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Generalized Pell Equation and Primes

  1. Dec 22, 2004 #1
    I have a conjecture that the equation X^2 - 2Y^2 = P has solutions in odd integers if P is a prime of the form 8*N+1. I know of a paper that requires one to find Q such that Q^2 = 2 mod P inorder to solve these equations using continued fractions. To get to first base in proving my conjecture, is there a proof that 2 is a quadratic residue of P where P is a prime of the form 8*N+1?
     
  2. jcsd
  3. Dec 22, 2004 #2
    When p is of the form 8k +-1, then X^2 ==2 Mod (p) is always the case! The theory of quadratic residues was developed by Gauss.
     
    Last edited: Dec 22, 2004
  4. Dec 23, 2004 #3
    Thanks. Also it is to be noted that I stated my conjecture wrongly. I should have wrote X^2 - 2Y^2 = - P has solutions in odd integers where P is a prime of the form 8*n +1. It is impossible for the other way around to be correct.
     
  5. Dec 24, 2004 #4
    If p is for the form 4k+1, which satisfies 8k+1, then there is a solution to X^2==-1, Mod P. So for p==1 Mod 8, there is a solution to X^2 ==-2 Mod p.
     
  6. Dec 24, 2004 #5
    Ramsey 2879: Thanks. Also it is to be noted that I stated my conjecture wrongly. I should have wrote X^2 - 2Y^2 = - P has solutions in odd integers where P is a prime of the form 8*n +1. It is impossible for the other way around to be correct.

    Well, there is a question here: 5^2-2(2)^2 = 25-8 = 17,

    But: 17 = 2(3^2)-(1)^2 = 18-1.

    So you' ll have to look at that again.
     
  7. Dec 25, 2004 #6
    Note that for odd integers, X^2- 2Y^2 always equals 7 mod 8. Sorry for the miscommunication.
     
    Last edited: Dec 25, 2004
  8. Dec 25, 2004 #7
    All odd squaes are congruent to 1 Mod 8. (2x+1)^2 = 4x^2+4x+1 =
    4x(x+1) +1 ==1 Mod 8.
     
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