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I know the method of continued fractions to solve the Pell's equation. I need help turning $2b(b-1) = t(t-1)$, with $b, t$ as integers, into the form $x^2 - ny^2 = 1$, if possible.

My attempt is $(t-1/2)^2 - 2(b - 1/2)^2 = -1/4$, but those aren't integer solutions anymore.

qwr
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  • I wish you wouldn't have identified the problem number. I'm not sure you can get it in the exact form you want, but at least you should be able to eliminate the fractions easily enough. – Mike Jul 11 '15 at 02:03
  • @Mike removed problem number – qwr Jul 11 '15 at 02:33

1 Answers1

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Your completing the square procedure is good. Multiply through by $4$. We get $(2t-1)^2-2(2b-1)^2=-1$. Not quite the Pell equation, but a close relative, $a^2-2b^2=-1$. The fundamental solution is $a=1$, $b=1$, and all positive solutions can be obtained as for the Pell equation. One way to generate them is that they are $(a_k,b_k)$, where $(1+\sqrt{2})^{2k+1}=a_k+b_k\sqrt{2}$.

Alternately, one can get a recurrence for the "next" solution $(a_{k+1},b_{k+1})$ in terms of the current solution $(a_k,b_k)$.

Added: For reference, the recurrence turns out to be $a_{k+1}=3a_k+4b_k$, $b_{k+1}=2a_k+3b_k$. From this it is easily checked that all the solutions are odd, so they all generate a solution of your original equation. The first few solutions $(a,b)$ of the equation $a^2-2b^2=-1$ are: $$(1,1),\quad (7,5),\quad (41,29),\quad (239,169),\quad (1393,985).$$

qwr
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André Nicolas
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  • Isn't the recurrence $a_{k+1} = a_k + 2b_k, b_{k+1} = a_k + b_k$, since $(a_1, b_1) = (1, 1)$? Or is this the recurrence for odd solutions only? – qwr Jul 11 '15 at 05:18
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    I used $(a_0,b_0)$ for $(1,1)$. Also, in order to get solutions of $a^2-2b^2=-1$ only, and not solutions of $a^2-2b^2=1$, I am using only odd powers of $1+\sqrt{2}$. So we are multiplying $a_k+b_k\sqrt{2}$ by $3+2\sqrt{2}$. Note that $(3,2)$ is the fundamental solution of the Pell equation $u^2-2v^2=1$. – André Nicolas Jul 11 '15 at 05:24