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Let $x, y$ be a positive integers. I want to know when $3 x^2 + 2 x = y^2$ has a solution.

Through some enumeration of all $x$, and trial and error, I have found the following recursion which appears to include all the solutions:

Initial conditions are:

$$\begin{array}{l} x_0 = 0, x_1 = 2\\ y_0 = 0, y_1 = 4 \end{array}$$

Recursion is:

$$\begin{array}{l} x_n = 8 y_{n - 1} + x_{n - 2}\\ y_n = 14 y_{n - 1} - y_{n - 2} \end{array}$$

This appears to be similar to Pell's equation, and here it seems that $x / y$ is some continued fraction approximation to $1 / \sqrt{3}$.

I'm not quite sure how to find all solutions mathematically though, and see that this indeed produces all solutions.

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    $\begingroup$ It is the same as $3y^2+1=z^2$ $\endgroup$ – Empy2 Aug 30 at 16:57
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Multuply by $3$ to get $9x^2+6x=3y^2$, add $1$ to get $(3x+1)^2=3y^2+1$, so setting $z=3x+1$ you get $z^2-3y^2=1$, a Pell equation.

Alternative you may solve for $x$ to get $\displaystyle x=\frac{-1\pm \sqrt{3y^2+1}}{3}$, so you need to $3y^2+1$ to be a square, $3y^2+1=z^2$.

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