# When is $3 x^2 + 2 x$ a square

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.

• It is the same as $3y^2+1=z^2$ Aug 30, 2020 at 16:57

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, $$3y^2+1=z^2$$.