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Old 2010-09-23, 21:47   #650
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Quote:
Originally Posted by science_man_88 View Post
you say I'm solving for t without proof of that and my statement of the opposite yet you still claim it hence you assume it's true there's the assumption.
I posted a problem for you which, with the variables you're using, requires you to solve for t. You don't have to solve for t -- you could solve for p, or play a game of baseball, or whatever -- but in that case you wouldn't be doing the proposed problem.
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Old 2010-09-23, 21:53   #651
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I don't see a problem asking for t at all actually but do as you want.
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Old 2010-09-23, 22:07   #652
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Quote:
Originally Posted by science_man_88 View Post
I don't see a problem asking for t at all actually but do as you want.
Uh... okay?

To clarify, if that's needed: insofar as a person is solving the problem I posed in #618 with your equation, they would solve for t.
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Old 2010-09-23, 22:10   #653
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Quote:
Originally Posted by CRGreathouse View Post
I was looking for a function that, given a probability p (and N, t, L), would give an n such that

estimatePrimes(N,t,n,L)

is 1 - p.
then this post is flawed as are all after it.
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Old 2010-09-23, 23:35   #654
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How would one of you go about proving or disproving the primality of 3379212930002668486657 using only hand calculations? It has no factors under 16777216.

Code:
(19:41) gp > Mod(13, 3379212930002668486657)^1689606465001334243328
%73 = Mod(1859858123868907490075, 3379212930002668486657)
Nevermind, the number is composite.

3379212930002668486657 = 30245153 * 111727420588769

Okay, 3773512084578210152449 is at least a PRP. How would you go about in proving it prime, with no computer assistance?

Last fiddled with by 3.14159 on 2010-09-23 at 23:48
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Old 2010-09-23, 23:44   #655
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I've heard theres a way with subtracting multiples of the divisor you want to check until you either prove or disprove the resultant.
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Old 2010-09-24, 00:01   #656
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@Sm88: Too impractical to do a few million times.

The shortest idea I have so far is manually using a few M-R tests.

Or applying Proth's theorem by hand, as the number in question (3773512084578210152449) is a Proth number.

Last fiddled with by 3.14159 on 2010-09-24 at 00:03
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Old 2010-09-24, 00:42   #657
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27009 * 10^20 + 1 = 7^2 * 55120408163265306122449;

55120408163265306122449 is a twin prime.
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Old 2010-09-24, 00:58   #658
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Quote:
Originally Posted by 3.14159 View Post
The shortest idea I have so far is manually using a few M-R tests.
I'm not sure how you could use this, by hand, to prove primality. Actually, considering that it's 72 bits, I'm not even sure of how I could use that to prove primality with a computer.

Quote:
Originally Posted by 3.14159 View Post
Or applying Proth's theorem by hand, as the number in question (3773512084578210152449) is a Proth number.
That's probably the most reasonable way. It would be a pain, though.
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Old 2010-09-24, 01:18   #659
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Quote:
Originally Posted by Charles
I'm not sure how you could use this, by hand, to prove primality. Actually, considering that it's 72 bits, I'm not even sure of how I could use that to prove primality with a computer.
About 20-25 tests prove that it is prime. Oh, right.. M-R's weakness = Proth numbers.

You can prove any non-Proth 72-bit number prime with M-R.

Ex:
Code:
(21:29) gp > ispseudoprime(8183202143983816686553, 30)
%130 = 1
That was a 72-bit number.

Quote:
Originally Posted by Charles
That's probably the most reasonable way. It would be a pain, though.
It depends on a. Suppose a = 23;

23^1886756042289105076224 modulo 3773512084578210152449?

Multiplication of 23 by itself has to happen at least 16 times for it to be ≥ 3773512084578210152449.

Also, a good idea would be to identify when the mod results repeat. (If they do repeat.)

Ex: Powers of 71, mod 23: 2, 4, 8, 16, 9, 18, 13, 3, 6, 12, 1, 2, 4, 8, 16, 9, 18, ...

Last fiddled with by 3.14159 on 2010-09-24 at 01:32
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Old 2010-09-24, 01:36   #660
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6^134217728 + 1 is divisible by 51808043009.

Two factors for 5^36893488147419103232 + 1: 221360928884514619393 and 2434970217729660813313 both divide 5^36893488147419103232 + 1.

Probably well-known at the moment;

I betcha no one can find the smallest divisor of 10000^(2^160) + 1, which = 10^(2^162) + 1.

Last fiddled with by 3.14159 on 2010-09-24 at 01:42
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