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I have tried several times to solve the following equation and finally, I was failed to complete. Help me to find the solutions of Pell equation $y^2-2x^2 = p^m$, where $p$ is prime and $8|(p-1)$ or $8|(p+1)$.

A waiting the reply.

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    @PantelisDamianou! How you got those values? I checked manually and all of them are correct. Is there any way to find such solutions. Plz explain.2012-10-11

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Since

$ (a^2-2b^2)(c^2-2d^2) = (ac-2bd)^2 - 2(bc-ad)^2 $

it is sufficient to find a solution to $x^2-2y^2 = p$. Since $p\equiv\pm 1\pmod{8}$, $2$ is a quadratic residue $\pmod{p}$, so there is a natural number $a<\frac{p}{2}$ such that $a^2\equiv 2\pmod{p}$ and

$ (\heartsuit)\quad a^2-2 = k p,$

with $k<\frac{p}{4}$. Moreover, every odd prime $q$ that divides $k$ is $\equiv\pm 1\pmod{8}$, because from $(\heartsuit)$ we have that $a$ is a square root of $2\pmod{q}$. We can get rid of the (possible) factor $2$ in $k$: if $k$ is even then $a$ is even too, and:

$ (a-1)^2 - 2\left(\frac{a}{2}-1\right)^2=\frac{1}{2}\left(a^2-2\right).$

So we have:

$ a^2-2b^2 = Qp, $ where $Q$ is a product of primes of the form $8n\pm1$. Let $q$ be one of them, and

$ a_q^2-2 = rq, $

with $r. By the initial identity we get:

$ (a a_q-2b)^2-2(b a_q-a)^2 = rqQp, $

but both $(a a_q-2b)$ and $(b a_q-a)$ are divisible by $q$, so:

$ a_*^2 - 2b_*^2 = \left(\frac{a a_q-2b}{q}\right)^2-2\left(\frac{b a_q-a}{q}\right)^2 = r(Q/q)p, $

and we can get rid of any prime factor in Q, simply starting from the greatest. This proves that any prime of the form $8n\pm1$ is represented by the quadratic form $x^2-2y^2$.

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    D Aurizio! Sir, I got solutions for $y^3$= $x^2$ + 1. Those are (x, y) = (-5, 2.96),(-4, 2.57), (-3, 2.15), (-2, 1.71), (-1, 1.26), (0,1), (1, 1.26), (2, 1.71), (3, 2.15), (4, 2.57), (5, 2.96) and so on. If we want this kind of solutions, what are the ways to find such solutions? please explain. because of your motivation, I found these and still I am learning more and more.2012-10-13