Quote:
Originally Posted by science_man_88
That the residues mod some mersenne in the LL test, are quadratic residues mod the next mersenne.
Reason I think this to be true:
Sqr(A*p+b)= A^2(p^2)+ b^2(1^2)b mod 2p+1 and two parts of that simplify to quadratic residues. Failure would only happen if the sums/ differences of quadratic residues wasn't a quadratic residue ( guess I may be wrong).

1 is a quadratic residue but 2 is a quadratic nonresidue mod 3.