Generalized Pell Equation and Primes

1. Dec 22, 2004

ramsey2879

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. Dec 22, 2004

robert Ihnot

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
3. Dec 23, 2004

ramsey2879

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.

4. Dec 24, 2004

robert Ihnot

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.

5. Dec 24, 2004

robert Ihnot

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.

6. Dec 25, 2004

ramsey2879

Note that for odd integers, X^2- 2Y^2 always equals 7 mod 8. Sorry for the miscommunication.

Last edited: Dec 25, 2004
7. Dec 25, 2004

robert Ihnot

All odd squaes are congruent to 1 Mod 8. (2x+1)^2 = 4x^2+4x+1 =
4x(x+1) +1 ==1 Mod 8.