20200928, 14:12  #1 
Jun 2019
2×17 Posts 
a quadratic residue modulo and( Mersenne numbre)
(2^{n2})7 is a quadratic residue modulo M(n)
x^{2} = ( 2^{n2})7 mod M(n) let n >5 example 5^{2} = (2^{72})7 mod 2^{7}1 3824^{2}= (2^{132})7 mod 2^{13}1 181^{2}= (2^{172})7 mod 2^{17}1 470621^{2}= (2^{192})7 mod 2^{19}1 1319207736^{2}= (2^{312})7 mod 2^{31}1 755860352310189931^{2}= (2^{612})7 mod 2^{61}1 537583923266350664493783973^{2}= (2^{892})7 mod 2^{89}1 56007396006218245557399356980543^{2}= (2^{1072})7 mod 2^{107}1 118014915194510929114799427731590821918^{2}= (2^{1272})7 mod 2^{127}1 
20200928, 14:23  #2 
Sep 2002
Database er0rr
3,533 Posts 
It is true for all odd prime p that kronecker(2^(p2)7,2^p1) == 1. What is your point?

20200928, 14:31  #3 
Jun 2019
2×17 Posts 
why is true Is there an order of Distribution of quadratic residues

20200928, 14:40  #4 
Sep 2002
Database er0rr
3,533 Posts 
2^(p2)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^24*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 20200928 at 14:51 
20200928, 14:58  #5 
Jun 2019
22_{16} Posts 
( 2n2)7 is quadratic residue
1 < a k= ( 2n2)7  ( ((a*a+1)/2)1 * 6 ) IF k > 0 is quadratic residue example M13(8191) ( 2n2)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) ( 2n2)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 20200928 at 15:16 
20200928, 15:26  #6 
Sep 2002
Database er0rr
3,533 Posts 
Rewriting your terrible use of brackets...
2^(p2) 7  ((a*(a+1)/2)  1)*6 has the same symbol over M_p as: 2^(p2)  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 20200928 at 15:31 
20200928, 15:48  #7  
Jun 2019
100010_{2} Posts 
Quote:
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