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2005-09-11, 15:00   #7
cyrix

Jul 2003
Thuringia; Germany

2×29 Posts

sorry Tony!
I messed something up. Now in a better form:

Quote:
 Originally Posted by cyrix Hi T.Rex! For prime $M_q$ with $q \equiv 1 \pmod 4$ (and q>1) your conjecture is true: Because of the quadratic reciprocity law (Gauß) 5 is a quadratic residue of $M_q$, iff $M_q$ is quadratic NON residue of 5 (because 5=4*1+1 and $M_q=4*n+3$).
$M_q \equiv 3 \pmod 4$, but since $5 \equiv 1 \pmod 4$ the reciprocity law works in the same way, So you can find a solution X with $X^2 \equiv 5 \pmod {M_q}$. This means, you can find the two solutions R_1 and R_2 with $R^2+R-1 \equiv 0 \pmod {M_q}$ because of the formel in my first post. (with the second solution $Y^2 \equiv 5 \pmod {M_q}$ you get the same solutions R_2, R_1)

Your conjecture reduces to: 5 is a quadratic residue of $M_q$ with q a prime and $q \equiv 1 \pmod 4$, iff $M_q$ is prime itself.

Yours,
Cyrix

Last fiddled with by cyrix on 2005-09-11 at 15:01