probably Lagrange or Legendre, Pell variant - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:47:16Z http://mathoverflow.net/feeds/question/101163 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101163/probably-lagrange-or-legendre-pell-variant probably Lagrange or Legendre, Pell variant Will Jagy 2012-07-02T17:51:59Z 2012-07-02T21:21:16Z <p>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 $x^2 -q y^2 = 2.$</p> <p>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. </p> <p>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> <p>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 $2x^2 -q y^2 = 1.$</p> http://mathoverflow.net/questions/101163/probably-lagrange-or-legendre-pell-variant/101181#101181 Answer by GH for probably Lagrange or Legendre, Pell variant GH 2012-07-02T21:12:31Z 2012-07-02T21:21:16Z <p>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".</p> <p>Actually Jour. für Math. stands for Crelle's journal, and Göpel's paper (which is his 1835 doctoral dissertation) is available online <a href="http://www.digizeitschriften.de/dms/img/?PPN=PPN243919689_0045&amp;DMDID=dmdlog4" rel="nofollow">here</a>.</p>