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.

  • 3
    $\begingroup$ It is the same as $3y^2+1=z^2$ $\endgroup$
    – Empy2
    Aug 30, 2020 at 16:57

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.