 mersenneforum.org a quadratic residue modulo and( Mersenne numbre)
 Register FAQ Search Today's Posts Mark Forums Read 2020-09-28, 14:12 #1 baih   Jun 2019 2×17 Posts 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   2020-09-28, 14:23 #2 paulunderwood   Sep 2002 Database er0rr 23·19·23 Posts It is true for all odd prime p that kronecker(2^(p-2)-7,2^p-1) == 1. What is your point?   2020-09-28, 14:31 #3 baih   Jun 2019 2·17 Posts why is true Is there an order of Distribution of quadratic residues   2020-09-28, 14:40 #4 paulunderwood   Sep 2002 Database er0rr 23·19·23 Posts 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   2020-09-28, 14:58 #5 baih   Jun 2019 1000102 Posts ( 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   2020-09-28, 15:26 #6 paulunderwood   Sep 2002 Database er0rr 66508 Posts 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   2020-09-28, 15:48   #7
baih

Jun 2019

3410 Posts Quote:
 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)
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