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.
Using recurrences to solve $3a^2=2b^2+1$
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0A new question about the same equation, although a solution using the hint from Engel's books has not been posted yet: [Solve the Diophantine equation $3x^2-2y^2=1$](http://math.stackexchange.com/questions/181758/solve-the-diophantine-equation-3x2-2y2-1) – 2012-08-18
2 Answers
Yes. See, for example, the pair of sequences https://oeis.org/A054320 and https://oeis.org/A072256, where the solutions are listed. The recurrence is defined by $a_0 = a_1 = 1; \qquad a_n = 10a_{n-1} - a_{n-2},\ n\ge 2.$
As to how to go about solving this, there are many good references on how to do this, including Wikipedia.
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.