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Old 2018-03-09, 04:05   #12
carpetpool
 
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Quote:
Originally Posted by JM Montolio A View Post

Solve: 29(5^m)-11 = 13*D
This can be done with basic algebra and modular math.

To get a set of basic solutions (m, D) write out:

29*5^m-11 = 0 (mod 13)

29*5^m = 11 (mod 13)

5^m = 8 (mod 13)

m = 3 (mod 4).

So for any integer n,

m = 3*n+4, D = (29*5^(3*n+4)-11)/13

will be a solution to

29(5^m)-11 = 13*D

as you put it.
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Old 2018-03-09, 05:00   #13
CRGreathouse
 
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Quote:
Originally Posted by JM Montolio A View Post
29(5^M)-11=13D.
The solutions are M = 4k + 3, D = (3625*625^k - 11)/13.

If b and n are relatively prime, you can find the general solutions to k(b^m) - z = nd by finding the order of b mod n, computing k(b^m) - z for m from 1 to the order, and taking any values which result in 0; these values, plus an arbitrary variable k times the order, are the possible values of m (and the d values can be computed from them).

The case where b and n have a common divisor is not essentially different; you check the small cases, where some p | b and p^e | n, but p^e does not divide b, individually, then look at the order of b with all the common primes divided out mod n with all the common primes divided out.

In the first case you could have 0 solutions or infinitely many; in the second case you could have finitely many or infinitely many.

Edit: see carpetpool's post above.

Last fiddled with by CRGreathouse on 2018-03-09 at 05:01
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Old 2018-03-09, 05:43   #14
axn
 
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OP sounds suspiciously like they're trying to sieve k*b^n+c form (variable n), in which case, just use newpgen or srsieve
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Old 2018-03-09, 09:12   #15
JM Montolio A
 
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Are the cumulative product of (B^g).

Last step not.
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Old 2018-03-09, 09:20   #16
JM Montolio A
 
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Are the cumulative product of (B^g).

Starting with 1.

And not the last step.

Thats gives "bits one of D"="number steps".

Rules are related to tserie used.

Mersenne is tserie "n+e = (2^g)(e')".
Most general: n+e=(B^g)(e')

Any tserie as a equation.

For Mersenne, (eLast)(B^M)-(eStart)=n*D.

Collatz is also a tserie. 1+3e=(2^g)(e').

But there are others tserie.

Thanks for your interest.

JMM
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Old 2018-03-09, 13:50   #17
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Quote:
Originally Posted by carpetpool View Post
This can be done with basic algebra and modular math.

To get a set of basic solutions (m, D) write out:

29*5^m-11 = 0 (mod 13)

29*5^m = 11 (mod 13)

5^m = 8 (mod 13)

m = 3 (mod 4).

So for any integer n,

m = 3*n+4, D = (29*5^(3*n+4)-11)/13

will be a solution to

29(5^m)-11 = 13*D

as you put it.
3 mod 4 is actually 4n+3.
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Old 2018-03-09, 15:29   #18
carpetpool
 
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Quote:
Originally Posted by science_man_88 View Post
3 mod 4 is actually 4n+3.
Too late to fix it but thanks for pointing out.
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Old 2018-03-09, 15:32   #19
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Quote:
Originally Posted by carpetpool View Post
Too late to fix it but thanks for pointing out.
In fact you can prove D is 18 mod 20 fairly quickly. Etc.
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