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Old 2020-09-28, 14:12   #1
baih
 
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Default a quadratic residue modulo and( Mersenne numbre)

(2n-2)-7 is a quadratic residue modulo M(n)


x2 = ( 2n-2)-7 mod M(n)

let n >5
example

52 --------------------------------= (27-2)-7 mod 27-1

38242----------------------------= (213-2)-7 mod 213-1

1812------------------------------= (217-2)-7 mod 217-1
4706212-------------------------= (219-2)-7 mod 219-1
13192077362------------------= (231-2)-7 mod 231-1
7558603523101899312------------------------------------------= (261-2)-7 mod 261-1
5375839232663506644937839732---------------------------= (289-2)-7 mod 289-1
560073960062182455573993569805432-------------------= (2107-2)-7 mod 2107-1
1180149151945109291147994277315908219182--------= (2127-2)-7 mod 2127-1
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Old 2020-09-28, 14:23   #2
paulunderwood
 
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It is true for all odd prime p that kronecker(2^(p-2)-7,2^p-1) == 1. What is your point?
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Old 2020-09-28, 14:31   #3
baih
 
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why is true Is there an order of Distribution of quadratic residues
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Old 2020-09-28, 14:40   #4
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2^(p-2)-7 has the same residue as 2^p - 7*4; same as 1 -7*4; same as -27; same as -3, and this is always true for odd Mersenne. LL is the same as x^(n+1)==1 mod (M_p, x^2-4*x+1) and the polynomial has discriminant 12 the kronecker symbol of which is -1; same as for 3. The kronecker symbol of -3 over M_p is 1 since kronecker(-1,M_p)==-1 because M_p==3 mod 4. That is how I see it.

Last fiddled with by paulunderwood on 2020-09-28 at 14:51
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Old 2020-09-28, 14:58   #5
baih
 
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( 2n-2)-7 is quadratic residue


1 < a


k= ( 2n-2)-7 - ( ((a*a+1)/2)-1 * 6 )
IF k > 0 is quadratic residue


example

M13(8191)

( 2n-2)-7 = 2041 is quadratic residue

2041 - ( ((2*3)/2)-1 * 6 ) = 2029 is quadratic residue
2041 - ( ((3*4)/2)-1 * 6 ) = 2011 is quadratic residue
2041 - ( ((4*5)/2)-1 * 6 ) = 1987 is quadratic residue
2041 - ( ((5*6)/2)-1 * 6 ) = 1957 is quadratic residue


M17(131071)

( 2n-2)-7 = 32761is quadratic residue

32761- ( ((2*3)/2)-1 * 6 ) = 32749is quadratic residue
32761- ( ((3*4)/2)-1 * 6 ) = 32731is quadratic residue
32761- ( ((4*5)/2)-1 * 6 ) = 32707is quadratic residue
32761- ( ((5*6)/2)-1 * 6 ) = 32677is quadratic residue

Last fiddled with by baih on 2020-09-28 at 15:16
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Old 2020-09-28, 15:26   #6
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Re-writing your terrible use of brackets...

2^(p-2)- 7 - ((a*(a+1)/2) - 1)*6 has the same symbol over M_p as:

2^(p-2) - 1 - 3*a*(a+1) same as

2^p -4 - 12*a*(a+1) same as

-3 - 12*a*(a+1) same as

-3(4*a^2+4*a+1) same as

-3(2*a+1)^2 same as

-3

I.e. the symbol is 1.

Last fiddled with by paulunderwood on 2020-09-28 at 15:31
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Old 2020-09-28, 15:48   #7
baih
 
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Quote:
Originally Posted by paulunderwood View Post
Re-writing your terrible use of brackets...

2^(p-2)- 7 - ((a*(a+1)/2) - 1)*6 has the same symbol over M_p as:

2^(p-2) - 1 - 3*a*(a+1) same as

2^p -4 - 12*a*(a+1) same as

-3 - 12*a*(a+1) same as

-3(4*a^2+4*a+1) same as

-3(2*a+1)^2 same as

-3

I.e. the symbol is 1.
thanks I am not a pure mathematician I am (just a fake mathematician)
my work is a programmer (java android)
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