felt like? Spock: It would be impossible to discuss the subject without a common frame-of-reference. McCoy: You're joking! Spock: A joke [pause] Spock: is a story with a humorous climax. McCoy: You mean I have to die to discuss your insights on death? Spock: Forgive me, Doctor. I am receiving a number of distress calls. McCoy: I don't doubt it.

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