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Prime divisors of 4n^2+4n-1

  1. May 30, 2010 #1
    I need to prove that p congruent modulo 8 to [tex]\pm 1[/tex] for every prime divisor p of [tex]4n^2+4n-1[/tex].
    [tex]4n^2+4n-1[/tex] is odd so we have
    [tex]p \equiv \pm 1,3,5 \pmod{8}[/tex]

    I don't know how to continue from here... I need some hint.

    Last edited: May 31, 2010
  2. jcsd
  3. May 30, 2010 #2
    Congruent modulo what? [itex]7|(4.5^2+4.5-1)[/itex], but [itex]7\neq\pm 1(5)[/itex], so presumably the answer is not [itex]n[/itex].
  4. May 31, 2010 #3
    I need to prove that p congruent modulo 8 to [tex]\pm 1[/tex].
  5. May 31, 2010 #4
    [itex]4n^2+4n-1=(2n+1)^2-2=0(p)[/itex] only if [itex]2[/itex] is a quadratic residue mod [itex]p[/itex] which is true only if [itex]p=\pm 1(8)[/itex]. (This is proved in any number theory text in the section on quadratic reciprocity.)
  6. May 31, 2010 #5
    Thank you !
  7. May 31, 2010 #6
    [tex]p \equiv \pm 1,3,5 \pmod{8}[/tex]

    Boy! that was a confusing bunch of stuff. Lot of times, the trouble with a problem is the writer did not correctly organize what he wants to solve. But I see, TTob got it all correct!
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