We have NktA0Nk = Ak for k ≥ 0.
For some n > 0, NntA0Nn=A0, where Nn is a positive matrix of the form and (u,v) is the least positive solution of Pell's equation. In the factorization of Nn, the Mi form a palindromic sequence.
The negative Pell equation will be soluble if and only if there exists a k such that Ak = , in which case, Ak will be at the centre of symmetry and Nk will have the form . Then (u, v) = (u12 + Dv12, 2u1v1) and (u1, v1) will be the least positive solution of the negative Pell equation.
Note that in Wildberger's paper, on page 7, the equation N = MMt is not correct and should be replaced by N = MM+, where M+ = .
Wildberger does not prove that the solutions found are the least positive solutions, but this can be demonstrated by considering George Raney's L-R factorization algorithm applied to matrices and , where x2 - Dy2 = 1 and x12 - Dy12 = 1.
E = 0 prints the least positive solution of Pell's equation x2 - Dy2 = 1 and determines in the negative Pell equation x2 - Dy2 = -1 is soluble, in which case the least positive solution is found.
Last modified 21st February 2015
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