[QUOTE=Christenson;263723]Let's see:
You are hypothesizing that log2(M48+1) = ( X^Y )mod(M47) for some X and Y?

How would I effectively computer X and Y? Can you show me X and Y for M2 through M47?[/QUOTE]

well okay I haven't got a complete proof of the idea but with the results so far 2 comes up most often so far. by a long shot in fact the only counter to 2 before the 1445 exponent was a 5 ( to have a 32 residue). I'll see what i can do to make headway.

[QUOTE=science_man_88;263724]well okay I haven't got a complete proof of the idea but with the results so far 2 comes up most often so far. by a long shot in fact the only counter to 2 before the 1445 exponent was a 5 ( to have a 32 residue). I'll see what i can do to make headway.[/QUOTE]

my full use of PARI so far ( the early results I did by pocket calculator):

[CODE](13:36)>(8191^2)%127
%54 = 32
(19:50)>(131071^2)%8191
%55 = 225
(19:51)>ispower(%,,)
%56 = 2
(19:52)>ispower(%,,15)
*** expected character: '&': ispower(%,,15)
^---
(19:52)>15^2
%57 = 225
(19:52)>(524287^2)%131071
%58 = 9
(19:58)>(2147483647^2)%524287
%59 = 516128
(19:59)>ispower(%,,)
%60 = 0
(19:59)>(2147483647^3)%524287
%61 = 143263
(20:00)>ispower(%,,)
%62 = 0
(20:00)>(2147483647^4)%524287
%63 = 509119
(20:00)>ispower(%,,)
%64 = 0
(20:00)>(2147483647^5)%524287
%65 = 277193
(20:00)>ispower(%,,)
%66 = 0
(20:00)>(2147483647^6)%524287
%67 = 23980
(20:00)>ispower(%,,)
%68 = 0
(20:00)>(2147483647^7)%524287
%69 = 156431
(20:01)>ispower(%,,)
%70 = 0
(20:01)>(2147483647^8)%524287
%71 = 430518
(20:01)>ispower(%,,)
%72 = 0
(20:01)>(2147483647^9)%524287
%73 = 318316
(20:01)>ispower(%,,)
%74 = 0
(20:01)>(2147483647^10)%524287
%75 = 126538
(20:01)>ispower(%,,)
%76 = 0
(20:01)>(2147483647^11)%524287
%77 = 177554
(20:01)>ispower(%,,)
%78 = 0
(20:01)>solve(y=0,1000,!ispower((2147483647^y)%524287))
*** ispower: missing exponent.
(20:02)>solve(y=0,1000,!ispower((2147483647^y)%524287,,))
*** ispower: missing exponent.
(20:02)>solve(y=0,1000,!ispower(((2147483647^y)%524287),,))
*** ispower: missing exponent.
(20:02)>(2147483647^12)%524287
%79 = 421848
(20:02)>ispower(%,,)
%80 = 0
(20:02)>(2147483647^13)%524287
%81 = 466182
(20:03)>ispower(%,,)
%82 = 0
(20:03)>(2147483647^14)%524287
%83 = 86323
(20:03)>ispower(%,,)
%84 = 0
(20:03)>(2147483647^15)%524287
%85 = 123247
(20:03)>ispower(%,,)
%86 = 0
(20:03)>(2147483647^16)%524287
%87 = 332371
(20:03)>ispower(%,,)
%88 = 0
(20:03)>(2147483647^17)%524287
%89 = 10193
(20:03)>ispower(%,,)
%90 = 0
(20:03)>(2147483647^18)%524287
%91 = 321662
(20:03)>ispower(%,,)
%92 = 0
(20:03)>(2147483647^19)%524287
%93 = 196946
(20:03)>ispower(%,,)
%94 = 0
(20:03)>(2147483647^20)%524287
%95 = 140464
(20:04)>ispower(%,,)
%96 = 0
(20:04)>(2147483647^21)%524287
%97 = 57241
(20:04)>ispower(%,,)
%98 = 0
(20:04)>(2147483647^22)%524287
%99 = 45606
(20:04)>ispower(%,,)
%100 = 0
(20:04)>(2147483647^23)%524287
%101 = 110398
(20:04)>ispower(%,,)
%102 = 0
(20:04)>(2147483647^100)%524287
%103 = 226300
(20:04)>ispower(%,,)
%104 = 0
(20:04)>(2147483647^10)%524287
%105 = 126538
(20:04)>ispower(%,,)
%106 = 0
(20:04)>solve(y=0,1000,ispower((2147483647^y)%524287)-1)
*** ispower: missing exponent.
(20:11)>solve(y=0,1000,ispower(((2147483647^y)%524287))-1)
*** ispower: missing exponent.
(20:11)>for(y=0,1000,ispower(((2147483647^y)%524287)))
(20:11)>for(y=0,1000,if(ispower(((2147483647^y)%524287)),print(Y)))
(20:11)>for(y=0,1000,if(ispower(((2147483647^y)%524287)),print(y)))
(20:11)>for(y=0,100000,if(ispower(((2147483647^y)%524287)),print(y)))
1445
1533
2636
3066
3252
*** _^_: user interrupt after 5,515 ms.
(20:12)>for(y=0,100000,if(ispower(((2147483647^y)%524287)),print(y);break()))
1445
(20:12)>for(y=0,100000,if(ispower(((524287^y)%131071)),print(y);break()))
2
(20:13)>for(y=0,100000,if(ispower(((524287^2)%131071)),print(y);break()))
0
(20:13)>for(y=0,100000,if(ispower(((524287^y)%131071)),print(y);break()))
2
(20:14)>for(y=0,100000,if(ispower(((2305843009213693951^y)%2147483647)),print(y);break()))
2
(20:14)>for(y=0,100000,if(ispower(((618970019642690137449562111^y)%2305843009213693951)),print(y);break()))
2
(20:16)>for(y=0,100000,if(ispower(((162259276829213363391578010288127^y)%618970019642690137449562111)),print(y);break()))
2
(20:19)>for(y=0,100000,if(ispower(((170141183460469231731687303715884105727^y)%162259276829213363391578010288127)),print(y);break()))
2[/CODE]

[QUOTE=Christenson;263723]Let's see:
You are hypothesizing that log2(M48+1) = ( X^Y )mod(M47) for some X and Y?

