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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?

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    What is your quantifier? "For all primes p" or "For some prime p" ?2012-06-28
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    Assuming $d$ is a given integer and $x,y$ are unknown then $x^2-dy^2=1$ *always* has an integral solution, namely, $x=\pm1$, $y=0$. If $d$ is a square, that's the only solution. If $d\gt1$ is not a square, this is Pell's equation, and it is guaranteed to have infinitely many solutions.2012-06-28
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    I do not understand the question. If it is from a book or notes, perhaps you could give more detail.2012-06-28

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