It's given ($p$ is a prime): $ x^2-dy^2 \equiv 1\pmod p $ Using only this can we say $ x^2-dy^2 = 1 $ has always integral solution?
Integral solution given $x^2-dy^2 \equiv 1\pmod p$
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number-theory
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1I do not understand the question. If it is from a book or notes, perhaps you could give more detail. – 2012-06-28
1 Answers
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If $=1 \pmod p$ actually means $\equiv \pmod p$,
$\implies x^2-1 \equiv dy^2 \pmod p$
Observe that if $p$ |LHS, $x≡±1 \pmod p$, then $p|dy^2$.
- if $p∤d$ i.e., $(p,d)=1, p|y$ for accepting solution.
- if $p|d$, any integral $y$ will give us integral solution.