Other Bases?

[wblipp]

Originally Posted by Christenson

So, how do we efficiently determine if k*(4*128159)+1 divides 41^(4*128159)-1?

How do you do it right now for base 2? I thought you calculated the value of N mod TrialDivisor using the method of successive squarings with accumulations. I thought the same method would work here, you just must start with 41 instead of 2.

William

[science_man_88]

Originally Posted by wblipp

How do you do it right now for base 2? I thought you calculated the value of N mod TrialDivisor using the method of successive squarings with accumulations. I thought the same method would work here, you just must start with 41 instead of 2.

William

the one on I've been linked to for base 2 is the one I talked about and like I said I can get it to work for other bases or use a lower binary value to calculate on a higher exponent ( though I haven't changed the code I made in PARI).

[science_man_88]

Originally Posted by science_man_88

the one on I've been linked to for base 2 is the one I talked about and like I said I can get it to work for other bases or use a lower binary value to calculate on a higher exponent ( though I haven't changed the code I made in PARI).

all you do is instead of multplying by 2 for each 1 in binary of the exponent multiply by the base you want to test in.

[science_man_88]

Originally Posted by science_man_88

all you do is instead of multplying by 2 for each 1 in binary of the exponent multiply by the base you want to test in.

>binary(512636)
[1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0]

from all I can tell this is the exponent written in binary plug away with the base.

1^2*41 is 41 mod the number in question your turn.

[Christenson]

Originally Posted by wblipp

How do you do it right now for base 2? I thought you calculated the value of N mod TrialDivisor using the method of successive squarings with accumulations. I thought the same method would work here, you just must start with 41 instead of 2.

William

RDS calls me a weak student, so I admit I'm not exactly sure...if the method used for base 2 extends to base 41, then we'll use it.

[science_man_88]

Originally Posted by Christenson

RDS calls me a weak student, so I admit I'm not exactly sure...if the method used for base 2 extends to base 41, then we'll use it.

the TF part I can confirm look at it this way say b=2 what's it doing to 2 raising it to power B using the binary representation of B. so it b^B change B it's doing the same thing to that base. it's just the numbers get bigger quicker as i tried to point out in my attempt at an example.

[science_man_88]

Originally Posted by Christenson

So, how do we efficiently determine if k*(4*128159)+1 divides 41^(4*128159)-1?

I made a CPU PARI code I'm not sure how to do it on GPU.

TF(x,y,B) so I'd try TF(4*128159,y,41) where y is the number to divide by 41 is the base and the other is the exponent. I tried a few of these:

(21:09)>for(k=1,100,print1(TF(4*128159,k*(4*128159)+1,41)","))
465739,1025272,180591,1735966,1724252,2453406,1393945,1986482,23441,2340841,5468485,2877009,4702276,132291,6619257,6754759,1297338,4865974,1546861,2304212,1778672,9450541,1135108,4745516,11557960,6472376,3534466,11729473,2784906,13719988,8744965,14060327,14110251,7120816,17323056,10547602,8627875,16224431,3598701,10351894,13891127,295092,10724560,11477726,21875089,1961359,16876649,11907894,11230891,1430529,18030928,9600595,5471313,11558141,26773699,2646145,3973517,18388108,26065691,13709319,24260065,28756610,28782451,16612051,30578419,12090299,20384407,4024850,6536881,23978194,31132838,1353372,36339229,36992586,8042264,27731323,4308734,7564002,30410926,10663850,27856235,25656967,39950978,35574366,29158852,17089570,23871435,34801369,43553571,44196445,41027425,36853632,13691170,26561896,31899673,10026316,23183467,148799,43248981,25394260,
(21:09)>##
  ***   last result computed in 16 ms.
(21:09)>

[R.D. Silverman]

Originally Posted by wblipp

How do you do it right now for base 2? I thought you calculated the value of N mod TrialDivisor using the method of successive squarings with accumulations. I thought the same method would work here, you just must start with 41 instead of 2.

William

You are correct.

And understanding how to do modular exponentiations and multi-precision
modular arithmetic is SO fundamental to this subject, that (IMO) it ought
to be required reading before one is allowed to participate in these forums.

Too many people prattle in almost total ignorance. This entire forum
is becoming like sci.math; (I don't include you). It is becoming inundated
by willfully ignorant cranks who refuse to do the required background
reading. Dunning and Kruger is quite on point for these people.

[R.D. Silverman]

Originally Posted by Christenson

RDS calls me a weak student, so I admit I'm not exactly sure...if the method used for base 2 extends to base 41, then we'll use it.

I never called you a weak student. But the very very many elementary
mispronouncements that you continue to make, and the ignorance you
show (even about very elementary aspects of this subject) suggests that you are a lazy student; unwilling to do the required reading.

[science_man_88]

Originally Posted by LiquidNitrogen

Well 41^(4*128159)-1 always ends in 0.

see that didn't help think about this:

odd*even = even
even *even = even
odd*odd = odd

so k*even = even so k*even + 1 = odd

but you say that it ends in 0 okay so it's even, can you prove there's no even number that will work ? if not it's not much of an argument even in my low standards.

[science_man_88]

Originally Posted by science_man_88

see that didn't help think about this:

odd*even = even
even *even = even
odd*odd = odd

so k*even = even so k*even + 1 = odd

but you say that it ends in 0 okay so it's even, can you prove there's no even number that will work ? if not it's not much of an argument even in my low standards.

sorry I mean't as a co-factor to the number suggested.