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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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 $p|dy^2$.