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Old 2005-09-11, 15:00   #7
cyrix
 
Jul 2003
Thuringia; Germany

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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
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