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.

  • $\begingroup$ Just a tip about use of LaTeX: you will achieve more consistent layout if you put \$ signs round complete equations, rather than small parts of an equation. $\endgroup$ – Old John Oct 10 '12 at 12:23
  • $\begingroup$ For $p=7$ and $m=1$ you have the solution $y=3, x=1$. For $p=17$, $y=5$, $x=2$. $\endgroup$ – PAD Oct 10 '12 at 12:49
  • $\begingroup$ @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. $\endgroup$ – VMRFDU Oct 11 '12 at 7:19


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

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 $r<q/4$. 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 $x^2-2y^2$.

  • $\begingroup$ D Aurizio! the way of explanation is too good. Thank you so much. If we take in the pace of n = 3. we see an elliptic curve. What kind of solutions there for such equation does not matter. How to find solutions in the case on n = 3. plz explain. So that, I will try for n= 5 and so on. Once again thank you so much for your help. $\endgroup$ – VMRFDU Oct 11 '12 at 7:22
  • $\begingroup$ D Aurizio! could you explain the solutions of the equation $y^3$ -3$x^2$ = $p^m$. I know this is elliptic curve. I had some knowledge of elliptic curve. But, I was totally upset to determine the solutions of this kind equation. Please explain. $\endgroup$ – VMRFDU Oct 11 '12 at 11:10
  • $\begingroup$ The study of primes represented by cubic forms is an extremely hard problem, due to the lack of semigroup identities, that in the quadratic case are crucial and extremely powerful. However, I think that the subcase $m=3$ is solvable, and leads to the study of the diophantine equation $107 = 3u^4-v^2$. $\endgroup$ – Jack D'Aurizio Oct 11 '12 at 13:52
  • $\begingroup$ D aurizio! how you reduce to 107 = 3$u^4$-$v^2$, when m = 3? Also, give some hints to solve the case for m = 3. Please... $\endgroup$ – VMRFDU Oct 12 '12 at 4:34
  • $\begingroup$ In the case $m=3$, assume $p>3$. You get $3|(y^3-p^3)$, but since $u^3\equiv u\pmod{3}$, you get that $y$ and $p$ lie in the same residue class $\pmod{3}$, so $9$ divides $y^3-p^3$, $3$ divides $x^2$, $27$ divides $y^3-p^3$, $y$ and $p$ lie in the same residue class $\pmod{9}$ and so on. This proves that for $m=3$ the only solutions are given by $p=3$, with $y=3u$, $x=3v$ and $u^3-1=(u-1)(u^2+u+1)=v^2$. Since $\gcd(u-1,u^2+u+1)|3$, there are two cases: or $(u-1)$ and $(u^2+u+1)$ are both squares of two coprime numbers, either they are both three times the square of two coprime numbers. $\endgroup$ – Jack D'Aurizio Oct 12 '12 at 9:24

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