How would I effectively computer X and Y? Can you show me X and Y for M2 through M47?[/QUOTE]

What kind of nonsense is this??? Let P = M47. [b]EVERY[/b] integer
in Z/PZ* (i.e. every integer from 1 to P-1) is of the form x^y mod P
for some x and y.

This "conjecture" says nothing at all.

[QUOTE=R.D. Silverman;263726]What kind of nonsense is this??? Let P = M47. [b]EVERY[/b] integer
in Z/PZ* (i.e. every integer from 1 to P-1) is of the form x^y mod P
for some x and y.

This "conjecture" says nothing at all.[/QUOTE]

your conjecture seems to imply every integer under p is a power and hence every integer under a infinite p is a power. my modulus is below every number I'm trying. I don't see how the 2 are the same.

[QUOTE=R.D. Silverman;263726]What kind of nonsense is this??? Let P = M47. [B]EVERY[/B] integer
in Z/PZ* (i.e. every integer from 1 to P-1) is of the form x^y mod P
for some x and y.

This "conjecture" says nothing at all.[/QUOTE]

No, it is trivially true!
[someone forgot his number theory, and was simply decoding...trying to get SM88s thoughts straight]....let's add that X<1000...now it's a conjecture! (not that I actually believe it)

RDS, keep your cool...remember that this is like sci.math ...

sieve of atkins: speeding it up for mersenne's ?
 

I can prove that every mersenne with prime exponent>2 falls into 1 category in the sieve of Atkin the problem is speeding up the check the sieve does to make it worth it. anyone up for it ?

2^3-1 is the starting place I'll use:

1) note that there are 4 answers mersenne numbers could be mod 60
2) they either jump by 2 landing on one 2 from it or ,4 and that jumps back to the same one it's on so we are dealing with just 2 answers really for any prime exponent.
3) so after p=3 they jump by 2 and 4 so 7= 7 mod 31 and so with a jump of 2 we get 31 = 31 mod 60 , jump of 2 gives us back to 7 mod 60 then it alternates 4,2 through all the 6x + or -1 numbers so prime exponents can only be 31 or 7 mod 60.

the problem for me is proving the equations solutions without having to use the full number. if this can be done efficiently could it not be back to altered sieve of Atkin that might prove Mersenne primes faster ?

and it's the same equation used for both 7 and 31 they are in the same grouping.

[QUOTE=science_man_88;265291]the problem for me is proving the equations solutions without having to use the full number. if this can be done efficiently could it not be back to altered sieve of Atkin that might prove Mersenne primes faster ?[/QUOTE]

No. Using the sieve of Atkin or any other prime-generating sieve to prove primality is roughly as bad as trial division, extremely inefficient.

[QUOTE=CRGreathouse;265315]No. Using the sieve of Atkin or any other prime-generating sieve to prove primality is roughly as bad as trial division, extremely inefficient.[/QUOTE]

okay fine I'll stop trying to do anything .

question
 

[QUOTE=science_man_88;265316]okay fine I'll stop trying to do anything .[/QUOTE]

okay I'm a crappy liar:


I've been doing experimenting and I'm trying to figure out a fast way to solve if s is 0 mod (2^p-1) so far I've seen this:

1)s[SUB]1[/SUB]= i^2 yes I know i=2 I'm trying to make it so we'd only have to figure out a smaller number possibly,

2)s[SUB]k[/SUB] = i^(2^k)-...+2

3)the formula alters subtract then add

4) the sum of coefficients including those signs add to -4

5) the powers decrease by 2 along the way


I realize that [TEX]2^{p-1}+2 \equiv 2^{p-1}+2 \text { mod } (2^{p}-1 )[/TEX] so could we use that and knowledge about the coefficients to figure the problem ?

[QUOTE=science_man_88;265888]okay I'm a crappy liar:


I've been doing experimenting and I'm trying to figure out a fast way to solve if s is 0 mod (2^p-1) so far I've seen this:

1)s[SUB]1[/SUB]= i^2 yes I know i=2 I'm trying to make it so we'd only have to figure out a smaller number possibly,

2)s[SUB]k[/SUB] = i^(2^k)-...+2

3)the formula alters subtract then add

4) the sum of coefficients including those signs add to -4

5) the powers decrease by 2 along the way


I realize that [TEX]2^{p-1}+2 \equiv 2^{p-1}+2 \text { mod } (2^{p}-1 )[/TEX] so could we use that and knowledge about the coefficients to figure the problem ?[/QUOTE]

oh and the powers go down to i^4 and in this case i=2 and the rest of it is 16*f so [TEX]16*f \equiv 2^{p-1}-3 \text { mod } 2^p-1[/TEX] for some f I don't have the properties of yet.

[QUOTE=science_man_88;265889]oh and the powers go down to i^4 and in this case i=2 and the rest of it is 16*f so [TEX]16*f \equiv 2^{p-1}-3 \text { mod } 2^p-1[/TEX] for some f I don't have the properties of yet.[/QUOTE]

oops sorry mixed up [TEX]16*f+ 2^{p-1}-3 \equiv 0 \text { mod } 2^p-1[/TEX]