Pell equation

**namelessguy** · December 9th 2008, 10:55 PM

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.

**ThePerfectHacker** · December 10th 2008, 12:00 PM

Originally Posted by namelessguy

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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