Integral solution given $x^2-dy^2 \equiv 1\pmod p$

Question

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?

What is your quantifier? "For all primes p" or "For some prime p" ?
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.
I do not understand the question. If it is from a book or notes, perhaps you could give more detail.

Joe · Answer 1 · 2012-07-14 16:00:24Z

up vote 0 down vote

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 $p|dy^2$.

edited Jul 14 '12 at 16:00

Joe
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answered Jul 10 '12 at 16:53

lab bhattacharjee
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