# 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.

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Which part? Which page? Section? –  Guess who it is. Nov 13 '11 at 15:55
Page number 154, solution to Problem 137 in the number theory section. –  Eisen Nov 13 '11 at 16:01
See also page 147, solution of 92(b). Both occurrences are after Pell's equation. –  Martin Sleziak Nov 13 '11 at 16:04
If you have access to this book, Section 3.2 Solving Pell's Equation in the book An Introduction to Diophantine Equations: A Problem-Based Approach By Titu Andreescu, Dorin Andrica, Ion Cucurezeanu should contain more than enough information on Pell's equation. See p.121. Maybe you don't even have to go through the proofs, it should be enough if you understand what the results claim and practice them on a few concrete examples. –  Martin Sleziak Nov 13 '11 at 16:14
Other useful resources to learn about Pell's equation could be: imomath.com/tekstkut/pelleqn_ddj.pdf, artofproblemsolving.com/Wiki/index.php/Pell_equation and, of cousrse, wikipedia. And, very probably, for solution of this exercises the facts mentioned in Engel's book should be sufficient. –  Martin Sleziak Nov 13 '11 at 16:30

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.

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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.

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