probably Lagrange or Legendre, Pell variant

2012-07-02T17:51:59Z

Evidently Legendre showed that, for positive primes, if $p \equiv 3 \pmod 8$ there is an integral solution to $x^2 - p y^2 = -2.$ Next, if $q \equiv 7 \pmod 8$ there is an integral solution to $x^2 -q y^2 = 2.$

What I would like, and seems to be true, is $x^2 - 2 p y^2 = -2$ for $p \equiv 3 \pmod 8,$ and $x^2 - 2 q y^2 = 2$ for $q \equiv 7 \pmod 8.$ It is probably in Mordell's book, which I do not have here.

Mordell does $x^2 - r y^2 = -1$ for any prime $r \equiv 1 \pmod 4,$ I do remember that. Anyway, I am writing up something and this issue came up.

P.S. Note these are the same as $2x^2 - p y^2 = -1$ if $p \equiv 3 \pmod 8,$ while if $q \equiv 7 \pmod 8$ there is $2x^2 -q y^2 = 1.$

Answer by GH for probably Lagrange or Legendre, Pell variant

2012-07-02T21:12:31Z

According to Dickson (History of numbers Vol. 2, Ch. XII, p.376), Göpel (Jour. für Math. 45, 1853, 1-14) proved your conjectures "by use of continued fractions".

Actually Jour. für Math. stands for Crelle's journal, and Göpel's paper (which is his 1835 doctoral dissertation) is available online here.

probably Lagrange or Legendre, Pell variant - MathOverflow

probably Lagrange or Legendre, Pell variant

Answer by GH for probably Lagrange or Legendre, Pell variant