I'm trying to solve $y^2=3x^2+3x+1$ for integers, which transforms into $(2y)^2-3(2x+1)^2=1$. I know how to solve pell's equation, but how can we extract only (odd,even) pair from the solutions of the diophantine equation $y^2-3x^2=1$? Any help would be appreciated.
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$\begingroup$ see math.stackexchange.com/questions/1311735/… $\endgroup$– Will JagyJun 5, 2015 at 23:01
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1 Answer
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The fundamental solution is $2+\sqrt{3}.$ The general solution is $x_n+\sqrt{3}y_n=(2+\sqrt{3})^n,$ that is $$x_n=\frac{(2+\sqrt{3})^n+(2-\sqrt{3})^n}{2}.$$ This is sequence A001075 at https://oeis.org. Writing it as a recurrence sequence we have $x_0 = 1, x_1 = 2, x_n = 4x_{n-1} - x_{n-2}$. It follows that $x_{2k+1}$ is even for all $k\geq 0.$ In your case \begin{eqnarray*} 2y&=&\frac{(2+\sqrt{3})^{2k+1}+(2-\sqrt{3})^{2k+1}}{2},\\ 2x+1&=&\frac{(2+\sqrt{3})^{2k+1}-(2-\sqrt{3})^{2k+1}}{2\sqrt{3}} \end{eqnarray*} provides all the solutions with $k\geq 0.$
