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Math Help - Pell equation

  1. #1
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    Pell equation

    Can anyone take a look to see if my argument in this proof is correct?
    Show that x^2-dy^2=-1 has no solution in integer if p \mid d, p\equiv -1(4), p is prime.
    Argue by contradiction, suppose that the Pell's equation does have an integer solution. Rewrite the equation x^2+1=dy^2 \Rightarrow x^2+1 \equiv 0(dy^2) \Rightarrow x^2 +1\equiv 0(p) since p \mid d \Rightarrow p \mid dy^2. So,  x^2\equiv -1(p) has a solution. This is a contradiction because if p \equiv -1(4) then the Legendre symbol (\frac{-1}{p})=-1 implying the above congruence is not solvable.
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    Can anyone take a look to see if my argument in this proof is correct?
    Show that x^2-dy^2=-1 has no solution in integer if p \mid d, p\equiv -1(4), p is prime.
    Argue by contradiction, suppose that the Pell's equation does have an integer solution. Rewrite the equation x^2+1=dy^2 \Rightarrow x^2+1 \equiv 0(dy^2) \Rightarrow x^2 +1\equiv 0(p) since p \mid d \Rightarrow p \mid dy^2. So,  x^2\equiv -1(p) has a solution. This is a contradiction because if p \equiv -1(4) then the Legendre symbol (\frac{-1}{p})=-1 implying the above congruence is not solvable.
    Looks good to me.
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