a quadratic residue modulo and( Mersenne numbre)

[baih]

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


x² = ( 2^n-2)-7 mod M(n)

let n >5
example

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

3824²----------------------------= (2^13-2)-7 mod 2¹³-1

181²------------------------------= (2^17-2)-7 mod 2¹⁷-1
470621²-------------------------= (2^19-2)-7 mod 2¹⁹-1
1319207736²------------------= (2^31-2)-7 mod 2³¹-1
755860352310189931²------------------------------------------= (2^61-2)-7 mod 2⁶¹-1
537583923266350664493783973²---------------------------= (2^89-2)-7 mod 2⁸⁹-1
56007396006218245557399356980543²-------------------= (2^107-2)-7 mod 2¹⁰⁷-1
118014915194510929114799427731590821918²--------= (2^127-2)-7 mod 2¹²⁷-1

[paulunderwood]

It is true for all odd prime p that kronecker(2^(p-2)-7,2^p-1) == 1. What is your point?

[baih]

why is true Is there an order of Distribution of quadratic residues

[paulunderwood]

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.

[baih]

( 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

[paulunderwood]

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.

[baih]

Originally Posted by paulunderwood

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)