Using recurrences to solve $3a^2=2b^2+1$ Is it possible to solve the equation $3a^2=2b^2+1$ for positive, integral $a$ and $b$ using recurrences?I am sure it is, as Arthur Engel in his Problem Solving Strategies has stated that as a method, but I don't think I understand what he means.Can anyone please tell me how I should go about it?Thanks.
Edit:Added the condition that $a$ and $b$ are positive integers.

Although it should be mentioned, and the equation: $$aX^2-qY^2=f$$
If the root of the whole: $\sqrt{\frac{f}{a-q}}$
Using equation Pell:  $$p^2-aqs^2=1$$  solutions can be written:
$$Y=(2aps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$$
$$X=(2qps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$$
And for that decision have to find double formula.
$$Y_2=Y+2as(qsY-pX)$$
$$X_2=X+2p(qsY-pX)$$
We will use these formulas to solve equations:  $$3X^2-2Y^2=1$$
Decisions will be determined by the Pell equation: $$p^2-6s^2=1$$
Starting from the first solution: $(p_0,s_0)$ - $(5,2)$
You can find all the rest of the formula.
$$s_2=2p_1+5s_1$$
$$p_2=5p_1+12s_1$$
These numbers will need to substitute in:
$$Y=p^2\pm6ps+6s^2$$
$$X=p^2\pm4ps+6ps$$
Then you can consider and the twins. It is necessary to take into account that all of the substitution number can have any signs.
